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A286875 If n = Product (p_j^k_j) then a(n) = Sum (k_j >= 2, p_j^k_j). 2

%I

%S 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,

%T 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,

%U 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

%N If n = Product (p_j^k_j) then a(n) = Sum (k_j >= 2, p_j^k_j).

%C Sum of unitary, proper prime power divisors of n.

%H Antti Karttunen, <a href="/A286875/b286875.txt">Table of n, a(n) for n = 1..16384</a>

%H <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>

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

%F a(A246547(k)) = A246547(k).

%F a(A005117(k)) = 0.

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

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

%o (Python)

%o from sympy import primefactors, isprime, gcd, divisors

%o 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])

%o print map(a, range(1, 109)) # _Indranil Ghosh_, Aug 02 2017

%o (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

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

%K nonn

%O 1,4

%A _Ilya Gutkovskiy_, Aug 02 2017

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Last modified May 30 05:35 EDT 2020. Contains 334712 sequences. (Running on oeis4.)