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A091669
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a(n) = (2^(n-1)/n!) * Product_{k=1..n-1} (2^k-1).
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1
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1, 1, 2, 7, 42, 434, 7812, 248031, 14055090, 1436430198, 267176016828, 91151551074486, 57425477176926180, 67196011936600334340, 146782968474309770332296, 601204690999713530559792879
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OFFSET
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1,3
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COMMENTS
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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
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LINKS
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FORMULA
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MAPLE
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seq( (2^(n-1)/n!)*mul(2^j-1, j=1..n-1), n=1..20); # G. C. Greubel, Feb 05 2020
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MATHEMATICA
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Table[QFactorial[n-1, 2] 2^(n-1)/n!, {n, 20}]
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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