cp's OEIS Frontend

A187075 A Galton triangle: T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1).

%I A187075 #46 Nov 12 2022 02:13:16
%S A187075 1,2,3,4,18,15,8,84,180,105,16,360,1500,2100,945,32,1488,10800,27300,
%T A187075 28350,10395,64,6048,72240,294000,529200,436590,135135,128,24384,
%U A187075 463680,2857680,7938000,11060280,7567560,2027025,256,97920,2904000,26107200,105099120,220041360,249729480,145945800,34459425
%N A187075 A Galton triangle: T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1).
%C A187075 This is a companion triangle to A186695.
%C A187075 Let f(x) = (exp(2*x) + 1)^(-1/2); then the n-th derivative of f equals Sum_{k=1..n} (-1)^k*T(n,k)*(f(x))^(2*k+1). - _Groux Roland_, May 17 2011
%C A187075 Triangle T(n,k), 1 <= k <= n, given by (0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, ...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Oct 20 2013
%H A187075 G. C. Greubel, <a href="/A187075/b187075.txt">Table of n, a(n) for the first 50 rows</a>
%H A187075 Shi-Mei Ma, <a href="http://arxiv.org/abs/1204.4963">A family of two-variable derivative polynomials for tangent and secant</a>, arXiv:1204.4963v3 [math.CO]
%H A187075 Shi-Mei Ma, <a href="https://doi.org/10.37236/2344">A family of two-variable derivative polynomials for tangent and secant</a>, El J. Combinat. 20 (1) (2013) P11.
%H A187075 E. Neuwirth, <a href="https://doi.org/10.1016/S0012-365X(00)00373-3">Recursively defined combinatorial functions: Extending Galton's board,</a> Discrete Math. 239 (2001) 33-51.
%F A187075 T(n,k) = 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k).
%F A187075 Recurrence relation T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1) with boundary conditions T(1,1) = 1, T(1,k) = 0 for k >= 2.
%F A187075 G.f.: F(x,t) = 1/sqrt((1+x)-x*exp(2*t)) - 1 = Sum_{n >= 1} R(n,x)*t^n/n! = x*t + (2*x+3*x^2)*t^2/2! + (4*x+18*x^2+15*x^3)*t^3/3! + ....
%F A187075 The g.f. F(x,t) satisfies the partial differential equation dF/dt = 2*(x+x^2)*dF/dx + x*F.
%F A187075 The row polynomials R(n,x) satisfy the recursion R(n+1,x) = 2*(x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.
%F A187075 O.g.f. for column k: (2k-1)!!*x^k/Product_{m = 1..k} (1-2*m*x) (compare with A075497). T(n,k) = (2*k-1)!!*A075497(n,k).
%F A187075 The row polynomials R(n,x) = Sum_{k = 1..n} T(n,k)*x^k satisfy R(n,-x-1) = (-1)^n*(1+x)/x*P(n,x) where P(n,x) is the n-th row polynomial of A186695. We also have R(n,x/(1-x)) = (x/(1-x)^n)*Q(n-1,x) where Q(n,x) is the n-th row polynomial of A156919.
%F A187075 T(n,k) = 2^(n-k)*A211608(n,k). - _Philippe Deléham_, Oct 20 2013
%e A187075 Triangle begins
%e A187075 n\k.|...1.....2......3......4......5......6
%e A187075 ===========================================
%e A187075 ..1.|...1
%e A187075 ..2.|...2.....3
%e A187075 ..3.|...4....18.....15
%e A187075 ..4.|...8....84....180....105
%e A187075 ..5.|..16...360...1500...2100....945
%e A187075 ..6.|..32..1488..10800..27300..28350..10395
%e A187075 ..
%e A187075 Examples of recurrence relation:
%e A187075 T(4,3) = 6*T(3,3) + 5*T(3,2) = 6*15 + 5*18 = 180;
%e A187075 T(6,4) = 8*T(5,4) + 7*T(5,3) = 8*2100 + 7*1500 = 27300.
%p A187075 A187075 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 then 2^(n-1); else 2*k*procname(n-1, k) + (2*k-1)*procname(n-1, k-1) ; end if; end proc:seq(seq(A187075(n,k),k = 1..n),n = 1..10);
%t A187075 Flatten[Table[2^(n - 2*k)*Binomial[2 k, k]*k!*StirlingS2[n, k], {n, 10}, {k, 1, n}]] (* _G. C. Greubel_, Jun 17 2016 *)
%o A187075 (Sage) # uses[delehamdelta from A084938]
%o A187075 # Adds a first column (1,0,0,0, ...).
%o A187075 def A187075_triangle(n):
%o A187075     return delehamdelta([(i+1)*int(is_even(i+1)) for i in (0..n)], [i+1 for i in (0..n)])
%o A187075 A187075_triangle(4)  # _Peter Luschny_, Oct 20 2013
%Y A187075 Cf. A075497, A156919, A186695, A211402.
%K A187075 nonn,easy,tabl
%O A187075 1,2
%A A187075 _Peter Bala_, Mar 27 2011