

A026274


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


10



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.
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 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*(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)
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. Wellknown 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 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)


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Michel Dekking, Points of increase of the sum of digits function of the base phi expansion, arXiv:2003.14125 [math.CO], 2020.


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


MATHEMATICA

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*r*(n+2r3)];
Table[a[n], {n, 200}]


PROG

(Haskell)
a026274 n = a026274_list !! (n1)
a026274_list = map (subtract 1) $ tail $ filter ((== 1) . a035612) [1..]
 Reinhard Zumkeller, Jul 20 2015


CROSSREFS

Cf. A184117, A026273, A001950, A003622.
Sequence in context: A213732 A247909 A184659 * A137910 A022850 A008576
Adjacent sequences: A026271 A026272 A026273 * A026275 A026276 A026277


KEYWORD

nonn


AUTHOR

Clark Kimberling


EXTENSIONS

Extended by Clark Kimberling, Jan 14 2011


STATUS

approved



