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%I #35 Oct 14 2023 23:48:15
%S 1,1,2,2,2,2,4,4,2,2,6,6,6,6,2,2,2,2,10,10,10,10,2,2,12,12,4,4,4,4,12,
%T 12,2,2,14,14,14,14,6,6,14,14,6,7,7,7,14,14,14,14,4,4,4,4,14,14,4,4,
%U 12,12,12,12,8,8,2,2,16,16,16,16,12,12,16,16,7,7,7,7,16,16,16,16,4,4,4,4
%N a(1) = 1. For m >= 0 and 1 <= k <= 2^m, a(2^m +k) = number of earlier terms of the sequence which equal a(k).
%C Interpreted as a triangle with row lengths A011782, row m+1 is the frequency of each term in rows 1..m among terms in the sequence thus far (including the part of row m+1 itself thus far). - _Neal Gersh Tolunsky_, Oct 03 2023
%H Neal Gersh Tolunsky, <a href="/A119802/b119802.txt">Table of n, a(n) for n = 1..10000</a>
%H Neal Gersh Tolunsky, <a href="/A119802/a119802.png">Run length transform of 2^16 terms</a>
%e 8 = 2^2 + 4; so for a(8) we want the number of terms among terms a(1), a(2),... a(7) which equal a(4) = 2. So a(8) = 4.
%e As a triangle:
%e k=1 2 3 4 5 6 7 8 ...
%e m=1: 1;
%e m=2: 1;
%e m=3: 2, 2;
%e m=4: 2, 2, 4, 4;
%e m=5: 2, 2, 6, 6, 6, 6, 2, 2;
%e m=6: 2, 2, 10, 10, 10, 10, 2, 2, 12, 12, 4, 4, 4, 4, 12, 12;
%e ...
%o (PARI) A119802(mmax)= { local(a,ncopr); a=[1]; for(m=0,mmax, for(k=1,2^m, ncopr=0; for(i=1,2^m+k-1, if( a[i]==a[k], ncopr++; ); ); a=concat(a,ncopr); ); ); return(a); }
%o print(A119802(6)); \\ _R. J. Mathar_, May 30 2006
%Y Cf. A119803, A011782 (row lengths), A365882.
%K easy,nonn
%O 1,3
%A _Leroy Quet_, May 24 2006
%E More terms from _R. J. Mathar_, May 30 2006
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