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A057723
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Sum of positive divisors of n that are divisible by every prime that divides n.
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43
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1, 2, 3, 6, 5, 6, 7, 14, 12, 10, 11, 18, 13, 14, 15, 30, 17, 24, 19, 30, 21, 22, 23, 42, 30, 26, 39, 42, 29, 30, 31, 62, 33, 34, 35, 72, 37, 38, 39, 70, 41, 42, 43, 66, 60, 46, 47, 90, 56, 60, 51, 78, 53, 78, 55, 98, 57, 58, 59, 90, 61, 62, 84, 126, 65, 66, 67, 102, 69, 70
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OFFSET
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1,2
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LINKS
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FORMULA
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If n = Product p_i^e_i then a(n) = Product (p_i + p_i^2 + ... + p_i^e_i).
Dirichlet g.f.: zeta(s) * zeta(s-1) * Product(p prime, 1 - p^(-s) + p^(1-2*s)). - Robert Israel, May 13 2015
Sum_{k=1..n} a(k) ~ c * Pi^2 * n^2 / 12, where c = A330596 = Product_{primes p} (1 - 1/p^2 + 1/p^3) = 0.7485352596823635646442150486379106016416403430053244045... - Vaclav Kotesovec, Dec 18 2019
a(n) = Sum_{d|n, rad(d)=rad(n)} d. - R. J. Mathar, Jun 02 2020
Lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k = Product_{p prime}(1 + 1/(p*(p^2-1))) = 1.231291... (A065487). - Amiram Eldar, Jun 10 2020
a(n) = Sum_{d|n, gcd(d, n/d) = 1} (-1)^omega(n/d) * sigma(d). - Ilya Gutkovskiy, Apr 15 2021
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EXAMPLE
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The divisors of 12 that are divisible by both 2 and 3 are 6 and 12. So a(12) = 6 + 12 = 18.
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MAPLE
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seq(mul(f[1]*(f[1]^f[2]-1)/(f[1]-1), f = ifactors(n)[2]), n = 1 .. 100); # Robert Israel, May 13 2015
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MATHEMATICA
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Table[(b = Times @@ FactorInteger[n][[All, 1]])*DivisorSigma[1, n/b], {n, 70}] (* Ivan Neretin, May 13 2015 *)
f[p_, e_] := (p^(e+1)-1)/(p-1) - 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 15 2023 *)
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PROG
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(Magma) [&*PrimeDivisors(n)*SumOfDivisors(n div &*PrimeDivisors(n)): n in [1..70]]; // Vincenzo Librandi, May 14 2015
(PARI) a(n) = {my(f = factor(n)); for (i=1, #f~, f[i, 2]=1); my(pp = factorback(f)); sumdiv(n, d, if (! (d % pp), d, 0)); } \\ Michel Marcus, May 14 2015
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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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