A214595 T(n,k) = number of n X n X n triangular 0..k arrays with every horizontal row having the same average value.
2, 3, 2, 4, 5, 2, 5, 8, 23, 2, 6, 13, 62, 401, 2, 7, 18, 157, 1862, 20351, 2, 8, 25, 312, 10177, 187862, 2869211, 2, 9, 32, 601, 33352, 3330677, 63120962, 1127599139, 2, 10, 41, 986, 103651, 20608352, 5495329427, 71200442882, 1248252244661, 2, 11, 50, 1619
Offset: 1
Examples
Table starts .2.....3......4.......5........6.........7.........8..........9.........10 .2.....5......8......13.......18........25........32.........41.........50 .2....23.....62.....157......312.......601.......986.......1619.......2426 .2...401...1862...10177....33352....103651....250042.....589763....1199614 .2.20351.187862.3330677.20608352.121537201.493575042.1877543213.5767190924 Some solutions for n = k = 4: .....2........1........2........2........2........2........2........2 ....3.1......0.2......2.2......3.1......2.2......1.3......4.0......4.0 ...3.2.1....0.3.0....3.2.1....2.4.0....0.2.4....3.0.3....1.2.3....4.0.2 ..2.2.3.1..2.1.0.1..1.2.4.1..4.2.2.0..1.4.3.0..4.0.2.2..3.2.3.0..4.0.4.0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..1475
Crossrefs
Programs
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PARI
/* helper function mult() gives multiplicity of a composition */ mult(p, L=1, m=(#p)!)={for(k=2,#p, p[k]!=p[k-1] && m\=(-L+L=k)!); m\(#p-L+1)!} A214595(n, k)={sum(a=1,k, prod(L=2,n, my(c=0); forpart(p=L*a, c+=mult(p), [0,k], L); c))+1} \\ M. F. Hasler, Aug 21 2025
Formula
Empirical for row n:
n=1: a(k)=2*a(k-1)-a(k-2)
n=2: a(k)=2*a(k-1)-2*a(k-3)+a(k-4)
n=3: (order 12 antisymmetric)
n=4: (order 32 symmetric)
n=5: (order 84 symmetric)
T(n, k) = Sum_{s=0..k} Product_{L=2..n} NC(s*L, L, k), where NC(s, n, k) is the number of compositions of sum s with n parts between 0 and k. - M. F. Hasler, Aug 21 2025
Comments