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%I #17 Dec 30 2017 03:36:15
%S 1,0,0,1,1,0,0,2,0,0,1,2,0,0,2,1,0,0,3,0,0,2,2,0,0,3,0,0,1,3,0,0,3,2,
%T 0,0,4,0,0,2,3,0,0,3,1,0,0,4,0,0,3,3,0,0,5,0,0,2,4,0,0,4,2,0,0,5,0,0,
%U 3,3,0,0,4,0,0,1,4,0,0,4,3,0,0,6,0,0,3,5,0,0,5,2,0,0,6,0,0,4,4
%N Number of partitions of n into distinct Lucas parts (A000204) greater than 1.
%C Convolution of the sequences A003263 and A033999.
%C Positions of 0: 1, 2, 5, 6, 8, 9, 12, 13, ... = A287775(n) - 1 (conjecture).
%C From _Michel Dekking_, Dec 30 2017: (Start)
%C Proof of the 'positions of 0' conjecture: let (z(n))=1,2,5,6,8,9,12,… be the positions of 0. The crucial observation is that if a number n is the sum of distinct Lucas parts greater than 1, then n+1 is a sum of Lucas parts. This implies that (z(2n))=2,6,9,13,… is the sequence of numbers A054770 that are not a sum of Lucas numbers. We see there that Ian Agol proved that b(n):=A054770(n)=floor(phi*n)+2n-1. But then the sequence of first differences (b(n+1)-b(n)) equals the Fibonacci word on the alphabet {4,3}, yielding that (z(2n)-z(2n-1)) equals the Fibonacci word on {3,2}, and we already know that z(2n+1)-z(2n)=1 for all n. On the other hand, A287775 has the same first difference sequence given by A108103. Since A(287775(1))=2, the conjecture follows. (End)
%C Positions of 1: 0, 3, 4, 10, 15, 28, 44, 75, ... = A001350(n+1) - 1 (conjecture).
%H Ilya Gutkovskiy, <a href="/A294203/a294203.jpg">Scatter plot of a(n) up to n=50000</a>
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%F G.f.: Product_{k>=2} (1 + x^Lucas(k)).
%e a(7) = 2 because we have [7] and [4, 3].
%t CoefficientList[Series[Product[1 + x^LucasL[k], {k, 2, 15}], {x, 0, 100}], x]
%Y Cf. A000204, A003263, A033999, A067592, A239002, A294204, A054770.
%K nonn
%O 0,8
%A _Ilya Gutkovskiy_, Oct 24 2017
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