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).

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

Author

Cameron Musard, Aug 16 2019

Keywords

• nonn
• base

Comments

a(n) = 1 occurs at n = 2^k for nonnegative integers k.

Examples

			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.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..16384

Crossrefs

Cf. A306607.

Programs

PARI

a(n) = my (b=binary(n)); sum(k=1, #b, b[k]*binomial(#b-1,k-1)) \\ Rémy Sigrist, Aug 20 2019

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)

Extensions

Edited by N. J. A. Sloane, Sep 21 2019

Data corrected by Rémy Sigrist, Sep 22 2019