A126182 Let P be Pascal's triangle A007318 and let N be Narayana's triangle A001263, both regarded as lower triangular matrices. Sequence gives triangle obtained from P*N, read by rows.
1, 2, 1, 4, 5, 1, 8, 18, 9, 1, 16, 56, 50, 14, 1, 32, 160, 220, 110, 20, 1, 64, 432, 840, 645, 210, 27, 1, 128, 1120, 2912, 3150, 1575, 364, 35, 1, 256, 2816, 9408, 13552, 9534, 3388, 588, 44, 1, 512, 6912, 28800, 53088, 49644, 24822, 6636, 900, 54, 1
Offset: 0
Examples
The triangle P begins 1, 1, 1 1, 2, 1 1, 3, 3, 1, ... and T begins 1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, ... The product P*T gives 1, 2, 1, 4, 5, 1, 8, 18, 9, 1, 16, 56, 50, 14, 1, ...
Links
- E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203.
- M. Dziemianczuk, Enumerations of plane trees with multiple edges and Raney lattice paths, Discrete Mathematics 337 (2014): 9-24.
- F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.
Programs
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Maple
T:=proc(n,k) if k=0 then 2^n elif k<=n then binomial(n+1,k)*sum(binomial(k,n-k-j)*binomial(n+1-k,j)*2^j,j=n-2*k..n-k)/(n+1) else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
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Mathematica
t[n_, 0] := 2^n; t[n_, k_] := Binomial[n+1, k]*Sum[Binomial[k, n-k-j]*Binomial[n+1-k, j]*2^j, {j, n-2*k, n-k}]/(n+1); Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 12 2013 *) nmax = 10; n[x_, y_] := (1-x*(1+y) - Sqrt[(1-x*(1+y))^2 - 4*y*x^2])/(2*x); s = Series[(n[x/(1-x), y]-1)/x, {x, 0, nmax+1}, {y, 0, nmax+1}];t[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]; Table[t[n, k], {n, 0, nmax}, {k, 1, n+1}] // Flatten (* Jean-François Alcover, Apr 16 2015, after Vladimir Kruchinin *)
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PARI
tabl(nn) = {mP = matrix(nn, nn, n, k, binomial(n-1, k-1)); mN = matrix(nn, nn, n, k, binomial(n-1, k-1) * binomial(n, k-1) / k); mprod = mP*mN; for (n=1, nn, for (k=1, n, print1(mprod[n, k], ", ");); print(););} \\ Michel Marcus, Apr 16 2015
Formula
T(n,k) = (1/(n+1))*binomial(n+1,k)*Sum_{j=n-2k..n-k}2^j*binomial(k,n-k-j)*binomial(n+1-k,j) if 0 < k <= n; T(n,0) = 2^n.
G.f. G=G(t,z) satisfies G = 1 + (t+2)*z*G + t*z^2*G^2.
E.g.f.: exp((t+2)*x)*BesselI_{1}(2*sqrt(t)*x)/(sqrt(t)*x). - Peter Luschny, Oct 29 2014
G.f.: N(x/(1-x),y)-1)/x, where N(x,y) is the g.f. of Narayana's triangle A001263. - Vladimir Kruchinin, Apr 06 2015.
Extensions
New definition in terms of P and N from Philippe Deléham, Jun 30 2007
Edited by N. J. A. Sloane, Jul 22 2007
Comments