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A293806 a(0) = a(1) = 1; a(n) = [x^n] Product_{k=1..n-1} 1/(1 - x^a(k)). 4

%I #10 Dec 13 2017 16:51:20

%S 1,1,1,4,6,8,11,14,19,24,30,37,47,57,70,84,102,121,144,170,202,235,

%T 275,319,372,429,495,567,652,742,848,963,1095,1237,1399,1574,1775,

%U 1990,2235,2499,2795,3114,3473,3859,4292,4755,5271,5827,6444,7107,7840,8625,9493,10422,11444,12541

%N a(0) = a(1) = 1; a(n) = [x^n] Product_{k=1..n-1} 1/(1 - x^a(k)).

%C a(n) = number of partitions of n into preceding terms starting from a(1), a(2), a(3), ... (for n > 1).

%H Alois P. Heinz, <a href="/A293806/b293806.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>

%F 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)).

%e a(4) = 6 because we have [4], [1a, 1a, 1a, 1a], [1a, 1a, 1a, 1b], [1a, 1a, 1b, 1b], [1a, 1b, 1b, 1b] and [1b, 1b, 1b, 1b].

%e G.f. = -x - 2*x^2 + 1/((1 - x)*(1 - x)*(1 - x^4)*(1 - x^6)*(1 - x^8)*(1 - x^11)*(1 - x^14)*(1 - x^19)*...) = 1 + x + x^2 + 4*x^3 + 6*x^4 + 8*x^5 + 11*x^6 + 14*x^7 + 19*x^8 + ...

%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,

%p b(n, i-1)+`if`(a(i)>n, 0, b(n-a(i), i))))

%p end:

%p a:= n-> `if`(n<2, 1, b(n, n-1)):

%p seq(a(n), n=0..60); # _Alois P. Heinz_, Oct 16 2017

%t a[n_] := a[n] = SeriesCoefficient[Product[1/(1 - x^a[k]), {k, 1, n - 1}], {x, 0, n}]; a[0] = a[1] = 1; Table[a[n], {n, 0, 55}]

%Y Cf. A000041, A007279, A151945, A229362.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Oct 16 2017

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Last modified April 24 18:17 EDT 2024. Contains 371962 sequences. (Running on oeis4.)