A328211 Starts of runs of 4 consecutive Zeckendorf-Niven numbers (A328208).

1, 2, 3, 123543, 124242, 545502, 1367583, 1856349, 2431230, 2465110, 2593590, 2783709, 3247389, 3479229, 3917823, 3942909, 4174749, 4303428, 4494390, 4920640, 5143830, 5710383, 6261309, 6493149, 6552903, 6956829, 7420509, 7470880, 8970948, 9107790, 9507069, 10952928

Offset: 1

Author

Amiram Eldar, Oct 07 2019

Keywords

• nonn

Comments

Grundman proved that this sequence is infinite by showing the F(120k-6) + F(8) + F(6) + F(4) is a term for all k >= 1, where F(k) is the k-th Fibonacci number.

She also proved that the only starts of runs of 5 consecutive Zeckendorf-Niven numbers are 1 and 2.

Examples

			1 is in the sequence since 1, 2, 3 and 4 are in A328208: A007895(1) = 1 is a divisor of 1, A007895(2) = 1 is a divisor of 2, A007895(3) = 1 is a divisor of 3, and A007895(4) = 2 is a divisor of 4.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..216
• Helen G. Grundman, Consecutive Zeckendorf-Niven and lazy-Fibonacci-Niven numbers, Fibonacci Quarterly, Vol. 45, No. 3 (2007), pp. 272-276.

Crossrefs

Cf. A005349, A007895, A141769, A328208.

Programs

Mathematica

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; aQ[n_] := Divisible[n, z[n]]; c = 0; k = 1; s = {}; v = Table[-1, {4}]; While[c < 32, If[aQ[k], v = Join[Rest[v], {k}]; If[AllTrue[Differences[v], # == 1 &], c++; AppendTo[s, k - 3]]]; k++]; s (* after Alonso del Arte at A007895 *)