OFFSET
0,2
COMMENTS
The n-th column of the triangle is the binomial transform of the n-th row of A081277, followed by zeros. Example: column 3, (1, 6, 19, 44, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). A104698 = reversal by rows of A142978. - Gary W. Adamson, Jul 17 2008
This sequence is jointly generated with A210222 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) + 1 and v(n,x) = 2x*u(n-1,x) + v(n-1,x) + 1. See the Mathematica section at A210222. - Clark Kimberling, Mar 19 2012
This Riordan triangle T appears in a formula for A001100(n, 0) = A002464(n), for n >= 1. - Wolfdieter Lang, May 13 2025
LINKS
Reinhard Zumkeller, First 100 rows of triangle, flattened
FORMULA
The triangle is extracted from the product A * B; A = [1; 1, 1; 1, 1, 1; ...], B = [1; 1, 1; 1, 3, 1; 1, 5, 5, 1; ...] both infinite lower triangular matrices (rest of the terms are zeros). The triangle of matrix B by rows = A008288, Delannoy numbers.
From Paul Barry, Jul 18 2005: (Start)
Riordan array (1/(1-x)^2, x(1+x)/(1-x)) = (1/(1-x), x)*(1/(1-x), x(1+x)/(1-x)).
T(n, k) = Sum_{j=0..n} Sum_{i=0..j-k} C(j-k, i)*C(k, i)*2^i.
T(n, k) = Sum_{j=0..k} Sum_{i=0..n-k-j} (n-k-j-i+1)*C(k, j)*C(k+i-1, i). (End)
T(n, k) = binomial(n+1, k+1)*2F1([-k, k-n], [-n-1], -1) where 2F1 is a Gaussian hypergeometric function. - R. J. Mathar, Sep 04 2011
T(n, k) = T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) for 1 < k < n; T(n, 0) = n + 1; T(n, n) = 1. - Reinhard Zumkeller, Jul 17 2015
From Wolfdieter Lang, May 13 2025: (Start)
The Riordan triangle T = (1/(1 - x)^2, x*(1 + x)/(1 - x)) has the o.g.f. G(x, y) = 1/((1 - x)*(1 - x - y*x*(1+x))) for the row polynomials R(n, y) = Sum_{k=0..n} T(n, k)*y^k.
The o.g.f. for column k is G(k, x) = (1/(1 - x)^2)*(x*(1 + x)/(1 - x))^k, for k >= 0.
The o.g.f. for the diagonal m is D(m, x) = N(m, x)/(1 - x)^(m+1), with the numerator polynomial N(m, x) = Sum_{k=0..floor(m/2)} A034867(m, k)*x^(2*k) for m >= 0.
The row sums with o.g.f. R(x) = 1/((1 -x)*(1 - 2*x -x^2) give A048739.
The alternating row sums with o.g.f. 1/((1 - x)(1 + x^2)) give A133872.
The A-sequence for this Riordan triangle has o.g.f. A(x) = 1 + x + sqrt(1 + 6*x + x^2))/2 giving A112478(n). Hence T(n, k) = Sum_{j=0..n-k} A112478(j)*T(n-1, k-1+j), for n >= 1, k >= 1, T(n, k) = 0 for n < k, and T(0, 0) = 1.
The Z-sequence has o.g.f. (5 + x - sqrt(1 + 6*x + x^2))/2 = 3 + x - A(x) giving Z(n) = {2, -1, -A112478(n >= 2)}. Hence T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1. For A- and Z-sequences of Riordan triangles see a W. Lang link at A006232 with references.
The Boas-Buck sequences alpha and beta for the Riordan triangle T (see A046521 for the Aug 10 2017 comment and reference) are alpha(n) = A040000(n+1) = repeat{2} and beta(n) = A010673(n+1) = repeat{2,0}. Hence the recurrence for column T(n, k)_{n>=k}, with input T(k, k) = 1, for k >= 0, is T(n, k) = (1/(n-k)) * Sum_{j=k..n-1} (2 + k*(1 + (-1)^(n-1-j))) *T(j,k), for n >= k+1. (End)
EXAMPLE
The Riordan triangle T begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
----------------------------------------------------
0: 1
1: 2 1
2: 3 4 1
3: 4 9 6 1
4: 5 16 19 8 1
5: 6 25 44 33 10 1
6: 7 36 85 96 51 12 1
7: 8 49 146 225 180 73 14 1
8: 9 64 231 456 501 304 99 16 1
9: 10 81 344 833 1182 985 476 129 18 1
10: 11 100 489 1408 2471 2668 1765 704 163 20 1
... reformatted and extended by Wolfdieter Lang, May 13 2025
From Wolfdieter Lang, May 13 2025: (Start)
Zumkeller recurrence (adapted for offset [0,0]): 19 = T(4, 2) = T(2, 1) + T(3, 1) + T(3,3) = 4 + 9 + 6 = 19.
A-sequence recurrence: 19 = T(4, 2) = 1*T(3. 1) + 2*T(3. 2) - 2*T(3, 3) = 9 + 12 - 2 = 19.
Z-sequence recurrence: 5 = T(4, 0) = 2*T(3, 0) - 1*T(3, 1) + 2*T(3, 2) - 6*T(3, 3) = 8 - 9 + 12 + 6 = 5.
Boas-Buck recurrence: 19 = T(4, 2) = (1/2)*((2 + 0)*T(2, 2) + (2 + 2*2)*T(3, 2)) = (1/2)*(2 + 36) = 19. (End)
MAPLE
A104698 := proc(n, k) add(binomial(k, j)*binomial(n-j+1, n-k-j), j=0..n-k) ; end proc:
seq(seq(A104698(n, k), k=0..n), n=0..15); # R. J. Mathar, Sep 04 2011
# Alternative:
T := (n, k) -> binomial(n + 1, k + 1)*hypergeom([-k, k - n], [-n - 1], -1):
for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
# Alternative:
T := proc(n, k) option remember; if k = 0 then n + 1 elif k = n then 1 else T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) fi end: # Peter Luschny, May 13 2025
MATHEMATICA
u[1, _] = 1; v[1, _] = 1;
u[n_, x_] := u[n, x] = x u[n-1, x] + v[n-1, x] + 1;
v[n_, x_] := v[n, x] = 2 x u[n-1, x] + v[n-1, x] + 1;
Table[CoefficientList[u[n, x], x], {n, 1, 11}] // Flatten (* Jean-François Alcover, Mar 10 2019, after Clark Kimberling *)
PROG
(PARI) T(n, k)=sum(j=0, n-k, binomial(k, j)*binomial(n-j+1, k+1)) \\ Charles R Greathouse IV, Jan 16 2012
(Haskell)
a104698 n k = a104698_tabl !! (n-1) !! (k-1)
a104698_row n = a104698_tabl !! (n-1)
a104698_tabl = [1] : [2, 1] : f [1] [2, 1] where
f us vs = ws : f vs ws where
ws = zipWith (+) ([0] ++ us ++ [0]) $
zipWith (+) ([1] ++ vs) (vs ++ [0])
-- Reinhard Zumkeller, Jul 17 2015
CROSSREFS
Diagonal sums are A008937(n+1).
KEYWORD
AUTHOR
Gary W. Adamson, Mar 19 2005
STATUS
approved