%I #15 Oct 05 2018 11:17:15

%S 0,0,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,

%T 1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0

%N a(n) = 0 if n of form m(m+1)(m+2)/6, otherwise 1.

%C a(A145397(n))=1; a(A000292(n))=0; a(n)=1-A023533(n). - _Reinhard Zumkeller_, Oct 14 2008

%C Characteristic function of A145397.

%H Antti Karttunen, <a href="/A014306/b014306.txt">Table of n, a(n) for n = 0..65537</a>

%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>

%e From _David A. Corneth_, Oct 01 2018: (Start)

%e For n = 0, floor((6*0-1) ^ (1/3)) = -1. binomial(-1 + 2, 3) = n so a(0) = 0.

%e For n = 10, floor((6*n-1) ^ (1/3)) = 3. binomial(3 + 2, 3) = n so a(10) = 0.

%e For n = 11, floor((6*n-1) ^ (1/3)) = 3. binomial(3 + 2, 3) != n so a(11) = 1. (End)

%o (PARI) A014306(n) = { my(k=0); while(binomial(k+2,3)<n,k++); !(binomial(k+2,3)==n); }; \\ _Antti Karttunen_, Sep 30 2018

%o (PARI) a(n) = if(n==0, return(0)); my(t = sqrtnint(6*n-1, 3)); binomial(t+2, 3) != n \\ _David A. Corneth_, Oct 01 2018

%o (PARI) first(n) = my(res = vector(n+1, i, 1), ov = nv = [1,2,1,0]); while(nv[4]<=n, res[nv[4]+1] = 0; for(i = 2, 4, nv[i] = ov[i-1] + ov[i]); ov = nv); res \\ _David A. Corneth_, Oct 01 2018

%Y Cf. A000292, A023533, A145397.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_

%E Data section extended up to a(120) by _Antti Karttunen_, Sep 30 2018