OFFSET
0,2
COMMENTS
Conjecture: cases f(n) = n mod 2 and f(n) = [(n mod 3) > 0] both gives A006318.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
FORMULA
Recurrence: (n+1)*a(n) = 2*(7*n-2)*a(n-1) - (67*n-101)*a(n-2) + 32*(4*n-11)*a(n-3) - 8*(8*n-37)*a(n-4) - 2*(22*n-71)*a(n-5) + 4*(n-5)*a(n-6). - Vaclav Kotesovec, Sep 23 2024
PROG
(PARI) upto(n) = my(v1); v1 = vector(2*(n+1), i, 1); v2 = vector(n+1, i, 0); v2[1] = 1; for(i=1, n, for(j=i+1, 2*(n+1)-i, v1[j] = v1[j+(((j-i)%6)>0)] + v1[j-1]); v2[i+1] = v1[i+1]); v2
CROSSREFS
KEYWORD
nonn
AUTHOR
Mikhail Kurkov, Sep 22 2024
STATUS
approved