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A368235 Triangle read by rows: n-th row polynomial equals the numerator of the rational function (-1)^n*f(x) * (d/dx)^n (1/f(x)), where f(x) = sqrt(x + x^2).

%I A368235 #23 Mar 29 2025 00:23:37
%S A368235 1,1,2,3,8,8,15,54,72,48,105,480,864,768,384,945,5250,12000,14400,
%T A368235 9600,3840,10395,68040,189000,288000,259200,138240,46080,135135,
%U A368235 1018710,3333960,6174000,7056000,5080320,2257920,645120,2027025,17297280,65197440,142248960,197568000,180633600,108380160,41287680,10321920
%N A368235 Triangle read by rows: n-th row polynomial equals the numerator of the rational function (-1)^n*f(x) * (d/dx)^n (1/f(x)), where f(x) = sqrt(x + x^2).
%C A368235 Unsigned row reverse of A123516.
%C A368235 The row polynomials also occur on repeated integration of 1/sqrt(x + x^2). See the example section.
%F A368235 T(n, k) = n! * 2^(2*k-n) * binomial(n, k) * binomial(2*n-2*k, n-k).
%F A368235 k*T(n, k) = (2*n^2)*T(n-1, k-1) for k >= 1 with T(n, 0) = (2*n - 1)!! = A001147(n).
%F A368235 T(n, 1) = 2*A161120(n).
%F A368235 T(n, n) = 2^n * n! = A000165(n); T(n+1, n) = 2^n * n! * (n+1)^2 = A014479(n);
%F A368235 T(n+2, n) = 3 * 2^(n-1)*(n+2)!*binomial(n+2, 2) = 3 * A286725(n).
%F A368235 More generally, T(n+r, n) = (2*r - 1)!! * A286724(n+r, r).
%F A368235 E.g.f.: Sum_{k >= 0} (1/2^k)*binomial(2*k, k)*t^k/(1 - 2*t*x)^(k+1) = 1 + (1 + 2*x)*t + (3 + 8*x + 8*x^2)*t^2/2! + (15 + 54*x + 72*x^2 + 48*x^3)*t^3/3! + ....
%F A368235 n-th row polynomial R(n, x) = (-2)^n*(x + x^2)^(n+1/2)*(d/dx)^n (1/sqrt(x + x^2)).
%F A368235 Recurrence for row polynomials:
%F A368235 R(n+1, x) = (2*x + 1)*(2*n + 1)*R(n, x) - 4*x*(x + 1)*n^2*R(n-1, x), with R(0, x) = 1.
%F A368235 R'(n, x) = 2*n^2 * R(n-1, x) for n >= 1.
%F A368235 Functional equation: R(n, -1 - x) = (-1)^n * R(n, x).
%F A368235 Conjecture: the zeros of the polynomial R(n, -x) lie on the vertical line Re(x) = 1/2 in the complex plane.
%F A368235 (-1)^n * x^n * R(n, (- 1 - x)/x) equals the n-th row polynomial of A123516.
%F A368235 (1 - x)^n * R(n, x/(1 - x)) equals the n-th row polynomial of A059366.
%F A368235 Let D denote the operator (1/x)*d/dx. Then D^(n+1)( arcsinh(x) ) = (-1)^n*R(n, x^2)/(x*sqrt(1 + x^2))^(2*n+1).
%F A368235 R(n, 1/2) = A331817(n); R(n, -1/2) = A177145(n+1);
%F A368235 (2^n) * R(n, 1/4) = A098461(n).
%F A368235 Alternating row sums R(n, -1) = (-1)^n * A001147(n).
%e A368235 Triangle begins
%e A368235  n\k |     0       1        2        3        4        5       6
%e A368235 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e A368235   0  |     1
%e A368235   1  |     1       2
%e A368235   2  |     3       8        8
%e A368235   3  |    15      54       72       48
%e A368235   4  |   105     480      864      768      384
%e A368235   5  |   945    5250    12000    14400     9600     3840
%e A368235   6  | 10395   68040   189000   288000   259200   138240   46080
%e A368235  ...
%e A368235 Repeated integration of 1/f(x), where f(x) = sqrt(x + x^2):
%e A368235 Let I denote the integral operator h(x) -> Integral_{t = 0..x} h(t) dt.
%e A368235 Let g(x) = I(1/f(x)) = log(2*x + 1 + 2*f(x)). Then
%e A368235 (2^1 * 1!^2) * I^(2)(1/f(x)) = (2*x + 1)*g(x) - 2*f(x).
%e A368235 (2^2 * 2!^2) * I^(3)(1/f(x)) = (8*x^2 + 8*x + 3)*g(x) - 6*(2*x + 1)*f(x).
%e A368235 (2^3 * 3!^2) * I^(4)(1/f(x)) = (48*x^3 + 72*x^2 + 54*x + 15)*g(x) - 2*(44*x^2 + 44*x + 15)*f(x).
%e A368235 (2^4 * 4!^2) * I^(5)(1/f(x)) = (384*x^4 + 768*x^3 + 864*x^2 + 480*x + 105)*g(x) - 10*(2*x + 1)*(40*x^2 + 40*x + 21)*f(x).
%p A368235 # sequence in triangular form
%p A368235 T := (n, k) -> n! * 2^(2*k-n) * binomial(n, k)*binomial(2*n-2*k, n-k):
%p A368235 for n from 0 to 8 do seq(T(n, k), k = 0..n) od;
%Y A368235 Cf. A001147 (column 1), 2*A161120 (column 2), A000165 (main diagonal) A014479 (first subdiagonal), 3*A286725 (second subdiagonal).
%Y A368235 Cf. A059366, A098461, A123516, A156919, A177145, A286724, A331817.
%K A368235 sign,tabl,easy
%O A368235 0,3
%A A368235 _Peter Bala_, Dec 18 2023