OFFSET
0,3
COMMENTS
Double hyperfactorial (by analogy with the double factorial).
LINKS
Ilya Gutkovskiy, Table of n, a(n) for n = 0..33
Ilya Gutkovskiy, Double hyperfactorial
Eric Weisstein, Double Factorial
Eric Weisstein's World of Mathematics, Hyperfactorial
FORMULA
a(n) = n^n*(n - 2)^(n - 2)*...*5^5*3^3*1^1, for n>0 odd; a(n) = n^n*(n - 2)^(n - 2)*...*6^6*4^4*2^2, for n>0 even; a(n) = 1, for n = 0.
a(n) = n^n*a(n-2), a(0)=1, a(1)=1.
a(n) = (1/a(n-1))*sqrt(a(2n)/2^(n*(n+1))).
a(n)*a(n-1) = A002109(n).
a(n)*a(n-1)*sqrt(a(2n))/((n!)^n*sqrt(2^(n*(n+1)))) = A168510(n).
EXAMPLE
a(0) = 1;
a(1) = 1^1 = 1;
a(2) = 2^2 = 4;
a(3) = 1^1*3^3 = 27;
a(4) = 2^2*4^4 = 1024;
a(5) = 1^1*3^3*5^5 = 84375;
a(6) = 2^2*4^4*6^6 = 47775744;
a(7) = 1^1*3^3*5^5*7^7 = 69486440625;
a(8) = 2^2*4^4*6^6*8^8 = 801543976648704, etc.
MATHEMATICA
Table[Product[(n - 2 k)^(n - 2 k), {k, 0, Floor[(n - 1)/2]}], {n, 0, 13}]
RecurrenceTable[{a[0] == 1, a[1] == 1, a[n] == n^n a[n - 2]}, a, {n, 13}]
PROG
(PARI) a(n) = prod(k=0, (n-1)\2, (n-2*k)^(n-2*k)); \\ Michel Marcus, Apr 07 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Apr 06 2016
STATUS
approved