|
Search: id:A119803
|
|
|
| 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). |
|
+0
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
(list; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
LINKS
|
Leroy Quet, Home Page (listed in lieu of email address)
|
|
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.
|
|
PROGRAM
|
(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 (mathar(AT)strw.leidenuniv.nl), May 30 2006
|
|
CROSSREFS
|
Cf. A119802.
Sequence in context: A162883 A081446 A158440 this_sequence A110569 A140815 A134840
Adjacent sequences: A119800 A119801 A119802 this_sequence A119804 A119805 A119806
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Leroy Quet May 24 2006
|
|
EXTENSIONS
|
More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 30 2006
|
|
|
Search completed in 0.002 seconds
|
