A349080 Numbers k for which there exists only one integer m with 1 <= m <= k such that A000178(k) / m! is a square, where A000178(k) = k$ = 1!*2!*...*k! is the superfactorial of k.

1, 2, 4, 12, 18, 20, 24, 28, 34, 36, 40, 44, 52, 56, 60, 62, 64, 68, 76, 80, 84, 88, 92, 98, 100, 104, 108, 112, 116, 120, 124, 132, 136, 140, 142, 144, 148, 152, 156, 164, 168, 172, 176, 180, 184, 188, 192, 194, 196, 204, 208, 212, 216, 220, 224, 228, 232, 236, 244, 248, 252, 254, 256

Offset: 1

Author

Bernard Schott, Nov 20 2021

Keywords

• nonn

Comments

This sequence is the union of {1} and of three infinite and disjoint subsequences.

-> Numbers k divisible by 4 but not of the form 8q^2 or 8q(q+1) = {4, 12, 20, 24, 28, ...} (see A182834). For these numbers, the corresponding unique m = k/2 (see example for k = 4).

-> Even numbers k not divisible by 4 and of the form k = 2*A055792 = 2*q^2, q>1 in A001541 = {18, 578, ...}. For these numbers, the corresponding unique m = k/2 - 2 = q^2-2 (see example for k = 18)

-> Even numbers k not divisible by 4, that are in A060626 but not of the form k=2q^2-4 with q>1 in A001541 = {2, 34, 62, 98, 142, 194, ...} (A349496). For these numbers, the corresponding unique m = k/2 + 1 (see example for k = 2).

See A348692 for further information.

Examples

			For k = 2, 2$ / 2! = 1^2, hence 2 is a term.
For k = 4, 4$ /1! = 288, 4$ / 3! = 48, 4$ / 4! = 12 but for m = 2, 4$ / 2! = 12^2, hence 4 is a term.
For k = 18 and m = 7, we have 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2 and there is no other solution m, hence 18 is a term.
		

Links

• Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.

Crossrefs

Cf. A000178, A001541, A060626, A348692, A182834, A349496.

Subsequence of A349079.

Programs

Mathematica

q[n_] := Count[BarnesG[n + 2]/Range[n]!, ?(IntegerQ@Sqrt[#] &)] == 1; Select[Range[100], q] (* _Amiram Eldar, Nov 20 2021 *)

PARI

sf(n) = prod(k=2, n, k!); \\ A000178
isok(m) = my(s=sf(m)); #select(issquare, vector(m, k, s/k!), 1) == 1; \\ Michel Marcus, Nov 20 2021