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

0,4

Links

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

Example

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.

Prog

(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

Crossrefs

Cf. A119802.

Sequence in context: A081446 A274309 A158440 * A195916 A245554 A110569

Adjacent sequences: A119800 A119801 A119802 * A119804 A119805 A119806

Keyword

easy,nonn

Author

Leroy Quet, May 24 2006

Extensions

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

Status

approved