OFFSET
0,5
COMMENTS
By convention, a(0) = 0.
For any n > 0:
- let (b_1, b_2, ..., b_h) be the positions of the ones in the binary representation of n,
- h = A000120(n) and 0 <= b_1 < b_2 < ... < b_h,
- n = Sum_{k = 1..h} 2^b_k,
- a(n) is the unique value remaining after taking successively the first differences of (b_1, ..., b_h) h-1 times.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 0..16384
FORMULA
EXAMPLE
For n = 59:
- the binary representation of 59 is "111011",
- so h = 5 and b_1 = 0, b_2 = 1, b_3 = 3, b_4 = 4, b_5 = 5,
- the corresponding difference table is:
0 1 3 4 5
1 2 1 1
1 -1 0
-2 1
3
- hence a(59) = 3.
PROG
(PARI) a(n) = { my (h=hammingweight(n), o=0, v=0); forstep (k=h-1, 0, -1, my (w=valuation(n, 2)); o += w; v += (-1)^k * binomial(h-1, k) * o; o++; n\=2^(1+w)); v };
CROSSREFS
KEYWORD
sign,base
AUTHOR
Rémy Sigrist, Mar 08 2019
STATUS
approved