A026274 Greatest k such that s(k) = n, where s = A026272.

3, 5, 8, 11, 13, 16, 18, 21, 24, 26, 29, 32, 34, 37, 39, 42, 45, 47, 50, 52, 55, 58, 60, 63, 66, 68, 71, 73, 76, 79, 81, 84, 87, 89, 92, 94, 97, 100, 102, 105, 107, 110, 113, 115, 118, 121, 123, 126, 128, 131, 134, 136, 139, 141, 144

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

This is the upper s-Wythoff sequence, where s(n)=n+1.

See comments at A026273.

Conjecture: This sequence consists precisely of those numbers without a 1 or 2 in their Zeckendorf representation. [In other words, numbers which are the sum of distinct nonconsecutive Fibonacci numbers greater than 2.] - Charles R Greathouse IV, Jan 28 2015

A Beatty sequence with complement A026273. - Robert G. Wilson v, Jan 30 2015

A035612(a(n)+1) = 1. - Reinhard Zumkeller, Jul 20 2015

From Michel Dekking, Mar 12 2018: (Start)

One has r*r*(n-2*r+3) = n*r^2 -2r^3+3*r^2 = (n+1)*r^2 -2, where r = (1+sqrt(5))/2.

So a(n) = floor((n+1)*r^2)-2, and we see that this sequence is simply the Beatty sequence of the square of the golden ratio, shifted spatially and temporally. In other words, if w = A001950 = 2,5,7,10,13,15,18,20,... is the upper Wythoff sequence, then a(n) = w(n+1) - 2.

(End)

From Michel Dekking, Apr 05 2020: (Start)

Proof of the conjecture by Charles R Greathouse IV.

Let Z(n) = d(L)...d(1)d(0) be the Zeckendorf expansion of n. Well-known is:

d(0) = 1 if and only if n = floor(k*r^2) - 1

for some integer k (see A003622).

Then the same characterization holds for n with d(1)d(0) = 01, since 11 does not appear in a Zeckendorf expansion. But such an n has predecessor n-1 which always has an expansion with d(1)d(0) = 00. Combined with my comment from March 2018, this proves the conjecture (ignoring n = 0). (End)

It appears that these are the integers m for which A007895(m+1) > A007895(m) where A007895(m) is the number of terms in Zeckendorf representation of m. - Michel Marcus, Oct 30 2020

This follows directly from Theorem 4 in my paper "Points of increase of the sum of digits function of the base phi expansion". - Michel Dekking, Oct 31 2020

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Hung Viet Chu, Difference in the Number of Summands in the Zeckendorf Partitions of Consecutive Integers, arXiv:2010.15592 [math.NT], 2020.
• Michel Dekking, Points of increase of the sum of digits function of the base phi expansion, arXiv:2003.14125 [math.CO], 2020.
• F. Michel Dekking, The sum of digits functions of the Zeckendorf and the base phi expansions, Theoretical Computer Science (2021) Vol. 859, 70-79.

Crossrefs

Cf. A184117, A026273, A001950, A003622.

Programs

Haskell

a026274 n = a026274_list !! (n-1)
a026274_list = map (subtract 1) $ tail $ filter ((== 1) . a035612) [1..]
-- Reinhard Zumkeller, Jul 20 2015

Mathematica

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*r*(n+2r-3)];
Table[a[n],{n,200}]
Table[Floor[GoldenRatio^2 (n+2*GoldenRatio-3)],{n,60}] (* Harvey P. Dale, Dec 23 2022 *)

PARI

a(n)=my(w=quadgen(20),phi=(1+w)/2); phi^2*(n+2*phi-3)\1 \\ Charles R Greathouse IV, Nov 10 2021

Python

from math import isqrt
def A026274(n): return (n+1+isqrt(5*(n+1)**2)>>1)+n-1 # Chai Wah Wu, Aug 17 2022

Formula

a(n) = floor(r*r*(n+2r-3)), where r = (1+sqrt(5))/2 = A001622. [Corrected by Tom Edgar, Jan 30 2015]

a(n) = 3*n - floor[(n+1)/(1+phi)], phi = (1+sqrt(5))/2. - Joshua Tobin (tobinrj(AT)tcd.ie), May 31 2008

a(n) = A003622(n+1) - 1 for n>1 (conjectured). - Michel Marcus, Oct 30 2020

This conjectured formula follows directly from the formula a(n) = floor((n+1)*r^2)-2 in my Mar 12 2018 comment above. - Michel Dekking, Oct 31 2020

Extensions

Extended by Clark Kimberling, Jan 14 2011