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%I #18 Jun 06 2021 15:46:21
%S 1,1,1,1,2,1,1,1,3,3,1,3,3,1,4,6,4,1,6,12,6,3,1,5,10,10,5,1,10,30,30,
%T 10,15,15,1,6,15,20,15,6,1,15,60,90,60,15,45,90,45,15,1,7,21,35,35,21,
%U 7,1,21,105,210,210,105,21,105,315,315,105,105,105
%N Concatenated Bessel-scaled Pascal triangles. Irregular triangle read by rows, T(n,k) with n >= 0 and 0 <= k <= (2*n*(n + 4) - 1 + (-1)^n)/8.
%C Let p(n) = Sum_{k=0..n/2} Sum_{j=0..n-2*k} (n!/(2^k*k!*j!*(n-2*k-j)!))*x^j*y^(n-2*k-j). Row n of the triangle is defined as the coefficient list of the polynomials p(n), where the monomials are in degree-lexicographic order.
%C One observes: The triangle of the coefficients appears as a series of concatenated subtriangles. The first one is Pascal's triangle A007318. Appending the rows of triangle A094305 on the right side starts in row 2. In row 4, the next triangle is appended, which is A344565. This scheme goes on indefinitely.
%C This can be formalized as follows: Let C(n) denote row n of the binomial triangle, set c.C(n) = Seq_{j=0..n} c*binomial(n, j), and let B(n, k) denote the Bessel numbers A100861(n, k). Then T(n) = Seq_{k=0..n/2} B(n, k).C(n-2*k). Since B(n, k) = binomial(n, 2*k)*(2*k - 1)!! it follows that: T(n) = Seq_{k=0..n/2} Seq_{j=0..n-2*k} binomial(n, 2*k)*binomial(n-2*k, j)*(2*k-1)!!. This expression equals the coefficient list of p(n) since the monomials are in degree-lexicographic order.
%C The polynomials are also the unsigned, probabilist's Hermite polynomials H_n(x+y)
%C which are discussed in A344678. The coefficients are listed there in a different order which do not reveal the structure described above.
%F The bivariate e.g.f. exp(t^2/2)*exp(t*(x + y)) = Sum_{n>=0} H_n(x + y)*t^n/n!, where H_n(x) are the unsigned, modified Hermite polynomials A099174, is given by _Tom Copeland_ in A344678.
%e The triangle begins:
%e [0] [ 1 ]
%e [1] [ 1, 1 ]
%e [2] [ 1, 2, 1 ][ 1 ]
%e [3] [ 1, 3, 3, 1 ][ 3, 3 ]
%e [4] [ 1, 4, 6, 4, 1 ][ 6, 12, 6 ][ 3 ]
%e [5] [ 1, 5, 10, 10, 5, 1 ][ 10, 30, 30, 10 ][ 15, 15 ]
%e [6] [ 1, 6, 15, 20, 15, 6, 1 ][ 15, 60, 90, 60, 15 ][ 45, 90, 45][ 15 ]
%e .
%e With the notations in the comment row 7 concatenates:
%e B(7, 0).C(7) = 1.[1, 7, 21, 35, 35, 21, 7, 1] = [1, 7, 21, 35, 35, 21, 7, 1],
%e B(7, 1).C(5) = 21.[1, 5, 10, 10, 5, 1] = [21, 105, 210, 210, 105, 21],
%e B(7, 2).C(3) = 105.[1, 3, 3, 1] = [105, 315, 315, 105],
%e B(7, 3).C(1) = 105.[1, 1] = [105, 105].
%e .
%e p_6(x,y) = x^6 + 6*x^5*y + 15*x^4*y^2 + 20*x^3*y^3 + 15*x^2*y^4 + 6*x*y^5 + y^6 +
%e 15*x^4 + 60*x^3*y + 90*x^2*y^2 + 60*x*y^3 + 15*y^4 + 45*x^2 + 90*x*y + 45*y^2 + 15.
%p P := n -> add(add(n!/(2^k*k!*j!*(n-2*k-j)!)*y^(n-2*k-j)*x^j, j=0..n-2*k), k=0..n/2):
%p seq(seq(subs(x = 1, y = 1, m), m = [op(P(n))]), n = 0..7);
%p # Alternatively, without polynomials:
%p B := (n, k) -> binomial(n, 2*k)*doublefactorial(2*k-1):
%p C := n -> seq(binomial(n, j), j=0..n):
%p seq(seq(B(n, k)*C(n-2*k), k = 0..n/2), n = 0..7);
%p # Based on the e.g.f. of the polynomials:
%p T := proc(numofrows) local gf, ser, n, m;
%p gf := exp(t^2/2)*exp(t*(x + y)); ser := series(gf, t, numofrows+1);
%p for n from 0 to numofrows do [op(sort(n!*expand(coeff(ser, t, n))))];
%p print(seq(subs(x=1, y=1, m), m = %)) od end: T(7);
%t P[n_] := Sum[ Sum[n! / (2^k k! j! (n - 2k - j)!) y^(n - 2k - j) x^j, {j, 0, n-2k}], {k, 0, n/2}];
%t DegLexList[p_] := MonomialList[p, {x, y}, "DegreeLexicographic"] /. x->1 /. y->1;
%t Table[DegLexList[P[n]], {n, 0, 7}] // Flatten
%Y Cf. A005425 (row sums), A100861 (scaling factors).
%Y Cf. A007318, A094305, A099174, A344565, A344678.
%K nonn,tabf
%O 0,5
%A _Peter Luschny_, Jun 03 2021
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