A340862 - OEIS

%I #21 Mar 20 2021 16:06:16

%S 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,

%T 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,

%U 4,2,4,2,5,3,3,3,5,3,3,3,3,2,5,2,4,4,3,3

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

%C 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.

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

%H Jinyuan Wang, <a href="/A340862/b340862.txt">Table of n, a(n) for n = 1..10000</a>

%e For n=2, the single solution is the third term of the Fibonacci sequence (k=1), so a(2)=1.

%e 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.

%e For n=4, we have k=2 and k=3, so a(4) = 2.

%e For n=5, we have k=1, k=3, k=4.

%o (Python)

%o def get_val(n):

%o     res = 0

%o     for k in range(1, n):

%o         (a, b) = (k, 1)

%o         while b < n:

%o             (a, b) = (b, a+b)

%o             if b == n:

%o                 res += 1

%o     return res

%o (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

%Y Cf. A067148, A067149.

%K nonn

%O 1,3

%A _Robby Goetschalckx_, Jan 24 2021

%E Offset changed and a(1) inserted by _Jinyuan Wang_, Mar 20 2021