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A352729 The binary expansion of a(n) contains the runs of consecutive 1's that appear both in the binary expansions of n and n+1.

%I A352729 #10 Apr 01 2022 09:03:32
%S A352729 0,0,0,0,4,0,0,0,8,8,8,0,12,0,0,0,16,16,16,16,20,16,16,0,24,24,24,0,
%T A352729 28,0,0,0,32,32,32,32,36,32,32,32,40,40,40,32,44,32,32,0,48,48,48,48,
%U A352729 52,48,48,0,56,56,56,0,60,0,0,0,64,64,64,64,68,64,64,64
%N A352729 The binary expansion of a(n) contains the runs of consecutive 1's that appear both in the binary expansions of n and n+1.
%C A352729 We only consider runs of consecutive 1's that completely match in binary expansions of n and n+1, not simply single common 1's.
%H A352729 Rémy Sigrist, <a href="/A352729/b352729.txt">Table of n, a(n) for n = 0..8192</a>
%H A352729 <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F A352729 a(n) = A352727(n, n+1).
%e A352729 For n = 42:
%e A352729 - the binary expansion of 42 is "101010",
%e A352729 - the binary expansion of 43 is "101011",
%e A352729 - the first two runs of 1's are the same, the others differ,
%e A352729 - so the binary expansion of a(42) is "101000",
%e A352729 - and a(42) = 40.
%o A352729 (PARI) A352724(n) = { my (r=[], o=0); while (n, my (v=valuation(n+n%2, 2)); if (n%2, r=concat(r, (2^v-1)*2^o)); o+=v; n\=2^v); r }
%o A352729 a(n) = vecsum(setintersect(A352724(n), A352724(n+1)))
%Y A352729 Cf. A129760, A352727.
%K A352729 nonn,base
%O A352729 0,5
%A A352729 _Rémy Sigrist_, Mar 30 2022