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%I #11 Aug 16 2017 05:02:36
%S 1,1,1,1,0,1,1,1,1,0,0,0,1,1,1,1,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,
%T 1,1,1,1,0,0,0,1,1,1,1,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,1,1,1,1,0,1,1,1,1,0,0,0,1,1,1,1,0,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,0,1,1,1,1
%N Number of partitions of n into distinct Pell parts (A000129).
%C All terms are 0 or 1. - _Robert Israel_, Aug 16 2017
%H Robert Israel, <a href="/A290808/b290808.txt">Table of n, a(n) for n = 0..10000</a>
%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 + x^A000129(k)).
%e a(8) = 1 because we have [5, 2, 1].
%p N:= 200: # to get a(0) to a(N)
%p Pell:= gfun:-rectoproc({a(0)=0,a(1)=1,a(n+1)=2*a(n)+a(n-1)},a(n),remember):
%p G:= 1:
%p for k from 1 while Pell(k) <= N do G:= G*(1+x^Pell(k)) od:
%p seq(coeff(G,x,n),n=0..N); # _Robert Israel_, Aug 16 2017
%t CoefficientList[Series[Product[(1 + x^Fibonacci[k, 2]), {k, 1, 15}], {x, 0, 108}], x]
%Y Cf. A000119, A000121, A000129, A003263, A290807.
%K nonn
%O 0
%A _Ilya Gutkovskiy_, Aug 11 2017
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