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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).
2
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
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: A274309 A374630 A158440 * A195916 A245554 A110569
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, May 24 2006
EXTENSIONS
More terms from R. J. Mathar, May 30 2006
STATUS
approved