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A119802 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). 2
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, 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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
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
LINKS
EXAMPLE
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.
As a triangle:
k=1 2 3 4 5 6 7 8 ...
m=1: 1;
m=2: 1;
m=3: 2, 2;
m=4: 2, 2, 4, 4;
m=5: 2, 2, 6, 6, 6, 6, 2, 2;
m=6: 2, 2, 10, 10, 10, 10, 2, 2, 12, 12, 4, 4, 4, 4, 12, 12;
...
PROG
(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); }
print(A119802(6)); \\ R. J. Mathar, May 30 2006
CROSSREFS
Cf. A119803, A011782 (row lengths), A365882.
Sequence in context: A060632 A160407 A007457 * A237120 A060369 A179004
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, May 24 2006
EXTENSIONS
More terms from R. J. Mathar, May 30 2006
STATUS
approved

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Last modified March 28 16:34 EDT 2024. Contains 371254 sequences. (Running on oeis4.)