cp's OEIS Frontend

A344681 a(n) is the smallest k > n such that 2^(k-n) == 1 (mod k).

%I A344681 #29 Nov 11 2023 03:38:36
%S A344681 1,3,20737,9,7,25,31,15,127,17,73,15,23,33,3479,21,31,65,131071,51,
%T A344681 524287,31,127,33,47,69,31,39,49,43,233,87,1361567,45,89,51,71,73,223,
%U A344681 57,79,65,13367,51,431,63,73,69,2351,97,127,63,103,65,6361,73,89,63,721,87,179951
%N A344681 a(n) is the smallest k > n such that 2^(k-n) == 1 (mod k).
%C A344681 Smallest odd k > n such that 2^k == 2^n (mod k).
%C A344681 a(n) is the smallest odd k > n such that A002326((k-1)/2) divides k-n.
%C A344681 If a(n) is a prime p, then 2^(n-1) == 1 (mod p).
%C A344681 Note that a(2n+2) <= A002184(n) for n > 0.
%C A344681 If p is an odd prime, then a(p) <= p^2.
%H A344681 Michel Marcus, <a href="/A344681/b344681.txt">Table of n, a(n) for n = 0..149</a>
%t A344681 a[0] = 1; a[n_] := Module[{k = n + 1}, While[PowerMod[2, k - n, k] != 1, k++];
%t A344681 k]; Array[a, 60, 0] (* _Amiram Eldar_, Aug 17 2021 *)
%o A344681 (Python)
%o A344681 def a(n):
%o A344681     if n == 0: return 1
%o A344681     k = n + 1
%o A344681     while pow(2, k-n, k) != 1: k += 1
%o A344681     return k
%o A344681 print([a(n) for n in range(61)]) # _Michael S. Branicky_, Aug 17 2021
%o A344681 (PARI) a(n) = my(k=n+1); while(Mod(2, k)^(k-n) != 1, k++); k; \\ _Michel Marcus_, Aug 17 2021
%Y A344681 Cf. A002184, A002326, A346988.
%K A344681 nonn
%O A344681 0,2
%A A344681 _Thomas Ordowski_, Aug 17 2021
%E A344681 More terms from _Amiram Eldar_, Aug 17 2021