A119803 a(0) = 0. For m >= 0 and 0 <= k <= 2^m -1, a(2^m +k) = number of earlier terms of the sequence which equal a(k).

0, 1, 1, 2, 1, 3, 3, 1, 1, 5, 5, 1, 6, 2, 2, 6, 1, 7, 7, 3, 7, 3, 4, 7, 7, 2, 2, 7, 2, 6, 6, 4, 1, 8, 8, 6, 8, 4, 4, 8, 8, 2, 2, 8, 5, 8, 8, 5, 8, 6, 6, 4, 6, 4, 6, 6, 6, 8, 8, 6, 8, 12, 12, 6, 1, 9, 9, 8, 9, 4, 4, 9, 9, 4, 4, 9, 13, 8, 8, 13, 9, 6, 6, 4, 6, 4, 12, 6, 6, 8, 8, 6, 8, 19, 19, 12, 9, 18, 18, 19

Offset: 0

Author

Leroy Quet, May 24 2006

Keywords

• easy
• nonn

Examples

			8 = 2^3 + 0; so for a(8) we want the number of terms among terms a(1), a(2),... a(7) which equal a(0) = 0. So a(8) = 1.
		

Links

• John Tyler Rascoe, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A119802.

Programs

PARI

A119803(mmax)= { local(a,ncopr); a=[0]; for(m=0,mmax, for(k=0,2^m-1, ncopr=0; for(i=1,2^m+k, if( a[i]==a[k+1], ncopr++; ); ); a=concat(a,ncopr); ); ); return(a); }
{ print(A119803(6)); } \\  R. J. Mathar, May 30 2006

Python

from collections import Counter
def A119803_list(max_n):
    A,C = [0],Counter()
    for n in range(1,max_n+1):
        C.update({A[-1]})
        A.append(C[A[int('0'+bin(n)[3:],2)]])
    return(A) # John Tyler Rascoe, Nov 07 2023

Extensions

More terms from R. J. Mathar, May 30 2006