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A308551
Start with an empty stack S; for n = 1, 2, 3, ..., interpret the binary representation of n from left to right as follows: in case of bit 1, push the number 1 on top of S, in case of bit 0, replace the two numbers on top of S with their sum; a(n) gives the number on top of S after processing n.
4
1, 2, 1, 3, 1, 2, 1, 4, 1, 3, 1, 3, 1, 2, 1, 5, 1, 12, 1, 4, 1, 2, 1, 4, 1, 3, 1, 3, 1, 2, 1, 6, 1, 15, 1, 23, 1, 2, 1, 5, 1, 4, 1, 4, 1, 2, 1, 5, 1, 4, 1, 4, 1, 2, 1, 4, 1, 3, 1, 3, 1, 2, 1, 7, 1, 19, 1, 30, 1, 2, 1, 47, 1, 57, 1, 5, 1, 2, 1, 6, 1, 20, 1, 5
OFFSET
1,2
COMMENTS
After processing n, S has A268289(n) elements, the sum of which is A000788(n).
Every positive integer appears infinitely many times in the sequence.
The sequence has the same shape when represented at different scales.
LINKS
Sean A. Irvine, Java program (github)
Rémy Sigrist, PARI program
FORMULA
a(n) = 1 iff n is odd.
a(A020989(k)) = k + 1 for any k >= 0.
If n is in A014486, then a(n) = a(n-1) + A000120(n) = 1 + A000120(n). - Charlie Neder, Jun 07 2019
EXAMPLE
The first terms, alongside the binary representation of n and the evolution of stack S, are:
n a(n) bin(n) S
- ---- ------ -------------------------------------------------
1 1 1 () -> (1)
2 2 10 (1) -> (1,1) -> (2)
3 1 11 (2) -> (2,1) -> (2,1,1)
4 3 100 (2,1,1) -> (2,1,1,1) -> (2,1,2) -> (2,3)
5 1 101 (2,3) -> (2,3,1) -> (2,4) -> (2,4,1)
6 2 110 (2,4,1) -> (2,4,1,1) -> (2,4,1,1,1) -> (2,4,1,2)
PROG
(Java) See Links section.
(PARI) See Links section.
CROSSREFS
KEYWORD
nonn,base,look
AUTHOR
Rémy Sigrist, Jun 07 2019
STATUS
approved