OFFSET
1,1
COMMENTS
Conjecture: Number of (n+1) X (k+1) 0..p arrays with every 2 X 2 subblock summing to 2p is Sum_{i=1..(p+1)} i^(n+1)*(p+2-i)^(k+1) - 2*Sum_{i=1..p} i^(n+1)*(p+1-i)^(k+1) + Sum_{i=1..(p-1)} i^(n+1)*(p-i)^(k+1). - Zhuorui He, Jun 20 2026
LINKS
Zhuorui He, Table of n, a(n) for n = 1..1000 (first 200 terms from R. H. Hardin)
Christian Krause, Proof of formula, Jun 20 2026
Index entries for linear recurrences with constant coefficients, signature (10,-35,50,-24).
FORMULA
a(n) = 10*a(n-1) - 35*a(n-2) + 50*a(n-3) - 24*a(n-4).
From Colin Barker, Mar 31 2018: (Start)
G.f.: 4*x*(34 - 255*x + 578*x^2 - 384*x^3) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 4*x)).
a(n) = 18 + 3*2^(3+n) + 2*3^(2+n) + 4^(1+n).
(End) [proved by Christian Krause, Jun 20 2026]
EXAMPLE
Some solutions for 5 X 3:
2 2 0 0 0 1 1 2 1 1 2 0 0 3 0 3 0 1 3 1 2
0 2 2 3 3 2 0 3 0 2 1 3 1 2 1 0 3 2 1 1 2
2 2 0 0 0 1 1 2 1 1 2 0 2 1 2 2 1 0 3 1 2
0 2 2 3 3 2 3 0 3 0 3 1 3 0 3 0 3 2 1 1 2
3 1 1 0 0 1 0 3 0 1 2 0 3 0 3 3 0 1 1 3 0
MATHEMATICA
A183635[n_] := 18 + 3*2^(n+3) + 2*3^(n+2) + 4^(n+1);
Array[A183635, 25] (* Paolo Xausa, Jun 22 2026 *)
PROG
(PARI) f(n, k, p)=sum(i=1, p+1, i^(n+1)*(p+2-i)^(k+1))-2*sum(i=1, p, i^(n+1)*(p+1-i)^(k+1))+sum(i=1, p-1, i^(n+1)*(p-i)^(k+1)) \\ f(n, k, p) is conjectured to be number of (n+1) X (k+1) 0..p arrays with every 2 X 2 subblock summing to 2p.
A183635(n) = f(n, 2, 3) \\ This is proved using the proved formula above.
\\ Zhuorui He, Jun 20 2026
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. H. Hardin, Jan 06 2011
STATUS
approved