OFFSET
1,3
COMMENTS
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
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
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..86
FORMULA
a(n) = 2^(n-1)*A005329(n-1)/n!.
a(n) = Product_{k=2..n} (2^k-2)/k = Product_{k=2..n} A225101(k)/A159353(k). - Thomas Ordowski, Jan 16 2020
MAPLE
seq( (2^(n-1)/n!)*mul(2^j-1, j=1..n-1), n=1..20); # G. C. Greubel, Feb 05 2020
MATHEMATICA
Table[QFactorial[n-1, 2] 2^(n-1)/n!, {n, 20}]
PROG
(PARI) a(n) = (2^(n-1)/n!) * prod(k=1, n-1, 2^k-1); \\ Michel Marcus, Jan 16 2020
(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
(Sage) from sage.combinat.q_analogues import q_factorial
[2^(n-1)*q_factorial(n-1, 2)/factorial(n) for n in (1..20)] # G. C. Greubel, Feb 05 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Karol A. Penson, Jan 27 2004
EXTENSIONS
Corrected and edited by Thomas Ordowski, Jan 16 2020
STATUS
approved