A344911 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.
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, 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, 7, 1, 21, 105, 210, 210, 105, 21, 105, 315, 315, 105, 105, 105
Offset: 0
Examples
The triangle begins: [0] [ 1 ] [1] [ 1, 1 ] [2] [ 1, 2, 1 ][ 1 ] [3] [ 1, 3, 3, 1 ][ 3, 3 ] [4] [ 1, 4, 6, 4, 1 ][ 6, 12, 6 ][ 3 ] [5] [ 1, 5, 10, 10, 5, 1 ][ 10, 30, 30, 10 ][ 15, 15 ] [6] [ 1, 6, 15, 20, 15, 6, 1 ][ 15, 60, 90, 60, 15 ][ 45, 90, 45][ 15 ] . With the notations in the comment row 7 concatenates: B(7, 0).C(7) = 1.[1, 7, 21, 35, 35, 21, 7, 1] = [1, 7, 21, 35, 35, 21, 7, 1], B(7, 1).C(5) = 21.[1, 5, 10, 10, 5, 1] = [21, 105, 210, 210, 105, 21], B(7, 2).C(3) = 105.[1, 3, 3, 1] = [105, 315, 315, 105], B(7, 3).C(1) = 105.[1, 1] = [105, 105]. . 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 + 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.
Crossrefs
Programs
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Maple
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): seq(seq(subs(x = 1, y = 1, m), m = [op(P(n))]), n = 0..7); # Alternatively, without polynomials: B := (n, k) -> binomial(n, 2*k)*doublefactorial(2*k-1): C := n -> seq(binomial(n, j), j=0..n): seq(seq(B(n, k)*C(n-2*k), k = 0..n/2), n = 0..7); # Based on the e.g.f. of the polynomials: T := proc(numofrows) local gf, ser, n, m; gf := exp(t^2/2)*exp(t*(x + y)); ser := series(gf, t, numofrows+1); for n from 0 to numofrows do [op(sort(n!*expand(coeff(ser, t, n))))]; print(seq(subs(x=1, y=1, m), m = %)) od end: T(7);
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Mathematica
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}]; DegLexList[p_] := MonomialList[p, {x, y}, "DegreeLexicographic"] /. x->1 /. y->1; Table[DegLexList[P[n]], {n, 0, 7}] // Flatten
Formula
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.
Comments