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Recurrence a(n+2) = (n+2)*a(n+1)*a(n), with a(0) = a(1) = 1.
2

%I #39 Jul 05 2021 10:26:07

%S 1,1,2,6,48,1440,414720,4180377600,13869489586176000,

%T 521817332305350780518400000,

%U 72373400562952038729626622187536384000000000,415422642927888257689749131592471020852730170822782196121600000000000000

%N Recurrence a(n+2) = (n+2)*a(n+1)*a(n), with a(0) = a(1) = 1.

%C Numbers of decimal digits in a(n) for 0 <= n <= 20: 1, 1, 1, 1, 2, 4, 6, 10, 17, 27, 44, 72, 117, 190, 307, 498, 806, 1305, 2112, 3417, 5530. - _Robert Israel_, Oct 09 2017

%H Robert Israel, <a href="/A230053/b230053.txt">Table of n, a(n) for n = 0..16</a>

%F a(n) = Product_{k=0..n-1} (n-k+1)^Fibonacci(k).

%F a(n) ~ c^(phi^n) / n, where c = 2.32072822997682611701924627353608916645018... and phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Jul 05 2021

%p f:= proc(n) option remember; n*procname(n-1)*procname(n-2) end proc:

%p f(0):= 1: f(1):= 1:

%p map(f, [$0..12]); # _Robert Israel_, Oct 08 2017

%t RecurrenceTable[{a[n + 2] == (n + 2) a[n + 1] a[n], a[0] == a[1] == 1}, a, {n, 0, 12}] (* or *)

%t Table[Product[(n - k + 1)^Fibonacci[k], {k, 0, n - 1}], {n, 0, 12}]

%Y Cf. A000045.

%K nonn

%O 0,3

%A _Emanuele Munarini_, Oct 08 2017

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Last modified September 21 10:11 EDT 2024. Contains 376084 sequences. (Running on oeis4.)