A225101 -id:A225101 - cp's OEIS frontend

A064535 a(n) = (2^prime(n)-2)/prime(n); a(0) = 0 by convention.

0, 1, 2, 6, 18, 186, 630, 7710, 27594, 364722, 18512790, 69273666, 3714566310, 53634713550, 204560302842, 2994414645858, 169947155749830, 9770521225481754, 37800705069076950, 2202596307308603178, 33256101992039755026, 129379903640264252430

Offset: 0

Shane Findley, Oct 09 2001

nonn

As a corollary to Fermat's little theorem, (2^p - 2)/p is always an integer for p prime. - Alonso del Arte, May 04 2013

			a(3) = 6, because prime(3) = 5, and (2^5 - 2)/5 = 30/5 = 6.
a(4) = 18, because prime(4) = 7, and (2^7  - 2)/7 = 126/7 = 18.

Harry J. Smith, Table of n, a(n) for n = 0..100

Cf. A007663, A056743, A225101 (superset).

[0] cat [(2^NthPrime(n)-2)/NthPrime(n): n in [1..25]]; // Vincenzo Librandi, Sep 14 2018

A064535 := proc(n) ( 2^ithprime(n) - 2 )/ithprime(n); end;

Table[(2^Prime[n] - 2)/Prime[n], {n, 50}] (* Alonso del Arte, Apr 28 2013 *)

{ for (n=0, 100, if (n, a=(2^prime(n) - 2)/prime(n), a=0); write("b064535.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 17 2009

a(n) = A001037(prime(n)) for n >= 1. - Hilko Koning, Sep 10 2018

a(n) = 2*A007663(n) for n > 1. - Jeppe Stig Nielsen, May 16 2021

A330718 a(n) = numerator(Sum_{k=1..n} (2^k-2)/k).

Original entry on oeis.org

0, 1, 3, 13, 25, 137, 245, 871, 517, 4629, 8349, 45517, 83317, 1074679, 1992127, 7424789, 13901189, 78403447, 147940327, 280060651, 531718651, 11133725681, 21243819521, 40621501691, 15565330735, 388375065019, 248882304985, 479199924517, 923951191477, 2973006070891

Offset: 1

Amiram Eldar and Thomas Ordowski, Dec 28 2019

If p > 3 is prime, then p^2 | a(p).
Note the similarity to Wolstenholme's theorem.
Conjecture: for n > 3, if n^2 | a(n), then n is prime.
Are there the weak pseudoprimes m such that m | a(m)?
Primes p such that p^3 | a(p) are probably A088164.
If p is an odd prime, then a(p+1) == A330719(p+1) (mod p).
If p > 3 is a prime, then p^2 | numerator(Sum_{k=1..p+1} F(k)), where F(n) = Sum_{k=1..n} (2^(k-1)-1)/k. Cf. A027612 (a weaker divisibility).

			Numerators of 0, 1, 3, 13/2, 25/2, 137/6, 245/6, ...

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Wolstenholme's Theorem

Cf. A000265, A001008, A007663, A088164, A159353, A225101, A279683, A290347, A330719.

[Numerator( &+[(2^k -2)/k: k in [1..n]] ): n in [1..30]]; // G. C. Greubel, Dec 28 2019

seq(numer(add((2^k -2)/k, k = 1..n)), n = 1..30); # G. C. Greubel, Dec 28 2019

Numerator @ Accumulate @ Array[(2^# - 2)/# &, 30]
Table[Numerator[Simplify[-(2^(n+1)*LerchPhi[2,1,n+1] +Pi*I +2*HarmonicNumber[n])]], {n,30}] (* G. C. Greubel, Dec 28 2019 *)

a(n) = numerator(sum(k=1, n, (2^k-2)/k)); \\ Michel Marcus, Dec 28 2019

[numerator( sum((2^k -2)/k for k in (1..n)) ) for n in (1..30)] # G. C. Greubel, Dec 28 2019

a(n) = numerator(Sum_{k=1..n} (2^(k-1)-1)/k).
a(n+1) = numerator(a(n)/A330719(n) + A225101(n+1)/(2*A159353(n+1))).
a(p) = a(p-1) + A007663(n)*A330719(p-1) for p = prime(n) > 2.
a(n) = numerator(-(2^(n+1)*LerchPhi(2,1,n+1) + Pi*i + 2*HarmonicNumber(n))). - G. C. Greubel, Dec 28 2019
a(n) = numerator(A279683(n)/n!) for n > 0. - Amiram Eldar and Thomas Ordowski, Jan 15 2020
For n > 1, a(n) = A000265(A290347(n)). - Thomas Ordowski, Mar 29 2025

A091669 a(n) = (2^(n-1)/n!) * Product_{k=1..n-1} (2^k-1).

Original entry on oeis.org

1, 1, 2, 7, 42, 434, 7812, 248031, 14055090, 1436430198, 267176016828, 91151551074486, 57425477176926180, 67196011936600334340, 146782968474309770332296, 601204690999713530559792879

Offset: 1

Karol A. Penson, Jan 27 2004

nonn

Primes p such that 2^p-2 divides a(p) are A216838. - Amiram Eldar and Thomas Ordowski, Jan 16 2020
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

Amiram Eldar, Table of n, a(n) for n = 1..86

Cf. A000142, A001122, A001567, A005329, A006935, A159353, A216838, A225101.

[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

seq( (2^(n-1)/n!)*mul(2^j-1, j=1..n-1), n=1..20); # G. C. Greubel, Feb 05 2020

Table[QFactorial[n-1, 2] 2^(n-1)/n!, {n, 20}]

a(n) = (2^(n-1)/n!) * prod(k=1, n-1, 2^k-1); \\ Michel Marcus, Jan 16 2020

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

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

Corrected and edited by Thomas Ordowski, Jan 16 2020

cp's OEIS Frontend

A064535 a(n) = (2^prime(n)-2)/prime(n); a(0) = 0 by convention.

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A330718 a(n) = numerator(Sum_{k=1..n} (2^k-2)/k).

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A091669 a(n) = (2^(n-1)/n!) * Product_{k=1..n-1} (2^k-1).

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