3

The first few numbers in the function, represented as (k, f(k)) are as follows: (0, 1/2) (1, 3/4) (2, 3/2) (3, 33/8) (4, 15) (5, 273/4). I found these using the identity $\sum_{n=0}^\infty \frac{(n+m-1)!}{m!(n-1)!a^n} = \frac{a^m}{(a-1)^{m+1}}$ and plugging in a=3 then isolating the desired exponent after expanding the factorial. It doesn't seem like there would be infinitely many integer results, but I'm not sure; I was wondering if it would be provable one way or another.

LogicCube
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  • Just by brute-force computation it appears that you have $f(4) = 15, f(8) = 17295, f(12) = 141045450$ - if I had to guess I'd say $f(k)$ is an integer when $k$ is a multiple of 4, although that might just be dumb luck. Can you give an example of how you get these results? – Michael Lugo Jan 23 '24 at 16:45
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    After computing the first 1000 values, I think there are probably infinitely many, but the "multiple of 4" pattern doesn't quite work - the counterexamples are 0, 28, 60, 92, 124, 156, 188, 220, 252, 284, 316, 348, 380, 412, 444, 476, 508, 540, 572, 604, 636, 668, 700, 732, 764, 796, 828, 860, 892, 924, 956, 988. There are also non-multiples of 4 that do give integers: 13, 77, 122, 141, 205, 269, 333, 378, 379, 397, 461, 525, 589, 634, 653, 717, 781, 845, 890, 909, 973. – paste bee Jan 23 '24 at 16:48
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    The sequence of $k$ that give integers is A261811 on the OEIS, which also contains a link to an existing question here. – paste bee Jan 23 '24 at 16:50
  • Oh, thank you, I didn't find that when looking for this question earlier; I'll go check it out. – LogicCube Jan 23 '24 at 16:53

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