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%I #17 Nov 08 2023 07:51:59
%S 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,
%T 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,
%U 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
%N 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).
%H John Tyler Rascoe, <a href="/A119803/b119803.txt">Table of n, a(n) for n = 0..10000</a>
%e 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.
%o (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); }
%o { print(A119803(6)); } \\ _R. J. Mathar_, May 30 2006
%o (Python)
%o from collections import Counter
%o def A119803_list(max_n):
%o A,C = [0],Counter()
%o for n in range(1,max_n+1):
%o C.update({A[-1]})
%o A.append(C[A[int('0'+bin(n)[3:],2)]])
%o return(A) # _John Tyler Rascoe_, Nov 07 2023
%Y Cf. A119802.
%K easy,nonn
%O 0,4
%A _Leroy Quet_, May 24 2006
%E More terms from _R. J. Mathar_, May 30 2006
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