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A269255
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a(n) = (2^(2*n+1) - 1)*(3^(n+1) - 1)/2.
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1
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1, 28, 403, 5080, 61831, 745108, 8952763, 107475760, 1289869711, 15479049388, 185750955523, 2229020652040, 26748283770391, 320979546636868, 3851755118036683, 46221063628493920, 554652772325571871, 6655833302847731548, 79869999773355124243, 958439997835247481400, 11501279976237683562151
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OFFSET
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0,2
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
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.
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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