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A371242
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The sum of the unitary divisors of n that are cubefree numbers (A004709).
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1
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1, 3, 4, 5, 6, 12, 8, 1, 10, 18, 12, 20, 14, 24, 24, 1, 18, 30, 20, 30, 32, 36, 24, 4, 26, 42, 1, 40, 30, 72, 32, 1, 48, 54, 48, 50, 38, 60, 56, 6, 42, 96, 44, 60, 60, 72, 48, 4, 50, 78, 72, 70, 54, 3, 72, 8, 80, 90, 60, 120, 62, 96, 80, 1, 84, 144, 68, 90, 96
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OFFSET
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1,2
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COMMENTS
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The number of these divisors is A365498(n).
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REFERENCES
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D. Suryanarayana, The number and sum of k-free intergers <= x which are prime to n, Indian J. Math., Vol. 11 (1969), pp. 131-139.
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LINKS
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Francesco Pappalardi, A survey on k-freeness, Number Theory, Ramanujan Math. Soc. Lect. Notes Ser., Vol. 1 (2003), pp. 71-88.
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FORMULA
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Multiplicative with a(p^e) = p^e + 1 for e <= 2, and a(p^e) = 1 for e >= 3.
a(n) = 1 if and only if n is cubefull (A036966).
a(n) <= A034448(n), with equality if and only if n is cubefree.
Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-2) - 1/p^(2*s-1) - 1/p^(3*s-2)).
Sum_{j=1..n} a(j) ~ c * n^2 / 2, where c = Product_{p prime} (1 - 1/p^3 + 1/(p^2 + p)) = 1.16545286600957717104.... .
In general, the formula above holds for the sum of unitary divisors of n that are k-free numbers (k >= 2) with c = Product_{p prime} (1 - 1/p^k + 1/(p^2 + p)) (Suryanarayana, 1969). If k = 2 then c = A065465. In the limit when k -> oo, c = A306633.
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MATHEMATICA
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f[p_, e_] := If[e < 3, p^e + 1, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
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PROG
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(PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] < 3, f[i, 1]^f[i, 2] + 1, 1)); }
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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STATUS
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approved
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