A035612 Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 1) contains n.

1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 7, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 8, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 9, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 7, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 10, 1, 2, 3, 1, 4, 1, 2, 5, 1, 2

Offset: 1

Author

J. H. Conway and N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

Ordinal transform of A003603. Removing all 1's from this sequence and decrementing the remaining numbers generates the original sequence. - Franklin T. Adams-Watters, Aug 10 2012

It can be shown that a(n) is the index of the smallest Fibonacci number used in the Zeckendorf representation of n, where f(0)=f(1)=1. - Rachel Chaiser, Aug 18 2017

The asymptotic density of the occurrences of k = 1, 2, ..., is (2-phi)/phi^(k-1), where phi is the golden ratio (A001622). The asymptotic mean of this sequence is 1 + phi (A104457). - Amiram Eldar, Nov 02 2023

Examples

			After the first 6 we see "1 2 3 1 4 1 2" then 7.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Paul Curtz, Comments on A035612, Jan 25 2016.
• N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
• N. J. A. Sloane, Classic Sequences.

Crossrefs

Cf. A019586, A035513, A035614.

Cf. A000045, A003603.

Cf. A003714, A007814, A022340, A022342, A026274.

Cf. A000012, A000027, A001045, A001610, A003622, A023548, A255671, A268034.

Cf. A001622, A104457, A132338.

Programs

Haskell

a035612 = a007814 . a022340
-- Reinhard Zumkeller, Jul 20 2015, Mar 10 2013

Mathematica

f[1] = {1}; f[2] = {1, 2}; f[n_] := f[n] = Join[f[n-1], Most[f[n-2]], {n}]; f[11] (* Jean-François Alcover, Feb 22 2012 *)

Formula

The segment between the first M and the first M+1 is given by the segment before the first M-1.

a(A022342(n)) > 1; a(A026274(n) + 1) = 1. - Reinhard Zumkeller, Jul 20 2015

a(n) = v2(A022340(n)), where v2(n) = A007814(n), the dyadic valuation of n. - Ralf Stephan, Jun 20 2004. In other words, a(n) = A007814(A003714(n)) + 1, which is certainly true. - Don Reble, Nov 12 2005

From Rachel Chaiser, Aug 18 2017: (Start)

a(n) = a(p(n))+1 if n = b(p(n)) where p(n) = floor((n+2)/phi)-1 and b(n) = floor((n+1)*phi)-1 where phi=(1+sqrt(5))/2; a(n)=1 otherwise.

a(n) = 3 - n_1 + s_z(n-1) - s_z(n) + s_z(p(n-1)) - s_z(p(n)), where s_z(n) is the Zeckendorf sum of digits of n (A007895), and n_1 is the least significant digit in the Zeckendorf representation of n. (End)

Extensions

Formula corrected by Tom Edgar, Jul 09 2018