A172431 Even row Pascal-square read by antidiagonals.

1, 1, 2, 1, 4, 3, 1, 6, 10, 4, 1, 8, 21, 20, 5, 1, 10, 36, 56, 35, 6, 1, 12, 55, 120, 126, 56, 7, 1, 14, 78, 220, 330, 252, 84, 8, 1, 16, 105, 364, 715, 792, 462, 120, 9, 1, 18, 136, 560, 1365, 2002, 1716, 792, 165, 10

Offset: 1

Author

Mark Dols, Feb 02 2010

Keywords

• nonn
• tabl

Comments

Apart from signs identical to A053123. Mirror of A078812.

As a triangle, row n consists of the coefficients of Morgan-Voyce polynomial B(n,x); e.g., B(3,x)=x^3+6x^2+10x+4. As a triangle, rows 0 to 4 are as follows: 1 1...2 1...4...3 1...6...10...4 1...8...21...20...5 See A054142 for coefficients of Morgan-Voyce polynomial b(n,x).

Scaled version of A119900. - Philippe Deléham, Feb 24 2012

A172431 is jointly generated with A054142 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012

Subtriangle of the triangle given by (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012

Examples

			Array begins:
  1,  2,  3,  4,  5,  6, ...
  1,  4, 10, 20, 35, ...
  1,  6, 21, 56, ...
  1,  8, 36, ...
  1, 10, ...
  1, ...
  ...
Example:
Starting with 1, every entry is twice the one to the left minus the second one to the left, plus the one above.
For n = 9 the a(9) = 10 solution is 2*4 - 1 + 3.
From _Philippe Deléham_, Feb 24 2012: (Start)
Triangle T(n,k) begins:
  1;
  1,   2;
  1,   4,   3;
  1,   6,  10,   4;
  1,   8,  21,  20,   5;
  1,  10,  36,  56,  35,   6;
  1,  12,  55, 120, 126,  56,   7; (End)
From _Philippe Deléham_, Mar 22 2012: (Start)
(1, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, ...) begins:
  1;
  1,   0;
  1,   2,   0;
  1,   4,   3,   0;
  1,   6,  10,   4,   0;
  1,   8,  21,  20,   5,   0;
  1,  10,  36,  56,  35,   6,   0;
  1,  12,  55, 120, 126,  56,   7,   0; (End)
		

Links

• G. C. Greubel, Rows n = 1..100 of triangle, flattened

Crossrefs

Cf. A078812, A053123, A007318, A001906 (antidiagonals sums), A007685.

Cf. also A054142, A082985.

Programs

GAP

F:=Factorial;; Flat(List([1..15], n-> List([1..n], k-> Sum([0..Int((k-1)/2)], j-> (-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)) )))); # G. C. Greubel, Dec 15 2019

Magma

F:=Factorial; [ &+[(-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)): j in [0..Floor((k-1)/2)]]: k in [1..n], n in [1..15]]; // G. C. Greubel, Dec 15 2019

Maple

T := (n, k) -> simplify(GegenbauerC(k, n-k, 1)):
for n from 0 to 10 do seq(T(n,k), k=0..n-1) od; # Peter Luschny, May 10 2016

Mathematica

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A054142 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A172431 *)
(* Clark Kimberling, Mar 09 2012 *)
Table[GegenbauerC[k-1, n-k+1, 1], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Dec 15 2019 *)

PARI

T(n,k) = sum(j=0, (k-1)\2, (-1)^j*(n-j-1)!*2^(k-2*j-1)/(j!*(n-k)!*(k-2*j-1)!) );
for(n=1, 10, for(k=1, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 15 2019

Sage

[[gegenbauer(k-1, n-k+1, 1) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Dec 15 2019

Formula

As a decimal sequence: a(n)= 12*a(n-1)- a(n-2) with a(1)=1. [I interpret this remark as: 1, 12=1,2, 143=1,4,3, 1704=1,6,10,4,... taken from A004191 are decimals on the diagonal. - R. J. Mathar, Sep 08 2013]

As triangle T(n,k): T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2). - Philippe Deléham, Feb 24 2012

As DELTA-triangle T(n,k) with 0<=k<=n: G.f.: (1-y*x)^2/((1-y*x)^2-x). - Philippe Deléham, Mar 22 2012

T(n, k) = GegenbauerC(k, n-k, 1). - Peter Luschny, May 10 2016

As triangle T(n,k): Product_{k=1..n} T(n,k) = Product_{k=0..n-1} binomial(2*k,k) = A007685(n-1) for n >= 1. - Werner Schulte, Apr 26 2017

As triangle T(n,k) with 1 <= k <= n: T(n,k) = binomial(2*n-k, k-1). - Paul Weisenhorn, Nov 25 2019