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).

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

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

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Neal Gersh Tolunsky, Run length transform of 2^16 terms

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

Adjacent sequences: A119799 A119800 A119801 * A119803 A119804 A119805

Keyword

easy,nonn

Author

Leroy Quet, May 24 2006

Extensions

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

Status

approved