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%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
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