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A345936
Number of divisors d of n for which A002034(d) < A002034(n), where A002034(n) is the smallest positive integer k such that n divides k!.
5
0, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 2, 4, 1, 4, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 4, 1, 5, 2, 2, 2, 6, 1, 2, 2, 4, 1, 4, 1, 3, 4, 2, 1, 8, 2, 4, 2, 3, 1, 6, 2, 4, 2, 2, 1, 6, 1, 2, 3, 5, 2, 4, 1, 3, 2, 4, 1, 8, 1, 2, 4, 3, 2, 4, 1, 8, 3, 2, 1, 6, 2, 2, 2, 4, 1, 8, 2, 3, 2, 2, 2, 10, 1, 4, 3, 6, 1, 4, 1, 4, 4
OFFSET
1,4
FORMULA
a(n) = Sum_{d|n} [A002034(d) < A002034(n)], where [ ] is the Iverson bracket.
a(n) = A000005(n) - A345935(n).
EXAMPLE
36 has 9 divisors: 1, 2, 3, 4, 6, 9, 12, 18, 36. When A002034 is applied to them, one obtains values [1, 2, 3, 4, 3, 6, 4, 6, 6], thus there are six divisors that do not obtain the maximal value 6 obtained at 36 itself, therefore a(36) = 6.
MATHEMATICA
a[n_]:=(m=1; While[Mod[m!, n]!=0, m++]; m); Table[Length@Select[Divisors@k, a@#<a@k&], {k, 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.
A345936(n) = { my(x=A002034(n)); sumdiv(n, d, A002034(d)<x); };
CROSSREFS
Cf. also A344589.
Sequence in context: A223135 A084113 A323086 * A370814 A347050 A348383
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 02 2021
STATUS
approved