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%I #36 Jan 10 2024 15:58:49
%S 1,3,2,15,20,4,105,210,84,8,945,2520,1512,288,16,10395,34650,27720,
%T 7920,880,32,135135,540540,540540,205920,34320,2496,64,2027025,
%U 9459450,11351340,5405400,1201200,131040,6720,128
%N Triangle S(n, k) by rows: coefficients of 2^((n-1)/2)*(x^(1/2)*d/dx)^n, where n = 1, 3, 5, ...
%C Triangle S(n,n-k) by rows: coefficients of 2^n * |L(n,1/2,x)|, with L the generalized Laguerre polynomials.
%H U. N. Katugampola, <a href="http://authors.elsevier.com/a/1QhUNLvMg0Zs~">Mellin Transforms of Generalized Fractional Integrals and Derivatives</a>, Appl. Math. Comput. 257(2015) 566-580.
%H U. N. Katugampola, <a href="http://arxiv.org/abs/1411.5229">Existence and Uniqueness results for a class of Generalized Fractional Differential Equations</a>, arXiv preprint arXiv:1411.5229, 2014
%F T(n, k) = 2^n * n!/(n-k)! * C(n+1/2, k), n>=0, k<=n.
%e Triangle begins:
%e 1;
%e 3, 2;
%e 15, 20, 4;
%e 105, 210, 84, 8;
%e 945, 2520, 1512, 288, 16;
%e 10395, 34650, 27720, 7920, 880, 32;
%e 135135, 540540, 540540, 205920, 34320, 2496, 64;
%e .
%e .
%e Expansion takes the form:
%e 2^0 (x^(1/2)*d/dx)^1 = 1*x^(1/2)*d/dx.
%e 2^1 (x^(1/2)*d/dx)^3 = 3*x^(1/2)*d^2/dx^2 + 2*x^(3/2)*d^3/dx^3.
%e 2^2 (x^(1/2)*d/dx)^5 = 15*x^(1/2)*d^3/dx^3 + 20*x^(3/2)*d^4/dx^4 + 4*x^(5/2)*d^5/dx^5.
%p a[0]:= f(x):
%p for i from 1 to 20 do
%p a[i]:= simplify(2^((i+1)mod 2)*x^(1/2)*(diff(a[i-1],x$1)));
%p end do:
%p for j from 1 to 10 do
%p b[j]:=a[2j-1];
%p end do;
%Y Cf. A223168-A223172.
%Y Rows includes odd rows of A223168.
%Y Rows includes absolute values of A098503 from right to left of the triangular form.
%K nonn,tabl
%O 1,2
%A _Udita Katugampola_, Mar 21 2013