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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.)