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A286875 If n = Product (p_j^k_j) then a(n) = Sum (k_j >= 2, p_j^k_j). 2
0, 0, 0, 4, 0, 0, 0, 8, 9, 0, 0, 4, 0, 0, 0, 16, 0, 9, 0, 4, 0, 0, 0, 8, 25, 0, 27, 4, 0, 0, 0, 32, 0, 0, 0, 13, 0, 0, 0, 8, 0, 0, 0, 4, 9, 0, 0, 16, 49, 25, 0, 4, 0, 27, 0, 8, 0, 0, 0, 4, 0, 0, 9, 64, 0, 0, 0, 4, 0, 0, 0, 17, 0, 0, 25, 4, 0, 0, 0, 16, 81, 0, 0, 4, 0, 0, 0, 8, 0, 9, 0, 4, 0, 0, 0, 32, 0, 49, 9, 29, 0, 0, 0, 8, 0, 0, 0, 31 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

Sum of unitary, proper prime power divisors of n.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..16384

Index entries for sequences related to sums of divisors

FORMULA

a(n) = Sum_{d|n, d = p^k, p prime, k >= 2, gcd(d, n/d) = 1} d.

a(A246547(k)) = A246547(k).

a(A005117(k)) = 0.

EXAMPLE

a(360) = a(2^3*3^2*5) = 2^3 + 3^2 = 17.

MATHEMATICA

Table[DivisorSum[n, # &, CoprimeQ[#, n/#] && PrimePowerQ[#] && !PrimeQ[#] &], {n, 108}]

PROG

(Python)

from sympy import primefactors, isprime, gcd, divisors

def a(n): return sum([d for d in divisors(n) if gcd(d, n/d)==1 and len(primefactors(d))==1 and isprime(d)==0])

print map(a, range(1, 109)) # Indranil Ghosh, Aug 02 2017

(PARI) A286875(n) = { my(f=factor(n)); for (i=1, #f~, if(f[i, 2] < 2, f[i, 1] = 0)); vecsum(vector(#f~, i, f[i, 1]^f[i, 2])); }; \\ Antti Karttunen, Oct 07 2017

CROSSREFS

Cf. A005117, A008475, A222416, A023888, A023889, A034448, A063956, A077610, A092261, A246547, A284117.

Sequence in context: A252798 A169766 A003194 * A105570 A327054 A101419

Adjacent sequences:  A286872 A286873 A286874 * A286876 A286877 A286878

KEYWORD

nonn

AUTHOR

Ilya Gutkovskiy, Aug 02 2017

STATUS

approved

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Last modified March 29 21:32 EDT 2020. Contains 333117 sequences. (Running on oeis4.)