A345932 a(n) = A002034(n) / gcd(n, A002034(n)), where A002034(n) gives the smallest positive integer k such that n divides k!.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3

Offset

1,9

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences related to factorial numbers

Formula

a(n) = A002034(n) / A345931(n) = A002034(n) / gcd(n, A002034(n))

Mathematica

Table[s=(m=1; While[Mod[m!, n]!=0, m++]; m); s/GCD[n, s], {n, 100}] (* Giorgos Kalogeropoulos, Jul 02 2021 *)

Prog

(PARI)
A002034(n) = if(1==n, n, my(s=factor(n)[, 1], k=s[#s], f=Mod(k!, n)); while(f, f*=k++); (k)); \\ After code in A002034.
A345932(n) = { my(u=A002034(n)); (u/gcd(n, u)); };

Crossrefs

Cf. A002034, A345931, A345933.

Sequence in context: A261527 A292436 A184097 * A205399 A135303 A036065

Adjacent sequences: A345929 A345930 A345931 * A345933 A345934 A345935

Keyword

nonn

Author

Antti Karttunen, Jul 01 2021

Status

approved