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A368843 a(n) gives the number of triples of equally spaced 1's in the binary expansion of n.

%I A368843 #12 Jan 10 2024 12:15:15
%S A368843 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,
%T A368843 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,
%U A368843 0,0,0,1,0,1,0,1,0,1,1,3,0,0,0,0,1,2,1
%N A368843 a(n) gives the number of triples of equally spaced 1's in the binary expansion of n.
%H A368843 <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F A368843 a(2*n) = a(n).
%F A368843 a(2*n + 1) >= a(n).
%e A368843 For n = 277:
%e A368843 - the binary expansion of 277 is "100010101",
%e A368843 - we have the following triples:  1   1   1
%e A368843                                       1 1 1
%e A368843 - so a(277) = 2.
%o A368843 (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); }
%o A368843 (Python)
%o A368843 def A368843(n):
%o A368843     l = len(s:=bin(n)[2:])
%o A368843     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
%Y A368843 Cf. A368842, A368844.
%K A368843 nonn,base,easy
%O A368843 0,16
%A A368843 _Rémy Sigrist_, Jan 07 2024