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A230053 Recurrence a(n+2) = (n+2)*a(n+1)*a(n), with a(0) = a(1) = 1.

%I A230053 #41 Mar 16 2025 12:25:28
%S A230053 1,1,2,6,48,1440,414720,4180377600,13869489586176000,
%T A230053 521817332305350780518400000,
%U A230053 72373400562952038729626622187536384000000000,415422642927888257689749131592471020852730170822782196121600000000000000
%N A230053 Recurrence a(n+2) = (n+2)*a(n+1)*a(n), with a(0) = a(1) = 1.
%C A230053 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 A230053 Robert Israel, <a href="/A230053/b230053.txt">Table of n, a(n) for n = 0..16</a>
%F A230053 a(n) = Product_{k=0..n-1} (n-k+1)^Fibonacci(k).
%F A230053 a(n) ~ c^(phi^n) / n, where c = exp(Sum_{k>=1} log(k+1) / (sqrt(5)*phi^k) ) = 2.32072822997682611701924627353608916645018... and phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Jul 05 2021, updated Mar 16 2025
%p A230053 f:= proc(n) option remember; n*procname(n-1)*procname(n-2) end proc:
%p A230053 f(0):= 1: f(1):= 1:
%p A230053 map(f, [$0..12]); # _Robert Israel_, Oct 08 2017
%t A230053 RecurrenceTable[{a[n + 2] == (n + 2) a[n + 1] a[n], a[0] == a[1] == 1}, a, {n, 0, 12}] (* or *)
%t A230053 Table[Product[(n - k + 1)^Fibonacci[k], {k, 0, n - 1}], {n, 0, 12}]
%Y A230053 Cf. A000045.
%K A230053 nonn
%O A230053 0,3
%A A230053 _Emanuele Munarini_, Oct 08 2017