A026355 - OEIS

%I #16 Aug 26 2022 02:49:22

%S 1,2,3,5,6,8,10,11,13,14,16,18,19,21,23,24,26,27,29,31,32,34,35,37,39,

%T 40,42,44,45,47,48,50,52,53,55,57,58,60,61,63,65,66,68,69,71,73,74,76,

%U 78,79,81,82,84,86,87,89,90,92,94,95,97,99,100,102,103,105,107,108,110

%N a(n) = least k such that s(k) = n+1, where s = A026354.

%C Let f(1)=1, f(2)=q and f(k+2) = f(k+1)+f(k)-n; a(n) is the smallest positive integer q such that f(k) -> infinity as k -> infinity. - _Benoit Cloitre_, Aug 04 2002

%H J. H. Conway and N. J. A. Sloane, <a href="/A019586/a019586.pdf">Notes on the Para-Fibonacci and related sequences</a>

%F For n>0, a(n) = floor((n-1)*phi) + 2, where phi=(1+sqrt(5))/2.

%F Recurrences: a(n+1) = a(n)+(3 + sign(phi*n-a(n)))/2 for n>=0. Also a(n+1) = a(n) + 1 + A005614(n-2) for n>=2. - _Benoit Cloitre_, Aug 04 2002

%o (Python)

%o from math import isqrt

%o def A026355(n): return (n-1+isqrt(5*(n-1)**2)>>1)+2 if n else 1 # _Chai Wah Wu_, Aug 25 2022

%Y Cf. A000201, A005614, A026351. Different from A007067.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_