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A294203
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Number of partitions of n into distinct Lucas parts (A000204) greater than 1.
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3
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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, 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, 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
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OFFSET
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0,8
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COMMENTS
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Positions of 0: 1, 2, 5, 6, 8, 9, 12, 13, ... = A287775(n) - 1 (conjecture).
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)
Positions of 1: 0, 3, 4, 10, 15, 28, 44, 75, ... = A001350(n+1) - 1 (conjecture).
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LINKS
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FORMULA
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G.f.: Product_{k>=2} (1 + x^Lucas(k)).
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EXAMPLE
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a(7) = 2 because we have [7] and [4, 3].
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MATHEMATICA
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CoefficientList[Series[Product[1 + x^LucasL[k], {k, 2, 15}], {x, 0, 100}], x]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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