Computing only Prime Powers with NestList
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Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:
{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}
up to some specified prime power. Thanks so much!
recursion
asked Feb 17 at 21:11
user413587user413587
825
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add a comment |
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Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:
{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}
up to some specified prime power. Thanks so much!
recursion
asked Feb 17 at 21:11
user413587user413587
825
$endgroup$
add a comment |
$begingroup$
Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:
{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}
up to some specified prime power. Thanks so much!
recursion
asked Feb 17 at 21:11
user413587user413587
825
$endgroup$
Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:
{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}
up to some specified prime power. Thanks so much!
recursion
recursion
asked Feb 17 at 21:11
user413587user413587
825
asked Feb 17 at 21:11
user413587user413587
825
asked Feb 17 at 21:11
user413587user413587
825
asked Feb 17 at 21:11
user413587user413587
825
asked Feb 17 at 21:11
user413587user413587
825
825
add a comment |
add a comment |
7 Answers
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Lets say you have f(x) and you want results up to 10
n=10;
NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 22:01
J42161217J42161217
3,968323
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Here's a way using Compose:
n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:51
bill sbill s
54.4k377156
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Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:
primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]
primeNest[f, x, 5]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]],
f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
(Depth /@ primeNest[f, x, 20]) - 1
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
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One can in fact use NestWhile along with NestList, in the same spirit as this previous answer:
NestList[NestWhile[f, f[#], ! PrimeQ[Depth[#] - 1] &] &, x, 5]
{x, f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Here is another variation which yields the same result:
NestList[Nest[f, #, NextPrime[Depth[#] - 1] - Depth[#] + 1] &, x, 5]
The last few solutions have the flaw of relying on Depth, which is not very useful if the function to be iterated is even moderately complicated. Nevertheless, these can be modified like so:
NestPrimeList[f_, x_, n_Integer?NonNegative] := Module[{p = 0, r = x, $1},
Prepend[Reap[Do[Sow[r = Nest[f, r, -p + (p = NextPrime[p])], $1], {n}], $1][[2, 1]], x]]
Example:
Exponent[#, x] & /@ NestPrimeList[x # + 1 &, 1, 5]
{0, 2, 3, 5, 7, 11}
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
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This may not be an elegant and smart way to do it but here is one way.
exp = NestList[g, x, 12];
selector = PrimeQ[LeafCount /@ exp - 1];
Pick[exp, selector] /. {g -> f, x -> x0}
{f[f[x0]], f[f[f[x0]]], f[f[f[f[f[x0]]]]], f[f[f[f[f[f[f[x0]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x0]]]]]]]]]]]}
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
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ClearAll[primesNest0]
primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]
primesNest0[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
Also
ClearAll[primesNest1]
primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]
primesNest1[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest2]
primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]
primesNest2[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest3]
primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@
Prime[Range@PrimePi@n]
primesNest3[f, x, 10]
. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:47
kglrkglr
189k10206424
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After a For loop faux pas, and after receiving the splendid advice below came up with this ...
n = 5;
ls = {};
Table[(Q = f[x];
Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}];), {k, Range[n]}];
Print[ls]
Here is the result:
{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Perhaps it is time to Table this discussion:
n = 5;
ls = {};
Table[(Q = f[x];
Table[Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}]), {k, Range[n]}];
Print[ls];
answered Feb 17 at 22:07
mjwmjw
8859
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1
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Why should I avoid the For loop in Mathematica?
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– corey979
Feb 17 at 22:24
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Thank you for the tip, here it is withForreplaced byDo:n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
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– mjw
Feb 17 at 23:26
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
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active
oldest
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$begingroup$
Lets say you have f(x) and you want results up to 10
n=10;
NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 22:01
J42161217J42161217
3,968323
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add a comment |
$begingroup$
Lets say you have f(x) and you want results up to 10
n=10;
NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 22:01
J42161217J42161217
3,968323
$endgroup$
add a comment |
$begingroup$
Lets say you have f(x) and you want results up to 10
n=10;
NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 22:01
J42161217J42161217
3,968323
$endgroup$
Lets say you have f(x) and you want results up to 10
n=10;
NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 22:01
J42161217J42161217
3,968323
edited Feb 17 at 22:06
answered Feb 17 at 22:01
J42161217J42161217
3,968323
answered Feb 17 at 22:01
J42161217J42161217
3,968323
answered Feb 17 at 22:01
J42161217J42161217
3,968323
3,968323
add a comment |
add a comment |
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Here's a way using Compose:
n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:51
bill sbill s
54.4k377156
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add a comment |
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Here's a way using Compose:
n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:51
bill sbill s
54.4k377156
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add a comment |
$begingroup$
Here's a way using Compose:
n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:51
bill sbill s
54.4k377156
$endgroup$
Here's a way using Compose:
n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:51
bill sbill s
54.4k377156
answered Feb 17 at 23:51
bill sbill s
54.4k377156
answered Feb 17 at 23:51
bill sbill s
54.4k377156
answered Feb 17 at 23:51
bill sbill s
54.4k377156
54.4k377156
add a comment |
add a comment |
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Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:
primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]
primeNest[f, x, 5]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]],
f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
(Depth /@ primeNest[f, x, 20]) - 1
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
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Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:
primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]
primeNest[f, x, 5]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]],
f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
(Depth /@ primeNest[f, x, 20]) - 1
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
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add a comment |
$begingroup$
Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:
primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]
primeNest[f, x, 5]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]],
f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
(Depth /@ primeNest[f, x, 20]) - 1
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
$endgroup$
Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:
primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]
primeNest[f, x, 5]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]],
f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
(Depth /@ primeNest[f, x, 20]) - 1
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
answered Feb 18 at 0:21
Chip HurstChip Hurst
22.5k15892
22.5k15892
add a comment |
add a comment |
$begingroup$
One can in fact use NestWhile along with NestList, in the same spirit as this previous answer:
NestList[NestWhile[f, f[#], ! PrimeQ[Depth[#] - 1] &] &, x, 5]
{x, f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Here is another variation which yields the same result:
NestList[Nest[f, #, NextPrime[Depth[#] - 1] - Depth[#] + 1] &, x, 5]
The last few solutions have the flaw of relying on Depth, which is not very useful if the function to be iterated is even moderately complicated. Nevertheless, these can be modified like so:
NestPrimeList[f_, x_, n_Integer?NonNegative] := Module[{p = 0, r = x, $1},
Prepend[Reap[Do[Sow[r = Nest[f, r, -p + (p = NextPrime[p])], $1], {n}], $1][[2, 1]], x]]
Example:
Exponent[#, x] & /@ NestPrimeList[x # + 1 &, 1, 5]
{0, 2, 3, 5, 7, 11}
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
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add a comment |
$begingroup$
One can in fact use NestWhile along with NestList, in the same spirit as this previous answer:
NestList[NestWhile[f, f[#], ! PrimeQ[Depth[#] - 1] &] &, x, 5]
{x, f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Here is another variation which yields the same result:
NestList[Nest[f, #, NextPrime[Depth[#] - 1] - Depth[#] + 1] &, x, 5]
The last few solutions have the flaw of relying on Depth, which is not very useful if the function to be iterated is even moderately complicated. Nevertheless, these can be modified like so:
NestPrimeList[f_, x_, n_Integer?NonNegative] := Module[{p = 0, r = x, $1},
Prepend[Reap[Do[Sow[r = Nest[f, r, -p + (p = NextPrime[p])], $1], {n}], $1][[2, 1]], x]]
Example:
Exponent[#, x] & /@ NestPrimeList[x # + 1 &, 1, 5]
{0, 2, 3, 5, 7, 11}
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
$endgroup$
add a comment |
$begingroup$
One can in fact use NestWhile along with NestList, in the same spirit as this previous answer:
NestList[NestWhile[f, f[#], ! PrimeQ[Depth[#] - 1] &] &, x, 5]
{x, f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Here is another variation which yields the same result:
NestList[Nest[f, #, NextPrime[Depth[#] - 1] - Depth[#] + 1] &, x, 5]
The last few solutions have the flaw of relying on Depth, which is not very useful if the function to be iterated is even moderately complicated. Nevertheless, these can be modified like so:
NestPrimeList[f_, x_, n_Integer?NonNegative] := Module[{p = 0, r = x, $1},
Prepend[Reap[Do[Sow[r = Nest[f, r, -p + (p = NextPrime[p])], $1], {n}], $1][[2, 1]], x]]
Example:
Exponent[#, x] & /@ NestPrimeList[x # + 1 &, 1, 5]
{0, 2, 3, 5, 7, 11}
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
$endgroup$
One can in fact use NestWhile along with NestList, in the same spirit as this previous answer:
NestList[NestWhile[f, f[#], ! PrimeQ[Depth[#] - 1] &] &, x, 5]
{x, f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Here is another variation which yields the same result:
NestList[Nest[f, #, NextPrime[Depth[#] - 1] - Depth[#] + 1] &, x, 5]
The last few solutions have the flaw of relying on Depth, which is not very useful if the function to be iterated is even moderately complicated. Nevertheless, these can be modified like so:
NestPrimeList[f_, x_, n_Integer?NonNegative] := Module[{p = 0, r = x, $1},
Prepend[Reap[Do[Sow[r = Nest[f, r, -p + (p = NextPrime[p])], $1], {n}], $1][[2, 1]], x]]
Example:
Exponent[#, x] & /@ NestPrimeList[x # + 1 &, 1, 5]
{0, 2, 3, 5, 7, 11}
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
edited Feb 19 at 2:24
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
answered Feb 18 at 0:57
J. M. is slightly pensive♦J. M. is slightly pensive
98.1k10305465
98.1k10305465
add a comment |
add a comment |
$begingroup$
This may not be an elegant and smart way to do it but here is one way.
exp = NestList[g, x, 12];
selector = PrimeQ[LeafCount /@ exp - 1];
Pick[exp, selector] /. {g -> f, x -> x0}
{f[f[x0]], f[f[f[x0]]], f[f[f[f[f[x0]]]]], f[f[f[f[f[f[f[x0]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x0]]]]]]]]]]]}
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
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add a comment |
$begingroup$
This may not be an elegant and smart way to do it but here is one way.
exp = NestList[g, x, 12];
selector = PrimeQ[LeafCount /@ exp - 1];
Pick[exp, selector] /. {g -> f, x -> x0}
{f[f[x0]], f[f[f[x0]]], f[f[f[f[f[x0]]]]], f[f[f[f[f[f[f[x0]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x0]]]]]]]]]]]}
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
$endgroup$
add a comment |
$begingroup$
This may not be an elegant and smart way to do it but here is one way.
exp = NestList[g, x, 12];
selector = PrimeQ[LeafCount /@ exp - 1];
Pick[exp, selector] /. {g -> f, x -> x0}
{f[f[x0]], f[f[f[x0]]], f[f[f[f[f[x0]]]]], f[f[f[f[f[f[f[x0]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x0]]]]]]]]]]]}
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
$endgroup$
This may not be an elegant and smart way to do it but here is one way.
exp = NestList[g, x, 12];
selector = PrimeQ[LeafCount /@ exp - 1];
Pick[exp, selector] /. {g -> f, x -> x0}
{f[f[x0]], f[f[f[x0]]], f[f[f[f[f[x0]]]]], f[f[f[f[f[f[f[x0]]]]]]],
f[f[f[f[f[f[f[f[f[f[f[x0]]]]]]]]]]]}
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
edited Feb 17 at 22:16
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
answered Feb 17 at 21:59
Okkes DulgerciOkkes Dulgerci
5,4141919
5,4141919
add a comment |
add a comment |
$begingroup$
ClearAll[primesNest0]
primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]
primesNest0[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
Also
ClearAll[primesNest1]
primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]
primesNest1[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest2]
primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]
primesNest2[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest3]
primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@
Prime[Range@PrimePi@n]
primesNest3[f, x, 10]
. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:47
kglrkglr
189k10206424
$endgroup$
add a comment |
$begingroup$
ClearAll[primesNest0]
primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]
primesNest0[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
Also
ClearAll[primesNest1]
primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]
primesNest1[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest2]
primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]
primesNest2[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest3]
primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@
Prime[Range@PrimePi@n]
primesNest3[f, x, 10]
. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:47
kglrkglr
189k10206424
$endgroup$
add a comment |
$begingroup$
ClearAll[primesNest0]
primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]
primesNest0[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
Also
ClearAll[primesNest1]
primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]
primesNest1[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest2]
primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]
primesNest2[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest3]
primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@
Prime[Range@PrimePi@n]
primesNest3[f, x, 10]
. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:47
kglrkglr
189k10206424
$endgroup$
ClearAll[primesNest0]
primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]
primesNest0[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
Also
ClearAll[primesNest1]
primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]
primesNest1[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest2]
primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]
primesNest2[f, x, 10]
{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
and
ClearAll[primesNest3]
primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@
Prime[Range@PrimePi@n]
primesNest3[f, x, 10]
. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}
answered Feb 17 at 23:47
kglrkglr
189k10206424
edited Feb 18 at 0:09
answered Feb 17 at 23:47
kglrkglr
189k10206424
answered Feb 17 at 23:47
kglrkglr
189k10206424
answered Feb 17 at 23:47
kglrkglr
189k10206424
189k10206424
add a comment |
add a comment |
$begingroup$
After a For loop faux pas, and after receiving the splendid advice below came up with this ...
n = 5;
ls = {};
Table[(Q = f[x];
Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}];), {k, Range[n]}];
Print[ls]
Here is the result:
{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Perhaps it is time to Table this discussion:
n = 5;
ls = {};
Table[(Q = f[x];
Table[Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}]), {k, Range[n]}];
Print[ls];
answered Feb 17 at 22:07
mjwmjw
8859
$endgroup$
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Feb 17 at 22:24
$begingroup$
Thank you for the tip, here it is withForreplaced byDo:n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
Feb 17 at 23:26
add a comment |
$begingroup$
After a For loop faux pas, and after receiving the splendid advice below came up with this ...
n = 5;
ls = {};
Table[(Q = f[x];
Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}];), {k, Range[n]}];
Print[ls]
Here is the result:
{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Perhaps it is time to Table this discussion:
n = 5;
ls = {};
Table[(Q = f[x];
Table[Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}]), {k, Range[n]}];
Print[ls];
answered Feb 17 at 22:07
mjwmjw
8859
$endgroup$
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Feb 17 at 22:24
$begingroup$
Thank you for the tip, here it is withForreplaced byDo:n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
Feb 17 at 23:26
add a comment |
$begingroup$
After a For loop faux pas, and after receiving the splendid advice below came up with this ...
n = 5;
ls = {};
Table[(Q = f[x];
Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}];), {k, Range[n]}];
Print[ls]
Here is the result:
{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Perhaps it is time to Table this discussion:
n = 5;
ls = {};
Table[(Q = f[x];
Table[Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}]), {k, Range[n]}];
Print[ls];
answered Feb 17 at 22:07
mjwmjw
8859
$endgroup$
After a For loop faux pas, and after receiving the splendid advice below came up with this ...
n = 5;
ls = {};
Table[(Q = f[x];
Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}];), {k, Range[n]}];
Print[ls]
Here is the result:
{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}
Perhaps it is time to Table this discussion:
n = 5;
ls = {};
Table[(Q = f[x];
Table[Q = Map[f, Q], {j, Range[Prime[k] - 1]}];
ls = Join[ls, {Q}]), {k, Range[n]}];
Print[ls];
answered Feb 17 at 22:07
mjwmjw
8859
edited Feb 19 at 1:47
answered Feb 17 at 22:07
mjwmjw
8859
answered Feb 17 at 22:07
mjwmjw
8859
answered Feb 17 at 22:07
mjwmjw
8859
8859
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Feb 17 at 22:24
$begingroup$
Thank you for the tip, here it is withForreplaced byDo:n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
Feb 17 at 23:26
add a comment |
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Feb 17 at 22:24
$begingroup$
Thank you for the tip, here it is withForreplaced byDo:n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
Feb 17 at 23:26
1
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Feb 17 at 22:24
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Feb 17 at 22:24
$begingroup$
Thank you for the tip, here it is with
For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]$endgroup$
– mjw
Feb 17 at 23:26
$begingroup$
Thank you for the tip, here it is with
For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]$endgroup$
– mjw
Feb 17 at 23:26
add a comment |
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