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).
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
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Jul 13 2016
STATUS
approved