A094665 Another version of triangular array in A083061: triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, ...] where DELTA is the operator defined in A084938.

1, 0, 1, 0, 1, 3, 0, 4, 15, 15, 0, 34, 147, 210, 105, 0, 496, 2370, 4095, 3150, 945, 0, 11056, 56958, 111705, 107415, 51975, 10395, 0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 0, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490, 18918900, 2027025

Offset: 0

Author

Philippe Deléham, Jun 07 2004, Jun 12 2007

Keywords

• nonn
• tabl

Comments

Diagonals: A000007, A002105; A001147, A001880.

Define polynomials P(n,x) = x(2x+1)P(n-1,x+1) - 2x^2P(n-1,x), P(0,x) = 1. Sequence gives triangle read by rows, defined by P(n,x) = Sum_{k = 0..n} T(n,k)*x^k. - Philippe Deléham, Jun 20 2004

From Johannes W. Meijer, May 24 2009: (Start)

In A160464 we defined the coefficients of the ES1 matrix by ES1[2*m-1,n=1] = 2*eta(2*m-1) and the recurrence relation ES1[2*m-1,n] = ((2*n-2)/(2*n-1))*(ES1[2*m-1,n-1] - ES1[2*m-3,n-1]/(n-1)^2) for m the positive and negative integers and n >= 1. As usual eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. It is well-known that ES1[1-2*m,n=1] = (4^m-1)*(-bernoulli(2*m))/m for m >= 1. and together with the recurrence relation this leads to ES1[-1,n] = 0.5 for n >= 1.

We discovered that the n-th term of the row coefficients ES1[1-2*m,n] for m >= 1, can be generated with the rather simple polynomials RES1(1-2*m,n) = (-1)^(m+1)*ECGP(1-2*m, n)/2^m. This discovery was enabled by the recurrence relation for the RES1(1-2*m,n) which we derived from the recurrence relation for the ES1[2*m-1,n] coefficients and the fact that RES1(-1,n) = 0.5. The coefficients of the ECGP(1-2*m,n) polynomials led to this triangle and subsequently to triangle A083061. (End)

From David Callan, Jan 03 2011: (Start)

T(n,k) is the number of increasing 0-2 trees (A002105) on 2n edges in which the minimal path from the root has length k.

Proof. The number a(n,k) of such trees satisfies the recurrence a(0,0)=1, a(1,1)=1 and, counting by size of the subtree rooted at the smaller child of the root,

a(n,k) = Sum_{j=1..n-1} C(2n-1,j)*a(j,k-1)*a(n-1-j)

for 2<=k<=n, where a(n) = Sum_{k>=0} a(n,k) is the reduced tangent number A002105 (indexed from 0). The recurrence translates into the differential equation

F_x(x,y) = y*F(x,y)*G(x)

for the GF F(x,y) = Sum_{n,k>=0} a(n,k)x^(2n)/(2n)!*y^k, where G(x):=Sum_{n>=0} a(n)x^(2n+1)/(2n+1)! is known to be sqrt(2)*tan(x/sqrt(2)). The differential equation has solution F(x,y) = sec(x/sqrt(2))^(2y). (End)

Examples

			Triangle begins:
.1;
.0, 1;
.0, 1, 3;
.0, 4, 15, 15;
.0, 34, 147, 210, 105;
.0, 496, 2370, 4095, 3150, 945;
.0, 11056, 56958, 111705, 107415, 51975, 10395;
.0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135;
From _Johannes W. Meijer_, May 24 2009: (Start)
The first few ECGP(1-2*m,n) polynomials are: ECGP(-1,n) = 1; ECGP(-3,n) = n; ECGP(-5,n) = n + 3*n^2; ECGP(-7,n) = 4*n + 15*n^2+ 15*n^3 .
The first few RES1(1-2*m,n) are: RES1(-1,n) = (1/2)*(1); RES1(-3,n) = (-1/4)*(n); RES1(-5,n) = (1/8)*(n+3*n^2); RES1(-7,n) = (-1/16)*(4*n+15*n^2+15*n^3).
(End)
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• H. J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, 40(2):175-180, May 2002.

Crossrefs

Cf. A000364 A084938 A083061.

From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)

Cf. A160464, A083061 and A160468.

A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.

Appears in A162005, A162006 and A162007.

(End)

Programs

Maple

nmax:=7; imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2 * x^2 * T1(i-1, x)): dx:=degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx) od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do T(n+1, k+1) := A083061(n, k) od: od: T(0, 0):=1: for n from 1 to nmax do T(n, 0):=0 od: seq(seq(T(n, k), k=0..n), n=0..nmax);
# Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012

Mathematica

nmax = 8;
T[n_, k_] := SeriesCoefficient[Sec[x/Sqrt[2]]^(2y), {x, 0, 2n}, {y, 0, k}]* (2n)!;
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)

Formula

Sum_{k = 0..n} T(n, k) = A002105(n+1).

Sum_{k = 0..n} T(n, k)*2^(n-k) = A000364(n); Euler numbers.

Sum_{k = 0..n} T(n, k)*(-2)^(n-k) = 1.

RES1(1-2*m,n) = n^2*RES1(3-2*m,n)-n*(2*n+1)*RES1(3-2*m,n+1)/2 for m >= 2, with RES1(-1,n) = 0.5 for n >= 1. - Johannes W. Meijer, May 24 2009

G.f.: Sum_{n,k>=0} T(n,k)x^n/n!*y^k = sec(x/sqrt(2))^(2y).

Extensions

Term corrected by Johannes W. Meijer, Sep 23 2012