A120289 -id:A120289 - cp's OEIS frontend

A293457 Primes that divide the numerator of the sum of the reciprocals of all smaller primes.

2, 5, 19, 47, 79, 109, 3667387

Offset: 1

Logan J. Kleinwaks, Oct 09 2017

Exhaustive search finds no more terms among the first 10^7 primes.

Primes p that divide A024451(A000720(p)-1). - Antti Karttunen, Feb 08 2024

			Since 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 = 716167/510510 and 19 divides 716167, 19 is in the sequence.
Since there are no primes less than 2, the sum of their reciprocals is 0/1, and as 2 divides 0, it is therefore included as the first term of this sequence. - _Antti Karttunen_, Feb 08 2024

Cf. A000040, A000720, A024451, A120289, A369972, A369973.

lista(nn) = my(s = 0); forprime(p=2, nn, if (!(numerator(s) % p), print1(p, ", ")); s += 1/p; ); \\ Michel Marcus, Oct 09 2017, edited for the new, more inclusive definition by Antti Karttunen, Feb 08 2024

a(n) = A000040(1+A369972(n)). - Antti Karttunen, Feb 08 2024

Relaxed the definition to include 2 as the first term - Antti Karttunen, Feb 08 2024

A120347 Numerator of Sum_{k=1..n-1} 1/k^n.

Original entry on oeis.org

1, 9, 1393, 257875, 47463376609, 940908897061, 972213062238348973121, 7727182467755471289426059, 10338014371627802833957102351534201, 26038773205374138944970092886340352227, 205885410277133543091182509665217407908365393153956577

Offset: 2

Alexander Adamchuk, Aug 16 2006, Oct 31 2006

Prime p>2 divides a(p). p^3 divides a(p) for prime p>3. p divides a((p+1)/2) for prime p = {7,11,17,19,23,31,41,43,47,59,67,71,73,79,83,89,97,103,...} = all primes excluding 2 and 3 from A045323[n] Primes congruent to {1, 2, 3, 7} mod 8.

a(n) = Numerator( H(n-1,n) ), where H(k,r) = Sum_{i=1..k} 1/i^r is the generalized harmonic number.

Vincenzo Librandi, Table of n, a(n) for n = 2..49
Eric Weisstein's World of Mathematics, Wolstenholme's Theorem
Eric Weisstein's World of Mathematics, Harmonic Number

Cf. A045323, A120289, A120352 (a(prime(n))), A119722 (a(prime(n))/prime(n)^3).

Table[Numerator[Sum[1/k^n,{k,1,n-1}]],{n,2,15}]

a(n) = Numerator(Sum_{k=1..n-1} 1/k^n). a(n) = Numerator[Zeta[n] - Zeta[n,n]].

cp's OEIS Frontend

A293457 Primes that divide the numerator of the sum of the reciprocals of all smaller primes.

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