A035002 Square array read by antidiagonals: T(m,n) = Sum_{k=1..m-1} T(m-k,n) + Sum_{k=1..n-1} T(m,n-k).
1, 1, 1, 2, 2, 2, 4, 5, 5, 4, 8, 12, 14, 12, 8, 16, 28, 37, 37, 28, 16, 32, 64, 94, 106, 94, 64, 32, 64, 144, 232, 289, 289, 232, 144, 64, 128, 320, 560, 760, 838, 760, 560, 320, 128, 256, 704, 1328, 1944, 2329, 2329, 1944, 1328, 704, 256, 512, 1536, 3104, 4864, 6266
Offset: 1
Examples
Table begins: 1 1 2 4 8 16 32 64 ... 1 2 5 12 28 64 144 320 ... 2 5 14 37 94 232 560 1328 ... 4 12 37 106 289 760 1944 4864 ... Alternative construction as a triangle: 1 1 1 2 2 2 4 5 5 4 8 12 14 12 8 16 28 37 37 28 16
Links
- Peter Kagey, Table of n, a(n) for n = 1..10011 (first 141 antidiagonals, flattened)
- C. Coker, Enumerating a class of lattice paths, Discrete Math., 271 (2003), 13-28.
- M. Erickson, S. Fernando, and K. Tran, Enumerating Rook and Queen Paths, Bulletin of the Institute for Combinatorics and Its Applications, Volume 60 (2010), 37-48.
Programs
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Maple
A035002 := proc(m,n) option remember; if n = 1 and m= 1 then 1; elif m = 1 then 2^(n-2) ; elif n = 1 then 2^(m-2) ; else add( procname(m-k,n),k=1..m-1) + add( procname(m,n-k),k=1..n-1) ; end if; end proc: # R. J. Mathar, Jun 06 2013
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Mathematica
T[n_, 1] = 2^(n-2); T[1, n_] = 2^(n-2); T[1, 1] = 1; T[m_, n_] := T[m, n] = Sum[T[m-k, n], {k, 1, m-1}] + Sum[T[m, n-k], {k, 1, n-1}]; Flatten[Table[T[m-n+1 , n], {m, 1, 11}, {n, 1, m}]] (* Jean-François Alcover, Nov 04 2011 *) nMax = 11; T = (((x - 1)*y - x + 1)/((3*x - 2)*y - 2*x + 1) + O[x]^nMax // Normal // Expand) + O[y]^nMax // Normal // Expand // CoefficientList[#, {x, y}]&; Table[T[[n - k + 1, k]], {n, 1, nMax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 18 2018, after Vladimir Kruchinin *) T[ n_, m_] := SeriesCoefficient[ (1 - x)*(1 - y)/( 1 - 2*x - 2*y + 3*x*y), {x, 0, n}, {y, 0, m}]; (* Michael Somos, Oct 05 2023 *)
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Maxima
T(n,m):=sum(binomial(m-1,m-i)*sum(binomial(k+i,i)*binomial(n-1,n-k),k,0,n),i,0,m); /* Vladimir Kruchinin, Apr 14 2015 */
Formula
G.f. T(n; x) for n-th row satisfies: T(n; x) = Sum_{k=1..n} (1+x^k)*T(n-k; x), T(0; x) = 1. - Vladeta Jovovic, Sep 03 2002
T(m+1,n+1) = 2*T(m+1,n) + 2*T(m,n+1) - 3*T(m,n); T(n,1) = T(1,n) = A011782(n). - Francisco Santos, Oct 20 2005
G.f.: ((x-1)*y-x+1)/((3*x-2)*y-2*x+1). - Vladimir Kruchinin, Apr 14 2015
T(n,m) = Sum_{i=0..m} C(m-1,m-i)*Sum_{k=0..n} C(k+i,i)*C(n-1,n-k). - Vladimir Kruchinin, Apr 14 2015
T(n,m) = T(m,n) for all n and m. - Michael Somos, Oct 04 2023
T(n,2) = (n+2)*2^(n-3) for n>1; T(n,3) = (n^2+11*n+14)*2^(n-5) for n>1 - Erich Friedman, Jan 14 2025
Comments