OFFSET
1,3
COMMENTS
Column 2 of A199256.
LINKS
Manuel Kauers and Christoph Koutschan, Table of n, a(n) for n = 1..1000 (terms 1..56 from R. H. Hardin).
M. Kauers and C. Koutschan, Some D-finite and some possibly D-finite sequences in the OEIS, arXiv:2303.02793 [cs.SC], 2023.
FORMULA
From Manuel Kauers and Christoph Koutschan, Mar 01 2023: (Start)
a(2*n) = coefficient of x^n*y^n*z^n*t^(2*n) in t*x*y*(1 + t*z)/(2*(1 - t*(x + y + z + x*y + x*z + y*z) - 7*t^2*x*y*z)).
a(2*n+1) = coefficient of x^(n+1)*y^(n+1)*z^(n+1)*t^(2*n+1) in t*x*y*(1 + t*z)*(x + y + z + x*y + x*z + y*z)/(2*(1 - t*(x + y + z + x*y + x*z + y*z) - 7*t^2*x*y*z)) for n>0.
Recurrence of order 4 and degree 8 for even indices: (4 + n)^3*(3 + 2*n)*(-13 + 67*n + 529*n^2 + 440*n^3 + 96*n^4)*a(2*n + 8) - 2*(-47232 + 243564*n + 2728691*n^2 + 5650345*n^3 + 5266809*n^4 + 2637037*n^5 + 736180*n^6 + 108160*n^7 + 6528*n^8)*a(2*n + 6) + 2*(-151008 + 3194000*n + 25261108*n^2 + 53468052*n^3 + 53319121*n^4 + 29037852*n^5 + 8890558*n^6 + 1438672*n^7 + 95808*n^8)*a(2*n + 4) - 98*(23232 + 227996*n + 960783*n^2 + 1960439*n^3 + 2151893*n^4 + 1338307*n^5 + 470452*n^6 + 86848*n^7 + 6528*n^8)*a(2*n + 2) + 2401*n^3*(5 + 2*n)*(1119 + 2829*n + 2425*n^2 + 824*n^3 + 96*n^4)*a(2*n) = 0.
Recurrence of order 4 and degree 10 for odd indices: (5 + n)^3*(3 + 2*n)*(-736 + 2812*n + 35991*n^2 + 63072*n^3 + 38589*n^4 + 9720*n^5 + 864*n^6)*a(2*n + 9) - (3 + 2*n)*(-5010656 + 16627420*n + 251763403*n^2 + 561479353*n^3 + 541644308*n^4 + 281844117*n^5 + 85376223*n^6 + 15113172*n^7 + 1453248*n^8 + 58752*n^9)*a(2*n + 7) + (-156900576 + 635576668*n + 9349986451*n^2 + 24663169255*n^3 + 30687106706*n^4 + 21910345387*n^5 + 9644646333*n^6 + 2664337824*n^7 + 450289356*n^8 + 42566688*n^9 + 1724544*n^10)*a(2*n + 5) - 49*(7 + 2*n)*(1228128 + 12549268*n + 55318177*n^2 + 118911819*n^3 + 139678988*n^4 + 95529783*n^5 + 38777853*n^6 + 9129108*n^7 + 1143936*n^8 + 58752*n^9)*a(2*n + 3) + 2401*n^3*(7 + 2*n)*(150312 + 472150*n + 566901*n^2 + 331908*n^3 + 100149*n^4 + 14904*n^5 + 864*n^6)*a(2*n + 1) = 0. (End)
EXAMPLE
Some solutions for n=4
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
2 3 1 2 1 0 2 0 2 3 2 3 2 3 2 3 2 0 2 0
0 2 2 3 2 3 1 3 1 2 1 2 3 1 1 0 3 1 3 2
3 1 3 0 3 2 3 2 3 0 0 3 2 0 2 3 2 3 1 3
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Nov 04 2011
STATUS
approved