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 A279363 Sum of 4th powers of proper divisors of n. 3
 0, 1, 1, 17, 1, 98, 1, 273, 82, 642, 1, 1650, 1, 2418, 707, 4369, 1, 7955, 1, 10898, 2483, 14658, 1, 26482, 626, 28578, 6643, 41090, 1, 62644, 1, 69905, 14723, 83538, 3027, 133923, 1, 130338, 28643, 174994, 1, 236692, 1, 249170, 57893, 279858, 1, 423794, 2402, 401267, 83603, 485810, 1, 644372, 15267, 659842, 130403, 707298, 1, 1053636 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS a(n) = 1 if n is prime. a(p^k) = (p^(4*k) - 1)/(p^4 - 1) for p is prime. LINKS Eric Weisstein's World of Mathematics, Proper divisors FORMULA Dirichlet g.f.: zeta(s-4)*(zeta(s) - 1). a(n) = A001159(n) - A000583(n). G.f.: -x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5 + Sum_{k>=1} k^4 x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 18 2017 Sum_{k=1..n} a(k) ~ (Zeta(5) - 1)*n^5 / 5. - Vaclav Kotesovec, Feb 02 2019 EXAMPLE a(10) = 1^4 + 2^4 + 5^4 = 642, because 10 has 3 proper divisors {1,2,5}. a(11) = 1^4 = 1, because 11 has 1 proper divisor {1}. MATHEMATICA Table[DivisorSigma[4, n] - n^4, {n, 60}] PROG (PARI) for(n=1, 60, print1(sigma(n, 4) - n^4, ", ")) \\ Indranil Ghosh, Mar 18 2017 (Python) from sympy.ntheory import divisor_sigma print[divisor_sigma(n, 4) - n**4 for n in xrange(1, 61)] # Indranil Ghosh, Mar 18 2017 CROSSREFS Cf. A000583, A001159. Cf. A001065, A067558, A276634, A279364. Sequence in context: A189120 A102292 A264439 * A295576 A223519 A139804 Adjacent sequences:  A279360 A279361 A279362 * A279364 A279365 A279366 KEYWORD nonn,easy AUTHOR Ilya Gutkovskiy, Dec 10 2016 STATUS approved

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Last modified October 16 20:35 EDT 2019. Contains 328103 sequences. (Running on oeis4.)