A319197 All entries from a(3) to a(n) appear in addition to 2^n as factors in the conjectured factorization of Fibonacci(2^(n-2)*3*m) for n >= 3 and all m >= 0.

1, 9, 161, 51841, 6989569, 53156199689438143, 5581524253378492696105796918365541568492478783, 89171045849445921581733341920411050611581102638589828325078491812417901966295041

Offset: 3

Author

Wolfdieter Lang, Oct 09 2018

Keywords

• nonn

Comments

It appears that Fibonacci(2^(n-2)*3*m)/(2^n) is a nonnegative integer for n >= 3 and all m >= 0. The remaining factor for n >= 3 is given by A(n) := Product_{j=3..n} a(j). For n = 3 and 4 see A049660 and A253368, respectively. Thus the conjecture is that I(n; m) := Fibonacci(2^(n-2)*3*m) / ((2^n)*A(n)) is a nonnegative integer for all m >= 0. This is best possible for all m (because for m = 1 this becomes 1); special m may allow more factors. E.g., n = 3: 8 | Fibonacci(6*m), for all m >= 0, but for even m larger powers of 2 than 2^3 appear; for m = 0 any power of 2.

The factorizations of a(n) are: 1, 3^2, 7*23, 47*1103, 3167*2207, 4481*11862575248703, 127*383*769*1087*5662847*6803327*19073614849*186812208641, 885503*119809*1359361*1769526527*4698167634523379875583*74374487830388825730162393840383, ...

Examples

			n=3: I(3; m) = A049660(m), m >= 0.
n=4: I(4; m) = A253368(m), with A253368(0) := 0.
n = 5: I(5; m) =  F(24*m)/(2^5*9*161) = F(24*m)/(2^5*3^2*7*23) = [0, 1, 103682, 10749957123, 1114577054323204, ...]
		

Crossrefs

Cf. A000045, A049660, A253368.

Formula

I(n; m) := F(2^(n-2)*3*m) / ((2^n)* Product_{j=3..n} a(j)) is conjectured to be a nonnegative integer for n >= 3 and all m >= 0, where F = A000045. There are no more factors > 1 for all m >= 0 because I(n, 1) = 1.