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A325801
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Number of divisors of n minus the number of distinct positive subset-sums of the prime indices of n.
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6
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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
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OFFSET
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1,24
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COMMENTS
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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.
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LINKS
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FORMULA
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MATHEMATICA
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hwt[n_]:=Total[Cases[FactorInteger[n], {p_, k_}:>PrimePi[p] k]];
Table[DivisorSigma[0, n]-Length[Union[hwt/@Divisors[n]]], {n, 100}]
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PROG
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(PARI)
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
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CROSSREFS
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Positions of positive integers are A299729.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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