A223523 Triangle S(n, k) by rows: coefficients of 2^((n-1)/2)*(x^(1/2)*d/dx)^n, where n = 1, 3, 5, ...
1, 3, 2, 15, 20, 4, 105, 210, 84, 8, 945, 2520, 1512, 288, 16, 10395, 34650, 27720, 7920, 880, 32, 135135, 540540, 540540, 205920, 34320, 2496, 64, 2027025, 9459450, 11351340, 5405400, 1201200, 131040, 6720, 128
Offset: 1
Examples
Triangle begins: 1; 3, 2; 15, 20, 4; 105, 210, 84, 8; 945, 2520, 1512, 288, 16; 10395, 34650, 27720, 7920, 880, 32; 135135, 540540, 540540, 205920, 34320, 2496, 64; . . Expansion takes the form: 2^0 (x^(1/2)*d/dx)^1 = 1*x^(1/2)*d/dx. 2^1 (x^(1/2)*d/dx)^3 = 3*x^(1/2)*d^2/dx^2 + 2*x^(3/2)*d^3/dx^3. 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.
Links
- U. N. Katugampola, Mellin Transforms of Generalized Fractional Integrals and Derivatives, Appl. Math. Comput. 257(2015) 566-580.
- U. N. Katugampola, Existence and Uniqueness results for a class of Generalized Fractional Differential Equations, arXiv preprint arXiv:1411.5229, 2014
Crossrefs
Programs
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Maple
a[0]:= f(x): for i from 1 to 20 do a[i]:= simplify(2^((i+1)mod 2)*x^(1/2)*(diff(a[i-1],x$1))); end do: for j from 1 to 10 do b[j]:=a[2j-1]; end do;
Formula
T(n, k) = 2^n * n!/(n-k)! * C(n+1/2, k), n>=0, k<=n.
Comments