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A325801
Number of divisors of n minus the number of distinct positive subset-sums of the prime indices of n.
6
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 4, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0
OFFSET
1,24
COMMENTS
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798, with sum A056239(n). A positive subset-sum of an integer partition is any sum of a nonempty submultiset of it.
LINKS
FORMULA
a(n) = A000005(n) - A299701(n).
MATHEMATICA
hwt[n_]:=Total[Cases[FactorInteger[n], {p_, k_}:>PrimePi[p] k]];
Table[DivisorSigma[0, n]-Length[Union[hwt/@Divisors[n]]], {n, 100}]
PROG
(PARI)
A325801(n) = (numdiv(n) - A299701(n));
A299701(n) = { my(f = factor(n), pids = List([])); for(i=1, #f~, while(f[i, 2], f[i, 2]--; listput(pids, primepi(f[i, 1])))); pids = Vec(pids); my(sv=vector(vecsum(pids))); for(b=1, (2^length(pids))-1, sv[sumbybits(pids, b)] = 1); 1+vecsum(sv); }; \\ Not really an optimal way to count these.
sumbybits(v, b) = { my(s=0, i=1); while(b>0, s += (b%2)*v[i]; i++; b >>= 1); (s); }; \\ Antti Karttunen, May 26 2019
CROSSREFS
Positions of 0's are A299702.
Positions of 1's are A325802.
Positions of positive integers are A299729.
Sequence in context: A122841 A326075 A060862 * A325194 A066087 A294927
KEYWORD
nonn
AUTHOR
Gus Wiseman, May 23 2019
STATUS
approved