

A141612


Write down 0,1,2,3,...n each in binary. Total up the number of 1's in each bitposition (total number of 1's in 1's position, total number of 1's in 2's position, total number of 1's in 4's position, etc.). a(n) = the number of such totals that each do not equal any other such total.


1



0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 2, 0, 0, 1, 0, 0, 1, 2, 1, 1, 3, 3, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 2, 1, 1, 2, 2, 1, 1, 3, 4, 4, 2, 2, 3, 2, 0, 0, 1, 0, 0, 2, 2, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 3, 1, 1, 2, 1, 1, 3, 4, 3, 3, 5, 5, 4, 2, 2, 3, 3, 2, 2, 3, 2, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2
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OFFSET

0,10


LINKS

Table of n, a(n) for n=0..104.


EXAMPLE

For n= 9, we have (with leading zeros written) 0 through 9 in binary:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
The number of 1's in the ones position (the rightmost position) is 5. The number of 1's in the 2's position (second position from the right) is 4. The number of 1's in the 4's position is 4. The number of 1's in the 8's position (the leftmost position here) is 2. Now, the total 4 occurs twice. But the total 2 occurs once, as does the total 5. Since two totals occur once each, then a(9) = 2.


MAPLE

A070939 := proc(n) max(1, ilog2(n)+1) ; end: A141612aux := proc(L, c) local a, i ; a := 0 ; for i in L do a := a+1min(1, abs(ic)) ; od; a ; end: A141612 := proc(n) local a, k, p, bds, i; if n = 0 then RETURN(0) ; fi; a := [seq(0, i=1..A070939(n))]; for k from 0 to n do bds := convert(k, base, 2) ; for p from 1 to nops(bds) do a := subsop( p=op(p, a)+op(p, bds), a) ; od: od: bds := 0 ; for i in a do if A141612aux(a, i) = 1 then bds := bds+1; fi; od; bds; end: for n from 0 to 120 do printf("%d, ", A141612(n)) ; od: # R. J. Mathar, Sep 12 2008


CROSSREFS

Sequence in context: A058190 A055736 A006997 * A316342 A297814 A298177
Adjacent sequences: A141609 A141610 A141611 * A141613 A141614 A141615


KEYWORD

nonn,base


AUTHOR

Leroy Quet, Aug 22 2008


EXTENSIONS

Extended by R. J. Mathar, Sep 12 2008


STATUS

approved



