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%I #5 Aug 18 2019 12:13:17
%S 1,4,9,81,244,793,2316,65536,262145,1049600,4196353,17308657,68703188,
%T 273234809,1088123500,152587890625,762939453126,3814697527769,
%U 19073486852414,95370918425026,476847618556329,2384217176269538,11921023106645561,59886119752101281
%N If 2*n = Sum (2^e_k) then a(n) = Sum (e_k^n).
%C Replace 2^k with (k + 1)^n in binary representation of n.
%F a(n) = [x^n] (1/(1 - x)) * Sum_{k>=0} (k + 1)^n*x^(2^k)/(1 + x^(2^k)).
%e 14 = 2*7 = 2^1 + 2^2 + 2^3 so a(7) = 1^7 + 2^7 + 3^7 = 2316.
%t Table[Reverse[#].Range[Length[#]]^n &@IntegerDigits[n, 2], {n, 1, 24}]
%t Table[SeriesCoefficient[1/(1 - x) Sum[(k + 1)^n x^2^k/(1 + x^2^k), {k, 0, Floor[Log[2, n]] + 1}], {x, 0, n}], {n, 1, 24}]
%Y Cf. A008935, A029931, A104258.
%K nonn
%O 1,2
%A _Ilya Gutkovskiy_, Aug 17 2019
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