A334735 -id:A334735 - cp's OEIS frontend

A166602 Numbers k such that Sum_{i=1..k} i^2 divides Product_{i=1..k} i^2.

1, 7, 13, 17, 19, 24, 25, 27, 31, 32, 34, 37, 38, 43, 45, 47, 49, 55, 57, 59, 61, 62, 64, 67, 71, 73, 76, 77, 79, 80, 84, 85, 87, 91, 92, 93, 94, 97, 101, 103, 104, 107, 109, 110, 115, 117, 118, 121, 122, 123, 124, 127, 129, 132, 133, 137, 139, 142, 143, 144, 145, 147

Offset: 1

Alexander Adamchuk, Oct 18 2009

nonn

Product_{i=1..k} i^2 = (k!)^2 and Sum_{i=1..k} i^2 = k*(k+1)*(2*k+1)/6. - J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010

			a(2) = A125314(2) = 7.

J. Mulder, Table of n, a(n) for a(n) below 20000

Cf. A000330, A001044, A125294, A125314, A060462, A166604, A166605, A166606, A166607, A166608, A166609, A166610, A334735.

Cf. A067656. - R. J. Mathar, Oct 23 2009

q:= k-> is(irem(k!^2, k*(k+1)*(2*k+1)/6)=0):
select(q, [$1..200])[];  # Alois P. Heinz, May 09 2020

Cases[Range[2, 5000], k_ /; Divisible[Factorial[k - 1]^2, 1/6 (-1 + k) k (-1 + 2 k)]] - 1 (* J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010 *)

isok(k) = ((k!)^2 % (k*(k+1)*(2*k+1)/6)) == 0; \\ Michel Marcus, May 09 2020

Terms below 5000 by J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010

More terms copied from the b-file by R. J. Mathar, Feb 14 2010

A125294 Numerator of (Sum_{k=1..n} k^2) / (Product_{k=1..n} k^2).

Original entry on oeis.org

1, 5, 7, 5, 11, 91, 1, 17, 19, 11, 23, 13, 1, 29, 31, 17, 1, 703, 1, 41, 43, 23, 47, 1, 1, 53, 1, 29, 59, 1891, 1, 1, 67, 1, 71, 2701, 1, 1, 79, 41, 83, 43, 1, 89, 1, 47, 1, 97, 1, 101, 103, 53, 107, 109, 1, 113, 1, 59, 1, 61, 1, 1, 127, 1, 131, 67, 1, 137, 139, 71, 1, 73, 1, 149

Offset: 1

Alexander Adamchuk, Jan 17 2007

All a(n) are either 1, semiprime or prime.
a(n) = 1 for n = 1 and n = {7, 13, 17, 19, 24, 25, 27, 31, 32, 34, 37, 38, 43, 45, 47, 49, ...} = A067656 = numbers n such that n!*B(2*n) is an integer, where the B(2*n)'s are the Bernoulli numbers.
p divides a(p-1) for prime p > 3. p divides a((p-1)/2) for prime p > 3.
a(p-1) = p*(2p-1) is a semiprime hexagonal number for prime p = {7, 19, 31, 37, 79, 97, 139, 157, 199, 211, 229, 271, 307, 331, 337, 367, 379, 439, 499, ...} = A005382(n) for n > 2, where A005382(n) are the numbers n such that n and 2*n-1 are primes.
a(p-1) = p for prime p = {5, 11, 13, 17, 23, 29, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, ...} = primes that do not belong to A005382(n).
a((p-1)/2) = p for prime p = {5, 7, 11, 17, 19, 23, 29, 31, 41, 43, 47, 53, 59, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 163, 167, 173, 179, 181, 191, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 259, 271, 281, 283, 293, 307, 311, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 401, ...}, which is apparently the union of {5} and A034849(n).

			The first few fractions are 1, 5/4, 7/18, 5/96, 11/2880, 91/518400, 1/181440, 17/135475200, 19/8778792960, ... = A125294/A334735. - _Petros Hadjicostas_, May 09 2020

Cf. A005382, A034849, A067656, A166602, A334735 (denominators).

Table[Numerator[n(n+1)(2n+1)/6/(n!)^2],{n,1,500}]

a(n) = numerator(sum(k=1, n, k^2)/prod(k=1, n, k^2)); \\ Michel Marcus, May 09 2020

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

a(n) = numerator(n*(n+1)*(2*n+1)/6/(n!)^2).

cp's OEIS Frontend

A166602 Numbers k such that Sum_{i=1..k} i^2 divides Product_{i=1..k} i^2.

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