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%I #24 Oct 03 2018 03:57:53
%S 1,3,15,90,720,7200,79200,1029600,15444000,247104000,4447872000,
%T 88957440000,1868106240000,42966443520000,1074161088000000,
%U 27928188288000000,781989272064000000,23459678161920000000,727250023019520000000,23999250759644160000000
%N a(n) = Product_{i=1..n} floor(5*i/3).
%C If p > 3 and gcd(p,3)=1 then Product_{i=1..n} floor(i*p/3) ~ (p/3)^n * n! * 2*Pi * 3^(1/p - 1/2) / (c(p) * n^(1/p)), where
%C c(p) = Gamma(2/3 - 2/(3*p)) * Gamma(1/3 - 1/(3*p)) if mod(p, 3) = 1,
%C c(p) = Gamma(1/3 - 2/(3*p)) * Gamma(2/3 - 1/(3*p)) if mod(p, 3) = 2.
%C In general, if q > 1, p > q and gcd(p,q)=1, then Product_{i=1..n} floor(i*p/q) ~ c(p,q) * (p/q)^n * n! / n^((q-1)/(2*p)), where c(p,q) is a constant.
%F a(n) ~ (5/3)^n * n! * 2*Pi / (3^(3/10) * Gamma(1/5) * Gamma(3/5) * n^(1/5)).
%F Recurrence: 27*(15*n - 32)*a(n) = 675*(n-2)*a(n-1) + 15*(75*n^2 - 255*n + 194)*a(n-2) + 5*(n-2)*(5*n - 12)*(5*n - 11)*(15*n - 17)*a(n-3).
%t Table[Product[Floor[i*5/3], {i, 1, n}], {n, 1, 20}]
%t RecurrenceTable[{27*(15*n - 32)*a[n] == 675*(n-2)*a[n-1] + 15*(75*n^2 - 255*n + 194)*a[n-2] + 5*(n-2)*(5*n - 12)*(5*n - 11)*(15*n - 17)*a[n-3], a[1]==1, a[2]==3, a[3]==15}, a, {n, 1, 20}]
%o (PARI) a(n) = prod(i=1, n, (5*i)\3); \\ _Michel Marcus_, Oct 03 2018
%Y Cf. A010786, A047220, A180736, A275062, A319948, A319949, A317980.
%K nonn
%O 1,2
%A _Vaclav Kotesovec_, Oct 02 2018
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