|
| |
|
|
A132276
|
|
Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,k), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0) (0<=k<=n).
|
|
5
| |
|
|
1, 1, 1, 3, 2, 1, 6, 7, 3, 1, 16, 18, 12, 4, 1, 40, 53, 37, 18, 5, 1, 109, 148, 120, 64, 25, 6, 1, 297, 430, 369, 227, 100, 33, 7, 1, 836, 1244, 1146, 760, 385, 146, 42, 8, 1, 2377, 3656, 3519, 2518, 1391, 606, 203, 52, 9, 1, 6869, 10796, 10839, 8188, 4900, 2346, 903, 272
(list; table; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,4
|
|
|
COMMENTS
| T(n,0)=A128720(n).
Mirror image of A059397. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 18 2007
Row sums yield A059398.
Riordan matrix (g(x),x*g(x)), where g(x) = (1-x-x^2-sqrt(1-2*x-5*x^2+2*x^3+x^4))/(2*x^2). [Emanuele Munarini, May 5 2011]
|
|
|
REFERENCES
| W. Klostermeyer et al., A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
|
|
|
FORMULA
| G.f.=G(t,z)=g/(1-t*z*g), where g=1+z*g+z^2*g+z^2*g^2 or g=c(z^2/(1-z-z^2)^2)/(1-z-z^2), where c=((1-sqrt(1-4*z))/(2*z) is the Catalan function.
T(n,k)=T(n-1,k-1)+T(n-1,k)+T(n-1,k+1)+T(n-2,k). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 18 2007
Column k has g.f. z^k*g^(k+1), where g=1+z*g+z^2*g+z^2*g^2=(1-z-z^2-sqrt((1+z-z^2)*(1-3*z-z^2)))/(2*z^2).
T(n,k)= sum(binomial(2*i+k,i)*(k+1)/(i+k+1)*sum(binomial(i+j+k,i+k)*binomial(j,n-k-2*i-j),j=0..n-k-2*i),i=0..(n-k)/2). [Emanuele Munarini, May 5 2011]
|
|
|
EXAMPLE
| Triangle begins:
1,
1,1,
3,2,1,
6,7,3,1,
16,18,12,4,1,
40,53,37,18,5,1,
109,148,120,64,25,6,1,
T(3,2)=3 because we have UUh, UhU and hUU.
|
|
|
MAPLE
| g:=((1-z-z^2-sqrt((1+z-z^2)*(1-3*z-z^2)))*1/2)/z^2: G:=simplify(g/(1-t*z*g)): Gser:=simplify(series(G, z=0, 13)): for n from 0 to 10 do P[n]:=sort(coeff(Gser, z, n)) end do: for n from 0 to 10 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form
|
|
|
MATHEMATICA
| Flatten[Table[Sum[Binomial[2i+k, i](k+1)/(i+k+1)Sum[Binomial[i+j+k, i+k]Binomial[j, n-k-2i-j], {j, 0, n-k-2i}], {i, 0, (n-k)/2}], {n, 0, 15}, {k, 0, n}]] [Emanuele Munarini, May 5 2011]
|
|
|
PROG
| (Maxima) create_list(sum(binomial(2*i+k, i) * (k+1)/(i+k+1) * sum(binomial(i+j+k, i+k) * binomial(j, n-k-2*i-j), j, 0, n-k-2*i), i, 0, (n-k)/2), n, 0, 15, k, 0, n); [Emanuele Munarini, May 5 2011]
|
|
|
CROSSREFS
| Cf. A059397, A128720 (the leading diagonal).
Cf. A059398.
Sequence in context: A079513 A060408 A139624 * A202390 A114586 A052174
Adjacent sequences: A132273 A132274 A132275 * A132277 A132278 A132279
|
|
|
KEYWORD
| nonn,tabl
|
|
|
AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 16 2007, Sep 03 2007
|
| |
|
|