A340862 Number of times the number n turns up in pseudo-Fibonacci sequences starting with [k, 1] (with k >= 1), excluding the starting terms.

0, 1, 2, 2, 3, 2, 3, 3, 3, 2, 4, 2, 4, 3, 3, 2, 4, 3, 3, 3, 4, 2, 5, 2, 3, 3, 3, 3, 5, 2, 3, 3, 4, 3, 4, 2, 4, 4, 3, 2, 4, 2, 4, 3, 4, 2, 5, 3, 3, 3, 3, 2, 6, 2, 4, 3, 3, 3, 4, 3, 4, 3, 4, 2, 4, 2, 3, 4, 4, 2, 4, 2, 5, 3, 3, 3, 5, 3, 3, 3, 3, 2, 5, 2, 4, 4, 3, 3

Offset

1,3

Comments

In the first 100000 terms, this never exceeds 8. For any n > 2, a(n) will be at least 2, since k=n-1 and k=n-2 will both work.

Conjecture: for n > 2, a(n) appears to be equal to 1 + A067148(n).

Links

Jinyuan Wang, Table of n, a(n) for n = 1..10000

Example

For n=2, the single solution is the third term of the Fibonacci sequence (k=1), so a(2)=1.
For n=3, we observe the value as the fourth term for k=1, and the third term for k=2 for a total count of a(3) = 2.
For n=4, we have k=2 and k=3, so a(4) = 2.
For n=5, we have k=1, k=3, k=4.

Prog

(Python)
def get_val(n):
    res = 0
    for k in range(1, n):
        (a, b) = (k, 1)
        while b < n:
            (a, b) = (b, a+b)
            if b == n:
                res += 1
    return res
(PARI) a(n) = my(c, x, y=1); while(n>=x+=2*y, y=x-y; x-=y; if((n-y)%x==0, c++)); c; \\ Jinyuan Wang, Mar 20 2021

Crossrefs

Cf. A067148, A067149.

Sequence in context: A304333 A136510 A080071 * A202472 A235613 A322418

Adjacent sequences: A340859 A340860 A340861 * A340863 A340864 A340865

Keyword

nonn

Author

Robby Goetschalckx, Jan 24 2021

Extensions

Offset changed and a(1) inserted by Jinyuan Wang, Mar 20 2021

Status

approved