A339277 Number of partitions of 2*n into powers of 2 where every part appears at least 2 times.
1, 1, 2, 3, 5, 6, 9, 11, 16, 19, 25, 30, 39, 45, 56, 65, 81, 92, 111, 127, 152, 171, 201, 226, 265, 295, 340, 379, 435, 480, 545, 601, 682, 747, 839, 920, 1031, 1123, 1250, 1361, 1513, 1640, 1811, 1963, 2164, 2335, 2561, 2762, 3027, 3253, 3548, 3813, 4153, 4448, 4827, 5167
Offset: 0
Keywords
Examples
a(4) = 5 because we have [4, 4], [2, 2, 2, 2], [2, 2, 2, 1, 1], [2, 2, 1, 1, 1, 1] and [1, 1, 1, 1, 1, 1, 1, 1].
Programs
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Mathematica
nmax = 55; CoefficientList[Series[(1/(1 - x^3)) Product[1/(1 - x^(2^k)), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] A000123[0] = 1; A000123[n_] := A000123[n] = A000123[Floor[n/2]] + A000123[n - 1]; a[n_] := Sum[ChebyshevU[k, -1/2] A000123[n - k], {k, 0, n}]; Table[a[n], {n, 0, 55}]
Formula
G.f.: (1/(1 - x^3)) * Product_{k>=0} 1/(1 - x^(2^k)).
G.f.: (1/(1 - x)) * Product_{k>=0} (1 + x^(2^(k+1))/(1 - x^(2^k))).
a(n) = [x^(2*n)] Product_{k>=0} (1 + x^(2^(k+1))/(1 - x^(2^k))).
a(n) = Sum_{k=0..n} U(k,-1/2) * A000123(n-k), where U(k,x) is the Chebyshev U-polynomial.