%I #63 Oct 19 2022 11:19:28
%S 1,2,1,6,4,1,20,16,6,1,70,64,30,8,1,252,256,140,48,10,1,924,1024,630,
%T 256,70,12,1,3432,4096,2772,1280,420,96,14,1,12870,16384,12012,6144,
%U 2310,640,126,16,1,48620,65536,51480,28672,12012,3840,924,160,18,1
%N A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).
%C In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).
%C The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.
%C Riordan array (1/sqrt(1-4x),x/sqrt(1-4x)). - _Paul Barry_, May 06 2009
%C The matrix inverse is apparently given by deleting the leftmost column from A206022. - _R. J. Mathar_, Mar 12 2013
%H G. C. Greubel, <a href="/A054335/b054335.txt">Rows n = 0..100 of triangle, flattened</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 Milan Janjić, <a href="https://www.emis.de/journals/JIS/VOL21/Janjic2/janjic103.html">Pascal Matrices and Restricted Words</a>, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
%F a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n<m.
%F Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.
%F G.f. for column m: cbie(x)*((x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).
%F G.f.: 1/(1-xy-2x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - _Paul Barry_, May 06 2009
%F Sum_{k>=0} T(n,2k)*(-1)^k*A000108(k) = A000108(n+1). - _Philippe Deléham_, Jan 30 2012
%F Sum_{k=0..floor(n/2)} T(n-k,n-2k) = A098615(n). - _Philippe Deléham_, Feb 01 2012
%F T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - _Philippe Deléham_, Feb 02 2012
%F Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - _Peter Bala_, Oct 17 2015
%e Triangle begins:
%e 1;
%e 2, 1;
%e 6, 4, 1;
%e 20, 16, 6, 1;
%e 70, 64, 30, 8, 1;
%e 252, 256, 140, 48, 10, 1;
%e 924, 1024, 630, 256, 70, 12, 1; ...
%e Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
%e From _Paul Barry_, May 06 2009: (Start)
%e Production matrix begins
%e 2, 1;
%e 2, 2, 1;
%e 0, 2, 2, 1;
%e -2, 0, 2, 2, 1;
%e 0, -2, 0, 2, 2, 1;
%e 4, 0, -2, 0, 2, 2, 1;
%e 0, 4, 0, -2, 0, 2, 2, 1;
%e -10, 0, 4, 0, -2, 0, 2, 2, 1;
%e 0, -10, 0, 4, 0, -2, 0, 2, 2, 1; (End)
%p A054335 := proc(n,k)
%p if k <0 or k > n then
%p 0 ;
%p elif type(k,odd) then
%p kprime := floor(k/2) ;
%p binomial(n-kprime-1,kprime)*4^(n-k) ;
%p else
%p kprime := k/2 ;
%p binomial(2*n-k,n-kprime)*binomial(n-kprime,kprime)/binomial(k,kprime) ;
%p end if;
%p end proc: # _R. J. Mathar_, Mar 12 2013
%p # Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
%p PMatrix(10, n -> binomial(2*(n-1), n-1)); # _Peter Luschny_, Oct 19 2022
%t Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
%t a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Sep 08 2011, updated Jan 16 2014 *)
%o (PARI) T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
%o for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Jul 20 2019
%o (Magma)
%o T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2),Floor(k/2))/Binomial(k,Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
%o [[T(n,k): k in [0..n]]: n in [0..10]]; // _G. C. Greubel_, Jul 20 2019
%o (Sage)
%o def T(n, k):
%o if (mod(k,2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2)
%o else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
%o [[T(n,k) for k in (0..n)] for n in (0..10)] # _G. C. Greubel_, Jul 20 2019
%o (GAP)
%o T:= function(n, k)
%o if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2),Int(k/2))/Binomial(k,Int(k/2));
%o else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
%o fi;
%o end;
%o Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # _G. C. Greubel_, Jul 20 2019
%Y Cf. A000984, A035324, A054336.
%Y Row sums: A026671.
%K easy,nice,nonn,tabl
%O 0,2
%A _Wolfdieter Lang_, Mar 13 2000