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A158454 Riordan array (1/(1-x^2), x/(1+x)^2). 15
1, 0, 1, 1, -2, 1, 0, 4, -4, 1, 1, -6, 11, -6, 1, 0, 9, -24, 22, -8, 1, 1, -12, 46, -62, 37, -10, 1, 0, 16, -80, 148, -128, 56, -12, 1, 1, -20, 130, -314, 367, -230, 79, -14, 1, 0, 25, -200, 610, -920, 771, -376, 106, -16, 1, 1, -30, 295, -1106, 2083, -2232, 1444, -574, 137, -18, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

Coefficient table of the square of Chebyshev S-polynomials. For the S-polynomials see A049310, and for a proof see the array A181878, where the odd numbered rows are shifted by one to the left. - Wolfdieter Lang, Dec 15 2010

Image of the Catalan numbers A000108 by this matrix is the all 1's sequence.

Image of the central binomial numbers A000984 by this matrix is the counting numbers A000027.

Inverse array is the Riordan array (1-x^2*c(x)^4, xc(x)^2), where c(x) is the g.f. of A000108.

The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k belong to the class of Boas-Buck polynomials. Hence they satisfy the Boas-Buck identity: (E_x - n*1)*R(n, x) = -Sum_{p=0..n-1} ((1 - (-1)^p)*1 + 2*(-1)^(p+1)*E_x) R(n-1-p, x) for n >= 0. See the Boas-Buck comments and references in A046521. The ensuing recurrence for the column sequences is given in the formula section. - Wolfdieter Lang, Aug 10 2017

REFERENCES

Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.

LINKS

G. C. Greubel, Rows n=0..100 of triangle, flattened

Paul Barry and A. Hennessey, Notes on a Family of Riordan Arrays and Associated Integer Hankel Transforms , JIS 12 (2009) 09.5.3

Richard J. Mathar, Recurrence for the Atkinson-Steenwijk Integrals for Resistors in the Infinite Triangular Lattice, viXra:2208.0111, Aug. 2022.

FORMULA

Number triangle T(n, k) = Sum_{j=0..n} (-1)^(j-k)*binomial(k+j, 2*k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+k-j, 2*k).

O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1-x^2))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010

T(n, k) = (-1)^(n-k)*Sum_{j=0..floor(n/2)} binomial(n+k-1-2*j, 2*k-1), 0 <= k <= n, else 0. From the o.g.f. for column k after convolution. - Wolfdieter Lang, Dec 17 2010

O.g.f. row polynomials (rising powers in y):

((1+x)/(1-x))/(1+(2-y)*x+x^2) = Sum_{n>=0} (S(n,sqrt(y))^2*x^n, with Chebyshev S-polynomials from A049310. - Wolfdieter Lang, Dec 15 2010

Recurrences from the A- and Z-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.

T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), n >= 1.

T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), n >= k >= 1.

Here Z(0)=0 and Z(j) = A000108(j), j >= 1, (o.g.f. -1 + c(x), with the Catalan o.g.f. c(x)), and A(j) = A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0, with o.g.f. 1/c(x)^2. - Wolfdieter Lang, Dec 20 2010

T(n, k) = Sum_{m=0..n} A129818(m, k), 0 <= k <= n. - Wolfdieter Lang, Dec 15 2010

Boas-Buck recurrence for column k: R(n, k) = (1/(n-k))*Sum_{p=k..n-1}((-1)^(n-p)*(2*k+1) + 1) * R(p, k), for n > k >= 0, with input R(k, k) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017

G.f.: (1 + x)/((1 - x)*(1 + x)^2 - t*x*(1 - x)). - G. C. Greubel, Dec 15 2018

T(n, k) = (-1)^(n - k)*binomial(k + n - 1, 2*k-1)*hypergeom([1, (k - n)/2, (1 + k - n)/2], [(1 - k - n)/2, (2 - k - n)/2], 1) for k >= 1 . - Peter Luschny, Aug 20 2022

EXAMPLE

The triangle T(n,k) begins:

n\k 0 1 2 3 4 5 6 7 8 9 10...

0: 1

1: 0 1

2: 1 -2 1

3: 0 4 -4 1

4: 1 -6 11 -6 1

5: 0 9 -24 22 -8 1

6: 1 -12 46 -62 37 -10 1

7: 0 16 -80 148 -128 56 -12 1

8: 1 -20 130 -314 367 -230 79 -14 1

9: 0 25 -200 610 -920 771 -376 106 -16 1

10: 1 -30 295 -1106 2083 -2232 1444 -574 137 -18 1

... Reformatted and extended by Wolfdieter Lang, Nov 24 2012

Recurrences (from A- and Z-sequences):

1 = T(6,0) = 0*0 + 1*9 +2*(-24) + 5*22 + 14*(-8)+ 42*1.

-80 = T(7,2) = 1*(-12) -2*(46) -1*(-62) -2*37 -5*(-10) -14*1. - Wolfdieter Lang, Dec 20 2010

MAPLE

A158454 := proc(n, k) (-1)^(n+k)*add(binomial(n+k-1-2*j, 2*k-1), j=0..floor(n/2)) ; end proc;

seq(seq(A158454(n, k), k=0..n), n=0..10) ; # R. J. Mathar, Dec 17 2010

MATHEMATICA

nmax = 10; t[n_, k_] := (-1)^(n+k)* Sum[Binomial[n+k-1-2*j, 2*k-1], {j, 0, Floor[n/2]}]; t[n_?EvenQ, 0] = 1; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 08 2011, after Maple *)

With[{m = 15}, CoefficientList[CoefficientList[Series[(1+x)/((1-x)*(1 + x)^2 -t*x*(1-x)), {x, 0, m}, {t, 0, m}], x], t]]//Flatten (* G. C. Greubel, Dec 15 2018 *)

T[n_, 0] := Boole[EvenQ[n]]; T[n_, k_] := (-1)^(n - k) Binomial[k+n-1, 2*k-1] HypergeometricPFQ[{1, (k - n)/2, (1 + k - n)/2}, {(1 - k - n)/2, (2 - k - n)/2}, 1]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // TableForm (* Peter Luschny, Aug 20 2022 *)

PROG

(PARI) {T(n, k) = sum(j=0, n, (-1)^(j-k)*binomial(k+j, 2*k))};

for(n=0, 10, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Dec 15 2018

(Magma) [[(&+[(-1)^(j-k)*Binomial(k+j, 2*k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Dec 15 2018

(Sage) [[sum((-1)^(j-k)*binomial(k+j, 2*k) for j in range(n+1)) for k in range(n+1)] for n in range(10)] # G. C. Greubel, Dec 15 2018

(GAP) T:=Flat(List([0..10], n->List([0..n], k->Sum([0..n], j-> (-1)^(j-k)*Binomial(k+j, 2*k))))); # G. C. Greubel, Dec 15 2018

CROSSREFS

Cf. A000027, A000108, A000984, A006232, A049310, A046521 (Boas-Buck info), A115141, A181878.

From Wolfdieter Lang, Aug 10 2017: (Start)

Row sums A011655(n+1), alternating row sums A007598(n+1)*(-1)^(n+1).

Column sequences k=0..5: A059841, A002620(n+2)*(-1)^(n), A001752(n)*(-1)^n, A001769(n)*(-1)^n, A001780(n)*(-1)^n, A001786(n)*(-1)^n. (End)

Sequence in context: A296129 A276544 A214753 * A049243 A077908 A052922

Adjacent sequences: A158451 A158452 A158453 * A158455 A158456 A158457

KEYWORD

easy,sign,tabl

AUTHOR

Paul Barry, Mar 19 2009

STATUS

approved

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Last modified December 3 09:50 EST 2022. Contains 358517 sequences. (Running on oeis4.)