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 A187075 A Galton triangle: T(n,k) = 2*k*T(n-1,k) + (2*k-1)*T(n-1,k-1). 5
 1, 2, 3, 4, 18, 15, 8, 84, 180, 105, 16, 360, 1500, 2100, 945, 32, 1488, 10800, 27300, 28350, 10395, 64, 6048, 72240, 294000, 529200, 436590, 135135, 128, 24384, 463680, 2857680, 7938000, 11060280, 7567560, 2027025, 256, 97920, 2904000, 26107200, 105099120, 220041360, 249729480, 145945800, 34459425 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS This is a companion triangle to A186695. 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] 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 LINKS G. C. Greubel, Table of n, a(n) for the first 50 rows Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, arXiv:1204.4963v3 [math.CO] Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11. E. Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 (2001) 33-51. FORMULA T(n,k) = 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k). 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. G.f.: F(x,t) = 1/sqrt((1+x)-x*exp(2*t)) - 1 = sum {n = 1..inf} 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! + .... The g.f. F(x,t) satisfies the partial differential equation dF/dt = 2*(x+x^2)*dF/dx + x*F. 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. 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). 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 haveR(n,x/(1-x)) = x/(1-x)^n*Q(n-1,x) where Q(n,x) is the n-th row polynomial of A156919. T(n,k) = 2^(n-k)*A211608(n,k). - Philippe Deléham, Oct 20 2013 EXAMPLE Triangle begins n\k.|...1.....2......3......4......5......6 =========================================== ..1.|...1 ..2.|...2.....3 ..3.|...4....18.....15 ..4.|...8....84....180....105 ..5.|..16...360...1500...2100....945 ..6.|..32..1488..10800..27300..28350..10395 .. Examples of recurrence relation T(4,3) = 6*T(3,3)+5*T(3,2) = 6*15+5*18 = 180; T(6,4) = 8*T(5,4)+7*T(5,3) = 8*2100+7*1500 = 27300. MAPLE 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); MATHEMATICA 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 *) PROG (Sage) # uses[delehamdelta from A084938] # Adds a first column (1, 0, 0, 0, ...). def A187075_triangle(n):     return delehamdelta([(i+1)*int(is_even(i+1)) for i in (0..n)], [i+1 for i in (0..n)]) A187075_triangle(4)  # Peter Luschny, Oct 20 2013 CROSSREFS Cf. A075497, A156919, A186695, A211402. Sequence in context: A037394 A037430 A329545 * A154715 A077407 A273002 Adjacent sequences:  A187072 A187073 A187074 * A187076 A187077 A187078 KEYWORD nonn,easy,tabl AUTHOR Peter Bala, Mar 27 2011 STATUS approved

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Last modified August 1 07:45 EDT 2021. Contains 346384 sequences. (Running on oeis4.)