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 A269255 a(n) = (2^(2*n+1) - 1)*(3^(n+1) - 1)/2. 1
 1, 28, 403, 5080, 61831, 745108, 8952763, 107475760, 1289869711, 15479049388, 185750955523, 2229020652040, 26748283770391, 320979546636868, 3851755118036683, 46221063628493920, 554652772325571871, 6655833302847731548, 79869999773355124243, 958439997835247481400, 11501279976237683562151 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The sum of divisors of powers of 12 (A001021). The sum of divisors of powers of prime p are sigma_1(p^n) = Sum_{m=0}^n p^m = (p^(n+1) - 1)/(p - 1) (see examples in the links section). LINKS Ilya Gutkovskiy, The sum of the divisors of k^n Index entries for linear recurrences with constant coefficients, signature (20,-115,240,-144) FORMULA O.g.f.: (1 + 8*x - 42*x^2)/((1 - x)*(1 - 3*x)*(1 - 4*x)*(1 - 12*x)). E.g.f.: (1 - 3*exp(2*x) - 2*exp(3*x) + 6*exp(11*x))*exp(x)/2. a(n) = 20*a(n-1) - 115*a(n-2) + 240*a(n-3) - 144*a(n-4). a(n) = A000203(A001021(n)). a(n) = A000203(A000244(n))*A000203(A000302(n)). a(n) = A083420(n)*A003462(n+1). Sum_{n>=0} (-1)^n*a(n)/n! = (6 - 2*exp(8) - 3*exp(9) + exp(11))/(2*exp(12)) = 0.0909619117822510506... Lim_{n->infinity} a(n)/a(n+1) = 1/12 = A021016. EXAMPLE a(1) = 28, because 12^1 = 12 and 12 has 6 divisors (1, 2, 3, 4, 6, 12) -> 1 + 2 + 3 + 4 + 6 + 12 = 28. MATHEMATICA LinearRecurrence[{20, -115, 240, -144}, {1, 28, 403, 5080}, 21] Table[(2^(2 n + 1) - 1) ((3^(n + 1) - 1)/2), {n, 0, 20}] Table[DivisorSigma[1, 12^n], {n, 0, 20}] PROG (PARI) a(n)=(2^(2*n+1)-1)*(3^(n+1)-1)/2 \\ Charles R Greathouse IV, Jul 26 2016 CROSSREFS Cf. A000203, A000244, A000302, A001021, A003462, A021016, A083420. Sequence in context: A035476 A042518 A297179 * A125463 A161571 A161956 Adjacent sequences:  A269252 A269253 A269254 * A269256 A269257 A269258 KEYWORD nonn,easy AUTHOR Ilya Gutkovskiy, Jul 13 2016 STATUS approved

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Last modified June 20 04:43 EDT 2021. Contains 345157 sequences. (Running on oeis4.)