|
%I #57 Sep 08 2022 08:45:13
%S 1,1,2,7,42,434,7812,248031,14055090,1436430198,267176016828,
%T 91151551074486,57425477176926180,67196011936600334340,
%U 146782968474309770332296,601204690999713530559792879
%N a(n) = (2^(n-1)/n!) * Product_{k=1..n-1} (2^k-1).
%C Primes p such that 2^p-2 divides a(p) are A216838. - _Amiram Eldar_ and _Thomas Ordowski_, Jan 16 2020
%C For odd n > 1, if a(n-1) divides a(n) and n does not divide a(n), then n is a prime (for which 2 is a primitive root, A001122). Composite numbers m such that a(m-1) divides a(m) are the pseudoprimes A001567 and A006935. Numbers n > 1 such that a(m) divides a(n) for all m < n are primes 2, 3, 5, 7, and 13. These are the primes p for which gpf(2^p-2) = p. - _Thomas Ordowski_, Jan 17 2020
%C If p is a prime with primitive root 2, A001122, then p | a(p-1) + 2^(p-2). Conjecture: (for n > 2), if n | a(n-1) + 2^(n-2), then n is a prime (A001122). Note that if p is an odd prime for which 2 is not a primitive root, A216838, then p | a(p-1). - _Amiram Eldar_ and _Thomas Ordowski_, Jan 19 2020
%H Amiram Eldar, <a href="/A091669/b091669.txt">Table of n, a(n) for n = 1..86</a>
%F a(n) = 2^(n-1)*A005329(n-1)/n!.
%F a(n) = Product_{k=2..n} (2^k-2)/k = Product_{k=2..n} A225101(k)/A159353(k). - _Thomas Ordowski_, Jan 16 2020
%p seq( (2^(n-1)/n!)*mul(2^j-1, j=1..n-1), n=1..20); # _G. C. Greubel_, Feb 05 2020
%t Table[QFactorial[n-1, 2] 2^(n-1)/n!, {n, 20}]
%o (PARI) a(n) = (2^(n-1)/n!) * prod(k=1, n-1, 2^k-1); \\ _Michel Marcus_, Jan 16 2020
%o (Magma) [1] cat [2^(n-1)/Factorial(n)*&*[(2^k-1):k in [1..n-1]]:n in [2..16]]; // _Marius A. Burtea_, Jan 16 2020
%o (Sage) from sage.combinat.q_analogues import q_factorial
%o [2^(n-1)*q_factorial(n-1, 2)/factorial(n) for n in (1..20)] # _G. C. Greubel_, Feb 05 2020
%Y Cf. A000142, A001122, A001567, A005329, A006935, A159353, A216838, A225101.
%K nonn
%O 1,3
%A _Karol A. Penson_, Jan 27 2004
%E Corrected and edited by _Thomas Ordowski_, Jan 16 2020
|
