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A340862 Number of times the number n turns up in pseudo-Fibonacci sequences starting with [k, 1] (with k >= 1), excluding the starting terms. 1
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified May 17 12:43 EDT 2021. Contains 343971 sequences. (Running on oeis4.)