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Greatest k such that s(k) = n, where s = A026272.
15

%I #78 Jan 19 2023 02:05:58

%S 3,5,8,11,13,16,18,21,24,26,29,32,34,37,39,42,45,47,50,52,55,58,60,63,

%T 66,68,71,73,76,79,81,84,87,89,92,94,97,100,102,105,107,110,113,115,

%U 118,121,123,126,128,131,134,136,139,141,144

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

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

%C See comments at A026273.

%C 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

%C A Beatty sequence with complement A026273. - _Robert G. Wilson v_, Jan 30 2015

%C A035612(a(n)+1) = 1. - _Reinhard Zumkeller_, Jul 20 2015

%C From _Michel Dekking_, Mar 12 2018: (Start)

%C 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.

%C 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.

%C (End)

%C From _Michel Dekking_, Apr 05 2020: (Start)

%C Proof of the conjecture by _Charles R Greathouse IV_.

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

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

%C for some integer k (see A003622).

%C 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)

%C 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

%C 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

%H Reinhard Zumkeller, <a href="/A026274/b026274.txt">Table of n, a(n) for n = 1..10000</a>

%H Hung Viet Chu, <a href="https://arxiv.org/abs/2010.15592">Difference in the Number of Summands in the Zeckendorf Partitions of Consecutive Integers</a>, arXiv:2010.15592 [math.NT], 2020.

%H Michel Dekking, <a href="https://arxiv.org/abs/2003.14125">Points of increase of the sum of digits function of the base phi expansion</a>, arXiv:2003.14125 [math.CO], 2020.

%H F. Michel Dekking, <a href="https://doi.org/10.1016/j.tcs.2021.01.011">The sum of digits functions of the Zeckendorf and the base phi expansions</a>, Theoretical Computer Science (2021) Vol. 859, 70-79.

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

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

%F a(n) = A003622(n+1) - 1 for n>1 (conjectured). - _Michel Marcus_, Oct 30 2020

%F 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

%t r=(1+Sqrt[5])/2;

%t a[n_]:=Floor[r*r*(n+2r-3)];

%t Table[a[n],{n,200}]

%t Table[Floor[GoldenRatio^2 (n+2*GoldenRatio-3)],{n,60}] (* _Harvey P. Dale_, Dec 23 2022 *)

%o (Haskell)

%o a026274 n = a026274_list !! (n-1)

%o a026274_list = map (subtract 1) $ tail $ filter ((== 1) . a035612) [1..]

%o -- _Reinhard Zumkeller_, Jul 20 2015

%o (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

%o (Python)

%o from math import isqrt

%o def A026274(n): return (n+1+isqrt(5*(n+1)**2)>>1)+n-1 # _Chai Wah Wu_, Aug 17 2022

%Y Cf. A184117, A026273, A001950, A003622.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_

%E Extended by _Clark Kimberling_, Jan 14 2011