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%I #53 Nov 11 2019 14:02:00
%S 0,1,1,2,1,2,3,4,1,2,4,5,4,5,7,8,1,2,5,6,7,8,11,12,5,6,9,10,11,12,15,
%T 16,1,2,6,7,11,12,16,17,11,12,16,17,21,22,26,27,6,7,11,12,16,17,21,22,
%U 16,17,21,22,26,27,31,32,1,2,7,8,16,17,22,23,21,22
%N Form a triangle: first row is n in base 2, next row is sums of pairs of adjacent digits of previous row, repeat until get a single number which is a(n).
%C a(n) = 1 occurs at n = 2^k for nonnegative integers k.
%H Rémy Sigrist, <a href="/A309776/b309776.txt">Table of n, a(n) for n = 0..16384</a>
%F From _Bernard Schott_, Sep 22 2019: (Start)
%F a(2^k + 1) = 2 for k >= 1 where 2^k+1 = 1000..0001_2.
%F a(2^k - 1) = 2^(k-1) for k >= 2 where 2^k-1 = 111..111_2.
%F a((4^k-1)/3) = 2^(2*k-3) for k >= 2 where (4^k-1)/3 = 10101..0101_2.
%F (End)
%e For n=5 the triangle is
%e 1 0 1
%e 1 1
%e 2
%e so a(5)=2.
%e For n=14 we get
%e 1 1 1 0
%e 2 2 1
%e 4 3
%e 7
%e so a(14)=7.
%e For n=26=11010_2; (n1+n2, n2+n3, n3+n4, n4+n5) = 2111; (n1'+n2', n2'+n3', n3'+n4') = 322; (n1''+n2'', n2''+n3'') = 54; (n1'''+n2''') = 9; a(26)= 9.
%o (PARI) a(n) = my (b=binary(n)); sum(k=1, #b, b[k]*binomial(#b-1,k-1)) \\ _Rémy Sigrist_, Aug 20 2019
%Y Cf. A306607.
%K nonn,base
%O 0,4
%A _Cameron Musard_, Aug 16 2019
%E Edited by _N. J. A. Sloane_, Sep 21 2019
%E Data corrected by _Rémy Sigrist_, Sep 22 2019
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