A293807 a(0) = a(1) = 1; a(n) = [x^n] Product_{k=1..n-1} 1/(1 - x^a(k))^a(k).
1, 1, 1, 4, 9, 14, 19, 24, 39, 63, 87, 111, 155, 235, 329, 423, 552, 771, 1091, 1430, 1825, 2400, 3295, 4392, 5597, 7117, 9367, 12476, 16077, 20182, 25677, 33472, 43406, 54578, 68109, 86475, 111316, 140965, 174836, 217520, 275130, 348555, 433578, 533640, 662620, 831747
Offset: 0
Keywords
Examples
a(3) = 4 because we have [1a, 1a, 1a], [1a, 1a, 1b], [1a, 1b, 1b] and [1b, 1b, 1b]. G.f. = -x - 2*x^2 + 1/((1 - x)*(1 - x)*(1 - x^4)^4*(1 - x^9)^9*(1 - x^14)^14*(1 - x^19)^19*(1 - x^24)^24*(1 - x^39)^39*...) = 1 + x + x^2 + 4*x^3 + 9*x^4 + 14*x^5 + 19*x^6 + 24*x^7 + 39*x^8 + ...
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Programs
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Maple
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(b(n-a(i)*j, i-1)*binomial(a(i)+j-1, j), j=0..n/a(i)))) end: a:= n-> `if`(n<2, 1, b(n, n-1)): seq(a(n), n=0..60); # Alois P. Heinz, Oct 16 2017 -
Mathematica
a[n_] := a[n] = SeriesCoefficient[Product[1/(1 - x^a[k])^a[k], {k, 1, n - 1}], {x, 0, n}]; a[0] = a[1] = 1; Table[a[n], {n, 0, 45}] -
Python
from sympy import binomial from sympy.core.cache import cacheit @cacheit def b(n, i): return 1 if n==0 else 0 if i<1 else sum(b(n - a(i)*j, i - 1) * binomial(a(i) + j - 1, j) for j in range(n//a(i) + 1)) def a(n): return 1 if n<2 else b(n, n - 1) print([a(n) for n in range(51)]) # Indranil Ghosh, Dec 13 2017, after Maple code
Formula
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies A(x) = -x - 2*x^2 + Product_{n>=1} 1/(1 - x^a(n))^a(n).
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