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A374539
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The sum of the squares of the infinitary divisors of n.
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1
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1, 5, 10, 17, 26, 50, 50, 85, 82, 130, 122, 170, 170, 250, 260, 257, 290, 410, 362, 442, 500, 610, 530, 850, 626, 850, 820, 850, 842, 1300, 962, 1285, 1220, 1450, 1300, 1394, 1370, 1810, 1700, 2210, 1682, 2500, 1850, 2074, 2132, 2650, 2210, 2570, 2402, 3130, 2900
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OFFSET
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1,2
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COMMENTS
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Also the sum of the infinitary divisors of n^2.
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LINKS
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FORMULA
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Multiplicative with a(p^e) = Product{k>=1, e_k=1} (p^(2^(k+1)) + 1), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = Product_{P} (1 + 1/(P^2*(P+1))) = 1.14142906130350119631..., and P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376).
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MATHEMATICA
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f[p_, e_] := p^(2^Position[Reverse@IntegerDigits[e, 2], _?(# == 1 &)]); a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]
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PROG
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(PARI) a(n) = {my(f = factor(n), b); prod(i = 1, #f~, b = binary(2 * f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1))); }
(Python)
from math import prod
from sympy import factorint
def A374539(n): return prod(p**(1<<i)+1 for p, e in factorint(n).items() for i, j in enumerate(bin(e)[-1:1:-1], 1) if j=='1') # Chai Wah Wu, Jul 11 2024
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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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