|
%I #65 Dec 14 2022 16:33:34
%S 1,1,1,2,2,1,4,6,3,1,8,16,12,4,1,16,40,40,20,5,1,32,96,120,80,30,6,1,
%T 64,224,336,280,140,42,7,1,128,512,896,896,560,224,56,8,1,256,1152,
%U 2304,2688,2016,1008,336,72,9,1,512,2560,5760,7680,6720,4032,1680,480,90,10,1
%N Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(n,2j)*binomial(n-2j,k).
%C Product of Pascal's triangle A007318 and A119467. Row sums are A007051. Diagonal sums are A113225.
%C Variant of A080928, A115068 and A082137. - _R. J. Mathar_, Feb 09 2010
%C Matrix inverse of the Euler tangent triangle A081733. - _Peter Luschny_, Jul 18 2012
%C Central column: T(2*n,n) = A069723(n). - _Peter Luschny_, Jul 22 2012
%C Subtriangle of the triangle in A198792. - _Philippe Deléham_, Nov 10 2013
%F G.f.: (1 - x - xy)/(1 - 2x - 2x*y + 2x^2*y + x^2*y^2).
%F Number triangle T(n,k) = Sum_{j=0..n} binomial(n,j)*binomial(j,k)*(1+(-1)^(j-k))/2.
%F Define matrix: A(n,m,k) = If[m < n, 1, -1];
%F p(x,k) = CharacteristicPolynomial[A[n,m,k],x]; then t(n,m) = coefficients(p(x,n)). - _Roger L. Bagula_ and _Gary W. Adamson_, Jan 25 2009
%F E.g.f.: exp(x*z)/(1-tanh(x)). - _Peter Luschny_, Aug 01 2012
%F T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) for n >= 2, T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - _Philippe Deléham_, Nov 10 2013
%F E.g.f.: [(e^(2t)+1)/2] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x + 2/(e^(-2D)+1), i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x) where P_n(x) = [(x+2)^n + x^n]/2. Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x + 1 + D - 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, so the unsigned differential component 2/[e^(2D)+1] = 2 Sum_{n >= 0} Eta(-n) (-2D)^n/n!, where Eta(s) is the Dirichlet eta function, and 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers. The polynomials PI_n(x) of A081733 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Aside from the signs and the main diagonals, multiplying this triangle by 2 gives the face-vectors of the hypercubes A038207. - _Tom Copeland_, Sep 27 2015
%F T(n,k) = 2^(n-k-1+0^(n-k))*binomial(n, k). - _Peter Luschny_, Nov 10 2017
%e Triangle begins
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 4, 6, 3, 1;
%e 8, 16, 12, 4, 1;
%e 16, 40, 40, 20, 5, 1;
%e 32, 96, 120, 80, 30, 6, 1;
%e 64, 224, 336, 280, 140, 42, 7, 1;
%e 128, 512, 896, 896, 560, 224, 56, 8, 1;
%e 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 1;
%e 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1;
%p A119468_row := proc(n) local s,t,k;
%p s := series(exp(z*x)/(1-tanh(x)),x,n+2);
%p t := factorial(n)*coeff(s,x,n); seq(coeff(t,z,k), k=(0..n)) end:
%p for n from 0 to 7 do A119468_row(n) od; # _Peter Luschny_, Aug 01 2012
%p # Alternatively:
%p T := (n, k) -> 2^(n-k-1+0^(n-k))*binomial(n,k):
%p for n from 0 to 9 do seq(T(n,k), k=0..n) od; # _Peter Luschny_, Nov 10 2017
%t A[k_] := Table[If[m < n, 1, -1], {m, k}, {n, k}]; a = Join[{{1}}, Table[(-1)^n*CoefficientList[CharacteristicPolynomial[A[n], x], x], {n, 1, 10}]]; Flatten[a] (* _Roger L. Bagula_ and _Gary W. Adamson_, Jan 25 2009 *)
%t Table[Sum[Binomial[n,2j]Binomial[n-2j,k],{j,0,n-k}],{n,0,10},{k,0,n}]//Flatten (* _Harvey P. Dale_, Dec 14 2022 *)
%o (Sage)
%o R = PolynomialRing(QQ, 'x')
%o def p(n,x) :
%o return 1 if n==0 else add((-1)^n*binomial(n,k)*(x^(n-k)-1) for k in range(n))
%o def A119468_row(n):
%o x = R.gen()
%o return [abs(cf) for cf in list((p(n,x-1)-p(n,x+1))/2+x^n)]
%o for n in (0..8) : print(A119468_row(n)) # _Peter Luschny_, Jul 22 2012
%Y A082137 read as triangle with rows reversed.
%Y Cf. A000182, A009006, A038207, A081733, A155585.
%K easy,nonn,tabl
%O 0,4
%A _Paul Barry_, May 21 2006
|
