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A345933
a(n) = n / gcd(n, A002034(n)), where A002034(n) gives the smallest positive integer k such that n divides k!.
3
1, 1, 1, 1, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 8, 1, 3, 1, 4, 3, 2, 1, 6, 5, 2, 3, 4, 1, 6, 1, 4, 3, 2, 5, 6, 1, 2, 3, 8, 1, 6, 1, 4, 15, 2, 1, 8, 7, 5, 3, 4, 1, 6, 5, 8, 3, 2, 1, 12, 1, 2, 9, 8, 5, 6, 1, 4, 3, 10, 1, 12, 1, 2, 15, 4, 7, 6, 1, 40, 9, 2, 1, 12, 5, 2, 3, 8, 1, 15, 7, 4, 3, 2, 5, 12, 1, 7, 9, 10, 1, 6, 1, 8, 15
OFFSET
1,6
FORMULA
a(n) = n / A345931(n) = n / gcd(n, A002034(n)).
MATHEMATICA
Table[n/GCD[n, m=1; While[Mod[m!, n]!=0, m++]; m], {n, 100}] (* Giorgos Kalogeropoulos, Jul 03 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.
A345933(n) = (n/gcd(n, A002034(n)));
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 01 2021
STATUS
approved