[seqfan] b(m) = b(m-1)*r(m) + s(m)

[Leroy Quet]

Define a(n) as the number of ways to achieve n from the following procedure.

Let R={r(k)} and S={s(k)} each be some permutation of (1,2,3,...,j) for some positive integer j.

Define (b(0),b(1),...,b(j)) as follows.
b(0)=1.
b(m) = b(m-1)*r(m) + s(m), for 1<= m <= j.

Does b(j) = n? If so, add 1 to the count. Calculate the b(j)'s by taking j over all positive integers, and taking R and S over all permutations for a given j. The total count equals a(n).

I get the sequence beginning (offset 2):
1,0,0,2,1,1,0,0,0,0,0,2,...

These positive integers n have non-zero counts:
{c(k)}: (1?), 2, 5, 6, 7, 13,...


For example: For j = 3, R = (3,2,1) and S(1,2,3), we get b(3)=13.
And for j = 3, R = (3,2,1) and S(1,3,2), we get b(3)=13.
So, a(13) is at least 2.


Neither sequence {a(k)} nor {c(k)} is in the EIS yet.

Could someone please calculate some terms of these sequences, then either post the results here for me to submit, or submit these sequences yourself?
(Unless I made an error, which is likely, and the sequences already are in the EIS.)

PS: What to do about the n=1 case?

Thanks,
Leroy Quet

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