A141612 Write down 0,1,2,3,...n each in binary. Total up the number of 1's in each bit-position (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.

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

Offset

0,10

Links

John Tyler Rascoe, Table of n, a(n) for n = 0..8192

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+1-min(1, abs(i-c)) ; 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

Prog

(Python)
from collections import Counter
def bin_i(n): #binary indices
    return([(i) for i, x in enumerate(bin(n)[2:][::-1]) if x =='1'])
def A141612_list(n_max):
    A, C = [], Counter()
    for n in range(n_max+1):
        for i in bin_i(n): C.update({i})
        B, x = list(C.values()), 0
        for k in B:
            if B.count(k) == 1: x += 1
        A.append(x)
    return(A) # John Tyler Rascoe, Feb 09 2024

Crossrefs

Cf. A000120, A070939.

Sequence in context: A058190 A055736 A006997 * A316342 A297814 A298177

Adjacent sequences: A141609 A141610 A141611 * A141613 A141614 A141615

Keyword

nonn,base,easy

Author

Leroy Quet, Aug 22 2008

Extensions

Extended by R. J. Mathar, Sep 12 2008

Status

approved