A281000: Triangle read by rows: T(n,k) = binomial(2*n+1, 2*k+1)*binomial(2*n-2*k, n-k)/(n+1-k) for 0 <= k <= n.

OEIS matches ?

• A099250: row sums, 15 terms agree OEIS nonn (row-sum match: finite agreement only) · view at the OEIS · generated b-file 200 terms from n = 0 ?
• A281000: triangle, 55 terms agree, 10 full rows row-aligned (tabl) OEIS nonn in polytool catalog · view at the OEIS · generated b-file 990 terms from n = 0 ?

A generated b-file is computed from the node's definition along the row alignment of its OEIS link and starts at the entry's offset. It is an identification on a finite prefix, not a proof that the node and the OEIS entry agree for every n.

OEIS submission text ?

Fields to paste into an OEIS submission: the example %e, the formula %F and programs %t (Mathematica) and %o (PARI/GP) that list the same rows. For an OEIS link the row numbers, the offset and the orientation are those of the entry, as matched; the matched terms were compared with the published ones, which identifies the node with the entry on a finite prefix, not for every n. The PARI/GP program is not machine-checked here.

Triangle begins:
   1:     1;
   2:     3,       1;
   3:    10,      10,       1;
   4:    35,      70,      21,       1;
   5:   126,     420,     252,      36,       1;
   6:   462,    2310,    2310,     660,      55,       1;
   7:  1716,   12012,   18018,    8580,    1430,      78,      1;
   8:  6435,   60060,  126126,   90090,   25025,    2730,    105,    1;
   9: 24310,  291720,  816816,  816816,  340340,   61880,   4760,  136,   1;
  10: 92378, 1385670, 4988412, 6651216, 3879876, 1058148, 135660, 7752, 171, 1;

The row polynomials satisfy (2*n^2+n)*P(n,x) = (8*n^2-2+(6*n^2-8*n+2)*x)*P(n-1,x) + ((16*n-4)*x-(4*n-1)*x^2)*d/dx P(n-1,x) for n >= 2, with P(1,x) = 1.

P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((8*n^2 - 2 + (6*n^2 - 8*n + 2)*x)*P[n - 1, x] + ((16*n - 4)*x - (4*n - 1)*x^2)*D[P[n - 1, x], x])/(2*n^2 + n)]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten

P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = ((8*n^2 - 2 + (6*n^2 - 8*n + 2)*x)*v[i - 1] + ((16*n - 4)*x - (4*n - 1)*x^2)*deriv(v[i - 1], x))/(2*n^2 + n)); v};
row(n, p) = {my(c = Vecrev(p)); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};
T(10)

Triangle begins:
  0:     1;
  1:     3,       1;
  2:    10,      10,       1;
  3:    35,      70,      21,       1;
  4:   126,     420,     252,      36,       1;
  5:   462,    2310,    2310,     660,      55,       1;
  6:  1716,   12012,   18018,    8580,    1430,      78,      1;
  7:  6435,   60060,  126126,   90090,   25025,    2730,    105,    1;
  8: 24310,  291720,  816816,  816816,  340340,   61880,   4760,  136,   1;
  9: 92378, 1385670, 4988412, 6651216, 3879876, 1058148, 135660, 7752, 171, 1;

The row polynomials satisfy (2*n^2+5*n+3)*P(n,x) = (8*n^2+16*n+6+(6*n^2+4*n)*x)*P(n-1,x) + ((16*n+12)*x-(4*n+3)*x^2)*d/dx P(n-1,x) for n >= 1, with P(0,x) = 1.

P[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((8*n^2 + 16*n + 6 + (6*n^2 + 4*n)*x)*P[n - 1, x] + ((16*n + 12)*x - (4*n + 3)*x^2)*D[P[n - 1, x], x])/(2*n^2 + 5*n + 3)]];
Table[PadRight[CoefficientList[P[n, x], x], n + 1], {n, 0, 9}] // Flatten

P(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = ((8*n^2 + 16*n + 6 + (6*n^2 + 4*n)*x)*v[i - 1] + ((16*n + 12)*x - (4*n + 3)*x^2)*deriv(v[i - 1], x))/(2*n^2 + 5*n + 3)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};
T(9)

Proved properties

• Real-rooted. Every row P_n, n >= 1, is zero or real-rooted with all roots in (-inf, 0)
  Every polynomial of the sequence has only real roots, inside the stated interval.
  by the cited steps: e-8b9bb9a97ecefb47
• Coefficients of one sign. Every nonzero row, n >= 1: the coefficients of P_n have one sign
  All coefficients of each polynomial have the same sign (after x -> -x when alternating).
  derived from the cited steps by a stated theorem: e-8b9bb9a97ecefb47
• Hadamard kernel (Malo). Rows P_n, n >= 1, are Hadamard kernels (preserve real-rootedness)
  Multiplying the coefficients of any real-rooted polynomial termwise by those of a row (Hadamard product) gives a real-rooted polynomial again.
  derived from the cited steps by a stated theorem: e-8b9bb9a97ecefb47
• Interlacing of consecutive rows. Rows P_(m-1), P_m interlace (weak, previous interlaces next) for every m >= 2
  The roots of consecutive polynomials alternate on the real line.
  by the cited steps: e-8b9bb9a97ecefb47

Proof chains

1 incoming proof step. Each step is expanded back to the roots of the proof DAG, the theorem checks of a node's own rule. JSON

• theorem_check liu_wang e-8b9bb9a97ecefb47 root: theorem check

  Liu–Wang theorem [LW07, Thm 2.1, Cor 2.4] ?

Related sequences ?

No candidate equalities or related sequences.

References

1. [LW07] Lily L. Liu and Yi Wang. A unified approach to polynomial sequences with only real zeros. Advances in Applied Mathematics, 38(4):542–560, May 2007. doi:10.1016/j.aam.2006.02.003