A294203 Number of partitions of n into distinct Lucas parts (A000204) greater than 1.

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

Offset: 0

Author

Ilya Gutkovskiy, Oct 24 2017

Keywords

• nonn

Comments

Convolution of the sequences A003263 and A033999.

Positions of 0: 1, 2, 5, 6, 8, 9, 12, 13, ... = A287775(n) - 1 (conjecture).

From Michel Dekking, Dec 30 2017: (Start)

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 A287775(1)=2, the conjecture follows. (End)

Positions of 1: 0, 3, 4, 10, 15, 28, 44, 75, ... = A001350(n+1) - 1 (conjecture).

Examples

			a(7) = 2 because we have [7] and [4, 3].
		

Links

• Ilya Gutkovskiy, Scatter plot of a(n) up to n=50000
• Index entries for sequences related to partitions

Crossrefs

Cf. A000204, A003263, A033999, A067592, A239002, A294204, A054770.

Programs

Mathematica

CoefficientList[Series[Product[1 + x^LucasL[k], {k, 2, 15}], {x, 0, 100}], x]

Formula

G.f.: Product_{k>=2} (1 + x^Lucas(k)).