

A113504


a(0) = 1. For n >= 1, a(n) = number of earlier terms of the sequence that have the same number of ones in their binary representations as n.


5



1, 1, 2, 0, 3, 1, 1, 0, 5, 2, 2, 0, 2, 0, 0, 0, 8, 2, 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 13, 2, 2, 1, 2, 1, 1, 0, 2, 1, 1, 0, 1, 0, 0, 0, 2, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 28, 2, 2, 2, 2, 2, 2, 0, 2, 2, 2, 0, 2, 0, 0, 0, 2, 2, 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 0, 2, 0, 0, 0, 2
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OFFSET

0,3


COMMENTS

A115211(n) = A000120(a(n)) = number of ones in binary representation of a(n) for n>0; record values: A115212(n) = a(A115213(n)).  Reinhard Zumkeller, Jan 17 2006


LINKS

Michael De Vlieger, Table of n, a(n) for n = 0..16384
Michael De Vlieger, Records and First Positions of Records of A113504.


EXAMPLE

The first 8 terms (terms 0 through 7) written in binary are [1,1,10,0,11,1,1,0]. Term 8 gives the number of earlier terms with the same number of 1's in their binary representation as 8 (which is 1000 in binary, for one 1). a(8) = 5 because there are five terms among the earlier terms with one binary 1 (terms with one 1: 1, 1, 2, 1 and 1).


MATHEMATICA

Block[{a = {{1}}}, Do[AppendTo[a, IntegerDigits[#, 2]] &@ Count[a, k_ /; Count[k, 1] == DigitCount[i, 2, 1]], {i, 104}]; FromDigits[#, 2] & /@ a] (* Michael De Vlieger, Sep 05 2017 *)


PROG

(PARI) lista(nn) = {vh = [1]; print1(1, ", "); for (n=1, nn, nb = #select(x>(x==hammingweight(n)), vh); print1(nb, ", "); vh = concat(vh, hammingweight(nb)); ); } \\ Michel Marcus, Sep 06 2017


CROSSREFS

Cf. A000120, A113503, A115212, A115213.
Sequence in context: A342657 A002187 A124756 * A276165 A344618 A124754
Adjacent sequences: A113501 A113502 A113503 * A113505 A113506 A113507


KEYWORD

nonn,base


AUTHOR

Leroy Quet, Jan 10 2006


EXTENSIONS

More terms from Reinhard Zumkeller, Jan 17 2006


STATUS

approved



