

A340862


Number of times the number n turns up in pseudoFibonacci 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
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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=n1 and k=n2 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=xy; x=y; if((ny)%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



