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A014306
a(n) = 0 if n of form m(m+1)(m+2)/6, otherwise 1.
39
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, 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, 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
OFFSET
0,1
COMMENTS
a(A145397(n))=1; a(A000292(n))=0; a(n)=1-A023533(n). -Reinhard Zumkeller,Oct 14 2008
Characteristic function ofA145397.
EXAMPLE
FromDavid A. Corneth,Oct 01 2018: (Start)
For n = 0, floor((6*0-1) ^ (1/3)) = -1. binomial(-1 + 2, 3) = n so a(0) = 0.
For n = 10, floor((6*n-1) ^ (1/3)) = 3. binomial(3 + 2, 3) = n so a(10) = 0.
For n = 11, floor((6*n-1) ^ (1/3)) = 3. binomial(3 + 2, 3)!= n so a(11) = 1. (End)
PROG
(PARI)A014306(n) = { my(k=0); while(binomial(k+2, 3)<n, k++);!(binomial(k+2, 3)==n); }; \\Antti Karttunen,Sep 30 2018
(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
(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
KEYWORD
nonn,easy
EXTENSIONS
Data section extended up to a(120) byAntti Karttunen,Sep 30 2018
STATUS
approved