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%I #94 Sep 05 2021 05:35:43
%S 1,2,1,6,6,1,20,30,10,1,70,140,70,14,1,252,630,420,126,18,1,924,2772,
%T 2310,924,198,22,1,3432,12012,12012,6006,1716,286,26,1,12870,51480,
%U 60060,36036,12870,2860,390,30,1,48620,218790,291720,204204,87516,24310
%N Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.
%C Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
%C 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.
%C As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - _Paul Barry_, May 30 2005
%C 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
%C As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - _Paul Barry_, Apr 14 2010
%C From _Wolfdieter Lang_, Aug 10 2017: (Start)
%C 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):
%C (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)).
%C This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(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).
%C 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)
%C From _Peter Bala_, Mar 04 2018: (Start)
%C The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
%C 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! + ....
%C 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).
%C 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)
%D 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 -).
%D Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
%H Muniru A Asiru, <a href="/A046521/b046521.txt">Antidiagonals n = 0..50, flattened</a>
%H Peter Bala, <a href="/A264772/a264772_1.pdf">A 4-parameter family of embedded Riordan arrays</a>
%H Peter Bala, <a href="/A081577/a081577.pdf">A note on the diagonals of a proper Riordan Array</a>
%H Paul Barry, <a href="http://arxiv.org/abs/1312.0583">Embedding structures associated with Riordan arrays and moment matrices</a>, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
%H J. W. Bober, <a href="http://arxiv.org/abs/0709.1977">Factorial ratios, hypergeometric series, and a family of step functions</a>, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
%H Wolfdieter Lang, <a href="/A046521/a046521.txt">First 10 rows.</a>
%H Wolfdieter Lang, <a href="http://www.fq.math.ca/Scanned/40-4/lang.pdf">On polynomials related to derivatives of the generating function of Catalan numbers</a>, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there.
%H H. Prodinger, <a href="https://www.fq.math.ca/Scanned/32-5/prodinger.pdf">Some information about the binomial transform</a>, The Fibonacci Quarterly, 32, 1994, 412-415.
%F T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
%F G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
%F Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
%F 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.
%F 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
%F From _Peter Bala_, Apr 11 2012: (Start):
%F 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.
%F The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
%F Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
%F 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
%F 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
%F From _Peter Bala_, Aug 13 2021: (Start)
%F Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
%F Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
%F G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
%F The m-th power of this array has entries m^(n-k)*T(n,k). (End)
%e Array begins:
%e 1, 2, 6, 20, 70, ...
%e 1, 6, 30, 140, 630, ...
%e 1, 10, 70, 420, 2310, ...
%e 1, 14, 126, 924, 6006, ...
%e Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
%e Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
%e From _Paul Barry_, Apr 14 2010: (Start)
%e As a number triangle, T(n, m) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 2 1
%e 2: 6 6 1
%e 3: 20 30 10 1
%e 4: 70 140 70 14 1
%e 5: 252 630 420 126 18 1
%e 6: 924 2772 2310 924 198 22 1
%e 7: 3432 12012 12012 6006 1716 286 26 1
%e 8: 12870 51480 60060 36036 12870 2860 390 30 1
%e 9: 48620 218790 291720 204204 87516 24310 4420 510 34 1
%e 10: 184756 923780 1385670 1108536 554268 184756 41990 6460 646 38 1
%e ... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
%e Production matrix begins
%e 2, 1,
%e 2, 4, 1,
%e -4, 0, 4, 1,
%e 10, 0, 0, 4, 1,
%e -28, 0, 0, 0, 4, 1,
%e 84, 0, 0, 0, 0, 4, 1,
%e -264, 0, 0, 0, 0, 0, 4, 1,
%e 858, 0, 0, 0, 0, 0, 0, 4, 1,
%e -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
%e 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
%e From _Peter Bala_, Feb 15 2018: (Start)
%e With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
%e -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) ).
%e 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)
%t 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. *)
%o (PARI) T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)
%o (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
%Y Columns for m=0..10 are A000984, A002457, A002802, A020918-A020932 (only even numbers). Row sums: A046748. Cf. A007318, A068555.
%Y Cf. A001147, A006882, A283150, A283151, A111959.
%K nonn,tabl,easy
%O 0,2
%A _Wolfdieter Lang_
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