A060510 Alternating with hexagonal stutters: if n is hexagonal (2k^2 - k, i.e., A000384) then a(n)=a(n-1), otherwise a(n) = 1 - a(n-1).
0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1
Offset: 0
Examples
Hexagonal numbers start 1,6,15, ... so this sequence goes 0 0 (stutter at 1) 1 0 1 0 0 (stutter at 6) 1 0 1 0 1 0 1 0 0 (stutter at 15) 1 0, etc. As a triangle, sequence begins: 0; 0, 1; 0, 1, 0; 0, 1, 0, 1; 0, 1, 0, 1, 0; 0, 1, 0, 1, 0, 1; ...
Crossrefs
Programs
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Maple
T := proc(n, k): if k mod 2 = 1 then return(1) else return(0) fi: end: seq(seq(T(n, k), k=0..n), n=0..13); # Johannes W. Meijer, Aug 12 2015
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Mathematica
nxt[{n_,a_}]:={n+1,If[IntegerQ[(1+Sqrt[1+8(n+1)])/4],a,1-a]}; NestList[ nxt,{0,0},110][[All,2]] (* Harvey P. Dale, Jan 13 2022 *)
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Python
from math import isqrt def A060510(n): return n+1&1^1^((m:=isqrt(n+1<<3)+1>>1)*(m-1)>>1&1) # Chai Wah Wu, Oct 23 2024
Formula
G.f.: x/(1-x^2) - (1+x)^(-1)*Sum(n>=1, x^(n*(2*n-1))). The sum is related to Theta functions. - Robert Israel, Aug 12 2015
Comments