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Number of partitions of n into Pell parts (A000129).
2

%I #5 Aug 15 2017 20:37:23

%S 1,1,2,2,3,4,5,6,7,8,10,11,14,15,18,20,23,26,29,32,36,39,44,47,53,57,

%T 63,68,74,81,88,95,103,110,120,128,139,148,159,170,182,195,208,221,

%U 236,250,267,282,300,317,336,355,375,396,418,440,464,487,514,539,568,595,625,655,687,720,754,788

%N Number of partitions of n into Pell parts (A000129).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PellNumber.html">Pell Number</a>

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

%F G.f.: Product_{k>=1} 1/(1 - x^A000129(k)).

%e a(5) = 4 because we have [5], [2, 2, 1], [2, 1, 1, 1] and [1, 1, 1, 1, 1].

%t CoefficientList[Series[Product[1/(1 - x^Fibonacci[k, 2]), {k, 1, 15}], {x, 0, 67}], x]

%Y Cf. A000129, A007000, A003107, A067592, A067593, A290808.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Aug 11 2017