A368843 a(n) gives the number of triples of equally spaced 1's in the binary expansion of n.
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 1, 2, 2, 4, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 3, 0, 0, 0, 0, 0, 1, 0, 2, 1, 1, 2, 2, 2, 3, 4, 6, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 3, 0, 0, 0, 0, 1, 2, 1
Offset: 0
Examples
For n = 277: - the binary expansion of 277 is "100010101", - we have the following triples: 1 1 1 1 1 1 - so a(277) = 2.
Programs
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PARI
a(n, t = 1, base = 2) = { my (d = digits(n, base), v = 0); for (i = 1, #d-2, if (d[i]==t, forstep (j = i+2, #d, 2, if (d[i]==d[j] && d[i]==d[(i+j)/2], v++;);););); return (v); }
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Python
def A368843(n): l = len(s:=bin(n)[2:]) return sum(1 for i in range(l-2) for j in range(1,l-i+1>>1) if s[i:i+(j<<1)+1:j]=='111') # Chai Wah Wu, Jan 10 2024
Formula
a(2*n) = a(n).
a(2*n + 1) >= a(n).
Comments