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A269255 a(n) = (2^(2*n+1) - 1)*(3^(n+1) - 1)/2.

%I A269255 #20 Jul 26 2016 20:46:09
%S A269255 1,28,403,5080,61831,745108,8952763,107475760,1289869711,15479049388,
%T A269255 185750955523,2229020652040,26748283770391,320979546636868,
%U A269255 3851755118036683,46221063628493920,554652772325571871,6655833302847731548,79869999773355124243,958439997835247481400,11501279976237683562151
%N A269255 a(n) = (2^(2*n+1) - 1)*(3^(n+1) - 1)/2.
%C A269255 The sum of divisors of powers of 12 (A001021).
%C A269255 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).
%H A269255 Ilya Gutkovskiy, <a href="/A269255/a269255.pdf">The sum of the divisors of k^n</a>
%H A269255 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (20,-115,240,-144)
%F A269255 O.g.f.: (1 + 8*x - 42*x^2)/((1 - x)*(1 - 3*x)*(1 - 4*x)*(1 - 12*x)).
%F A269255 E.g.f.: (1 - 3*exp(2*x) - 2*exp(3*x) + 6*exp(11*x))*exp(x)/2.
%F A269255 a(n) = 20*a(n-1) - 115*a(n-2) + 240*a(n-3) - 144*a(n-4).
%F A269255 a(n) = A000203(A001021(n)).
%F A269255 a(n) = A000203(A000244(n))*A000203(A000302(n)).
%F A269255 a(n) = A083420(n)*A003462(n+1).
%F A269255 Sum_{n>=0} (-1)^n*a(n)/n! = (6 - 2*exp(8) - 3*exp(9) + exp(11))/(2*exp(12)) = 0.0909619117822510506...
%F A269255 Lim_{n->infinity} a(n)/a(n+1) = 1/12 = A021016.
%e A269255 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.
%t A269255 LinearRecurrence[{20, -115, 240, -144}, {1, 28, 403, 5080}, 21]
%t A269255 Table[(2^(2 n + 1) - 1) ((3^(n + 1) - 1)/2), {n, 0, 20}]
%t A269255 Table[DivisorSigma[1, 12^n], {n, 0, 20}]
%o A269255 (PARI) a(n)=(2^(2*n+1)-1)*(3^(n+1)-1)/2 \\ _Charles R Greathouse IV_, Jul 26 2016
%Y A269255 Cf. A000203, A000244, A000302, A001021, A003462, A021016, A083420.
%K A269255 nonn,easy
%O A269255 0,2
%A A269255 _Ilya Gutkovskiy_, Jul 13 2016