Data update: the sequence through n=30 is 1,3,9,27,75,189,447,951,1911,3621,6513,11103,18267,29013,44691,67251,98547,140865,197679,272799,370659,497403,658371,859863,1110453,1420527,1799373,2260161,2815401,3479235,4269279 Looks more n^5ish than n^6ish, at a casual glance. The last term took something like 11 hours to count, so I think it's time to give Mma a break. --Michael On Fri, Jul 23, 2010 at 5:35 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 7/23/10, Michael Kleber <michael.kleber@gmail.com> wrote:
On Fri, Jul 23, 2010 at 12:34 PM, Fred lunnon <fred.lunnon@gmail.com> wrote: ... I'm satisfied with the first 12 terms validating the correctness of the symmetry-aware version. Just made the change, and it reproduced those 12 terms in 67 seconds. During the time I spent composing this email, it's gotten up to n=16:
1, 3, 9, 27, 75, 189, 447, 951, 1911, 3621, 6513, 11103, 18267, 29013, 44691, 67251, 98547
Great stuff! (methodically unclenching teeth)
I'll let it run for a while and post here. Allan is welcome to own the OEIS entry :-).
Yes indeed.
Any ideas on functional recursions, explicit expressions, asymptotics?
Not I; see "didn't have time to think" above...
Hmm ... evalf(log(98547/67251)/log(16/15)); 5.920521497 evalf(log(67251/44691)/log(15/14)); 5.923217032 I wonder if perchance f_3(n) = O(n^6) ??
f_2(n) = O(n^3) must follow straightforwardly via standard results from the explicit formula in A005598; but I presently haven't a clue why that should hold either.
Fred Lunnon
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