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

Author

Robby Goetschalckx, Jan 24 2021

Keywords

• nonn

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).

Examples

			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.
		

Links

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

Crossrefs

Cf. A067148, A067149.

Programs

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

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

Extensions

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