OFFSET
0,2
COMMENTS
for n>=1 a(n) mod 2 = A059448(n) the parity of number of 0's in binary expansion of n.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..200
FORMULA
From Vaclav Kotesovec, Oct 30 2021: (Start)
Recurrence: n*(2*n - 1)*(1024*n^4 - 10688*n^3 + 40544*n^2 - 65368*n + 37617)*a(n) = (14336*n^6 - 169088*n^5 + 773760*n^4 - 1725328*n^3 + 1933390*n^2 - 999891*n + 175950)*a(n-1) - (14336*n^6 - 189568*n^5 + 983936*n^4 - 2540080*n^3 + 3358598*n^2 - 2079295*n + 445620)*a(n-2) - (38912*n^6 - 503424*n^5 + 2607232*n^4 - 6826768*n^3 + 9326358*n^2 - 6043703*n + 1296960)*a(n-3) - 2*(28672*n^6 - 362752*n^5 + 1804800*n^4 - 4416032*n^3 + 5425772*n^2 - 2994798*n + 432585)*a(n-4) + 8*(28672*n^6 - 403712*n^5 + 2232064*n^4 - 6111200*n^3 + 8528476*n^2 - 5498774*n + 1180095)*a(n-5) - 64*(n-5)*(2*n - 7)*(1024*n^4 - 6592*n^3 + 14624*n^2 - 12248*n + 3129)*a(n-6).
a(n) ~ 4^n / 3 * (1 + sqrt(3)/(2*sqrt(Pi*n))). (End)
PROG
(PARI) a(n)=sum(i=0, n, sum(j=0, i, sum(k=0, n, if(i+j+k-n, 0, (n+k)!/i!/j!/(2*k)!))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Mar 25 2004
STATUS
approved