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A309776
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).
1
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, 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, 16, 17, 21, 22, 26, 27, 31, 32, 1, 2, 7, 8, 16, 17, 22, 23, 21, 22
OFFSET
0,4
COMMENTS
a(n) = 1 occurs at n = 2^k for nonnegative integers k.
LINKS
FORMULA
From Bernard Schott, Sep 22 2019: (Start)
a(2^k + 1) = 2 for k >= 1 where 2^k+1 = 1000..0001_2.
a(2^k - 1) = 2^(k-1) for k >= 2 where 2^k-1 = 111..111_2.
a((4^k-1)/3) = 2^(2*k-3) for k >= 2 where (4^k-1)/3 = 10101..0101_2.
(End)
EXAMPLE
For n=5 the triangle is
1 0 1
1 1
2
so a(5)=2.
For n=14 we get
1 1 1 0
2 2 1
4 3
7
so a(14)=7.
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.
PROG
(PARI) a(n) = my (b=binary(n)); sum(k=1, #b, b[k]*binomial(#b-1, k-1)) \\ Rémy Sigrist, Aug 20 2019
CROSSREFS
Cf. A306607.
Sequence in context: A233782 A233972 A169778 * A255560 A328472 A046653
KEYWORD
nonn,base
AUTHOR
Cameron Musard, Aug 16 2019
EXTENSIONS
Edited by N. J. A. Sloane, Sep 21 2019
Data corrected by Rémy Sigrist, Sep 22 2019
STATUS
approved