

A026274


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


15



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
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OFFSET

1,1


COMMENTS

This is the upper sWythoff sequence, where s(n)=n+1.
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
One has r*r*(n2*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)
Let Z(n) = d(L)...d(1)d(0) be the Zeckendorf expansion of n. Wellknown is:
d(0) = 1 if and only if n = floor(k*r^2)  1
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 n1 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



FORMULA

a(n) = floor(r*r*(n+2r3)), 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
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


MATHEMATICA

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*r*(n+2r3)];
Table[a[n], {n, 200}]
Table[Floor[GoldenRatio^2 (n+2*GoldenRatio3)], {n, 60}] (* Harvey P. Dale, Dec 23 2022 *)


PROG

(Haskell)
a026274 n = a026274_list !! (n1)
a026274_list = map (subtract 1) $ tail $ filter ((== 1) . a035612) [1..]
(Python)
from math import isqrt


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



