A089401 a(n) = m - A089398(2^m + n) for m>=n.

1, 1, 3, 2, 4, 5, 6, 5, 7, 8, 11, 9, 11, 12, 13, 12, 14, 15, 18, 18, 19, 20, 21, 20, 22, 23, 26, 24, 26, 27, 28, 27, 29, 30, 33, 33, 36, 36, 37, 36, 38, 39, 42, 40, 42, 43, 44, 43, 45, 46, 49, 49, 50, 51, 52, 51, 53, 54, 57, 55, 57, 58, 59, 58, 60, 61, 64, 64, 67, 69, 69, 68, 70

Offset: 1

Author

Paul D. Hanna, Oct 30 2003

Keywords

• nonn

Comments

A089398(n) = n-th column sum of binary digits of k*2^(k-1), where summation is over all k>=1, without carrying from columns sums that may exceed 2.

Row sums of triangular arrays in A103582 and in A103583. - Philippe Deléham, Apr 04 2005

Examples

			a(6)=5 since 7 - A089398(2^7 + 6) = 7 - 2 = 5.
		

Crossrefs

Cf. A089398, A089400.

Programs

Mathematica

f[n_] := Block[{lg = Floor[Log[2, n]] + 1}, Sum[ Join[ Reverse[ IntegerDigits[n - i + 1, 2]], {0}][[i]], {i, lg}]]; Table[n - f[2^n + n] + 2, {n, 0, 72}] (* Robert G. Wilson v, Mar 29 2005 *)

PARI

a(n)=n/2+1/2*sum(k=1,n,(-1)^floor((n-k)/2^(k-1))) \\ Benoit Cloitre

PARI

{a(n)=if(n<=0,0,m=floor(log(n)/log(2)); if(n-2^m<=m,n-m+a(n-2^m),2^m-1+a(n-2^m)))} \\ Paul D. Hanna, Mar 28 2005

Formula

a(n) = n/2+1/2*sum(k=1, n, (-1)^floor((n-k)/2^(k-1))). - Benoit Cloitre, Mar 28 2005

Let a(0)=0; when n - 2^[log_2(n)] <= [log_2(n)] then a(n) = a(n - 2^[log_2(n)]) + n - [log_2(n)], else a(n) = a(n - 2^[log_2(n)]) + 2^[log_2(n)] - 1. Thus a(2^m) = 2^m - m for all m>=0; for 0<=k<=m: a(2^m + k) = a(k) + 2^m + k - m; for mPaul D. Hanna, Mar 28 2005

Extensions

More terms from Benoit Cloitre and Robert G. Wilson v, Mar 28 2005