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 A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down. 24
 1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4). This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306. As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005 The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007 As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010 From Wolfdieter Lang, Aug 10 2017: (Start) The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):   (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)). This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: R(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m). For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End) From Peter Bala, Mar 04 2018: (Start) The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)). 1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + .... 2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5). Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End) REFERENCES Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -). Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146. LINKS Muniru A Asiru, Antidiagonals n = 0..50, flattened P. Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013. J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444. W. Lang, First 10 rows. W. Lang, On polynomials related to derivatives of the generating function of Catalan numbers, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there. FORMULA T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0. G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x). Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1. Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1. As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010 One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555. The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere. Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). - Peter Bala, Apr 11 2012 T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016 Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017 EXAMPLE Array begins:   1,  2,   6,  20,   70, ...   1,  6,  30, 140,  630, ...   1, 10,  70, 420, 2310, ...   1, 14, 126, 924, 6006, ... Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30. Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1. From Paul Barry, Apr 14 2010: (Start) As a number triangle, T(n, m) begins: n\k       0      1       2       3      4      5     6    7   8  9 10 ... 0:        1 1:        2      1 2:        6      6       1 3:       20     30      10       1 4:       70    140      70      14      1 5:      252    630     420     126     18      1 6:      924   2772    2310     924    198     22     1 7:     3432  12012   12012    6006   1716    286    26    1 8:    12870  51480   60060   36036  12870   2860   390   30   1 9:    48620 218790  291720  204204  87516  24310  4420  510  34  1 10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1 ... [Reformatted and extended by Wolfdieter Lang, Aug 10 2017] Production matrix begins       2, 1,       2, 4, 1,      -4, 0, 4, 1,      10, 0, 0, 4, 1,     -28, 0, 0, 0, 4, 1,      84, 0, 0, 0, 0, 4, 1,    -264, 0, 0, 0, 0, 0, 4, 1,     858, 0, 0, 0, 0, 0, 0, 4, 1,   -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End) Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - Wolfdieter Lang, Aug 10 2017 From Peter Bala, Feb 15 2018: (Start) With C(x) = (1 - sqrt( 1 - 4*x))/(2*x), -x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ). x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End) MATHEMATICA t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *) PROG (PARI) T(i, j)=if(i<0 || j<0, 0, (2*i+2*j)!*i!/(2*i)!/(i+j)!/j!) (GAP) Flat(List([0..9], n->List([0..n], m->Binomial(2*n, n)*Binomial(n, m)/Binomial(2*m, m)))); # Muniru A Asiru, Jul 19 2018 CROSSREFS Columns for m=0..10 are A000984, A002457, A002802, A020918-A020932 (only even numbers). Row sums: A046748. Cf. A007318, A068555. Cf. A001147, A006882, A283150, A283151. Sequence in context: A269646 A269336 A300700 * A104684 A060538 A260848 Adjacent sequences:  A046518 A046519 A046520 * A046522 A046523 A046524 KEYWORD nonn,tabl,easy,changed AUTHOR STATUS approved

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Last modified August 14 06:01 EDT 2018. Contains 313748 sequences. (Running on oeis4.)