Ec numbers congruent to 7 mod 1063











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A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. ec numbers are introduced obtained by the concatenation of two consecutive Mersenne numbers (40952047 for example). Ec(7)=12763 and ec(8)=255127 are both congruent to 7 mod 1063. I did not find yet another example of ec(k) and ec(k+1) both congruent to 7 mod 1063. Is there any particolar mathematical reason, can be that ruled out or is it just coincidence?










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    A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. ec numbers are introduced obtained by the concatenation of two consecutive Mersenne numbers (40952047 for example). Ec(7)=12763 and ec(8)=255127 are both congruent to 7 mod 1063. I did not find yet another example of ec(k) and ec(k+1) both congruent to 7 mod 1063. Is there any particolar mathematical reason, can be that ruled out or is it just coincidence?










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      down vote

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      A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. ec numbers are introduced obtained by the concatenation of two consecutive Mersenne numbers (40952047 for example). Ec(7)=12763 and ec(8)=255127 are both congruent to 7 mod 1063. I did not find yet another example of ec(k) and ec(k+1) both congruent to 7 mod 1063. Is there any particolar mathematical reason, can be that ruled out or is it just coincidence?










      share|cite|improve this question













      A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. ec numbers are introduced obtained by the concatenation of two consecutive Mersenne numbers (40952047 for example). Ec(7)=12763 and ec(8)=255127 are both congruent to 7 mod 1063. I did not find yet another example of ec(k) and ec(k+1) both congruent to 7 mod 1063. Is there any particolar mathematical reason, can be that ruled out or is it just coincidence?







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      asked Nov 18 at 8:40









      paolo galli

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          The numbers



          $ec(289922)$ and $ec(289923)$ are both congruent to $7$ modulo $1063$






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          • $k=2268439$ is the third solution
            – Peter
            Nov 18 at 19:24










          • $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
            – paolo galli
            Nov 19 at 14:20











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote













          The numbers



          $ec(289922)$ and $ec(289923)$ are both congruent to $7$ modulo $1063$






          share|cite|improve this answer





















          • $k=2268439$ is the third solution
            – Peter
            Nov 18 at 19:24










          • $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
            – paolo galli
            Nov 19 at 14:20















          up vote
          1
          down vote













          The numbers



          $ec(289922)$ and $ec(289923)$ are both congruent to $7$ modulo $1063$






          share|cite|improve this answer





















          • $k=2268439$ is the third solution
            – Peter
            Nov 18 at 19:24










          • $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
            – paolo galli
            Nov 19 at 14:20













          up vote
          1
          down vote










          up vote
          1
          down vote









          The numbers



          $ec(289922)$ and $ec(289923)$ are both congruent to $7$ modulo $1063$






          share|cite|improve this answer












          The numbers



          $ec(289922)$ and $ec(289923)$ are both congruent to $7$ modulo $1063$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 19:22









          Peter

          46.2k1039125




          46.2k1039125












          • $k=2268439$ is the third solution
            – Peter
            Nov 18 at 19:24










          • $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
            – paolo galli
            Nov 19 at 14:20


















          • $k=2268439$ is the third solution
            – Peter
            Nov 18 at 19:24










          • $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
            – paolo galli
            Nov 19 at 14:20
















          $k=2268439$ is the third solution
          – Peter
          Nov 18 at 19:24




          $k=2268439$ is the third solution
          – Peter
          Nov 18 at 19:24












          $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
          – paolo galli
          Nov 19 at 14:20




          $289922=2*144961$. $144961$ is the concatenation of two squares 12^2 and 31^2.
          – paolo galli
          Nov 19 at 14:20


















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