%I #42 Oct 29 2021 05:19:12
%S 1,0,1,0,1,3,0,4,15,15,0,34,147,210,105,0,496,2370,4095,3150,945,0,
%T 11056,56958,111705,107415,51975,10395,0,349504,1911000,4114110,
%U 4579575,2837835,945945,135135,0,14873104,85389132,197722980,244909665,178378200,77567490,18918900,2027025
%N 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.
%C Diagonals: A000007, A002105; A001147, A001880.
%C 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
%C From _Johannes W. Meijer_, May 24 2009: (Start)
%C 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.
%C 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)
%C From _David Callan_, Jan 03 2011: (Start)
%C 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.
%C 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,
%C a(n,k) = Sum_{j=1..n-1} C(2n-1,j)*a(j,k-1)*a(n-1-j)
%C 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
%C F_x(x,y) = y*F(x,y)*G(x)
%C 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)
%H Alois P. Heinz, <a href="/A094665/b094665.txt">Rows n = 0..140, flattened</a>
%H H. J. H. Tuenter, <a href="https://arxiv.org/abs/math/0606080">Walking into an absolute sum</a>, arXiv:math/0606080 [math.NT], 2006. Published version on <a href="http://www.fq.math.ca/Scanned/40-2/tuenter.pdf">Walking into an absolute sum</a>, The Fibonacci Quarterly, 40(2):175-180, May 2002.
%F Sum_{k = 0..n} T(n, k) = A002105(n+1).
%F Sum_{k = 0..n} T(n, k)*2^(n-k) = A000364(n); Euler numbers.
%F Sum_{k = 0..n} T(n, k)*(-2)^(n-k) = 1.
%F 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
%F G.f.: Sum_{n,k>=0} T(n,k)x^n/n!*y^k = sec(x/sqrt(2))^(2y).
%e Triangle begins:
%e .1;
%e .0, 1;
%e .0, 1, 3;
%e .0, 4, 15, 15;
%e .0, 34, 147, 210, 105;
%e .0, 496, 2370, 4095, 3150, 945;
%e .0, 11056, 56958, 111705, 107415, 51975, 10395;
%e .0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135;
%e From _Johannes W. Meijer_, May 24 2009: (Start)
%e 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 .
%e 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).
%e (End)
%p 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);
%p # _Johannes W. Meijer_, Jun 27 2009, revised Sep 23 2012
%t nmax = 8;
%t T[n_, k_] := SeriesCoefficient[Sec[x/Sqrt[2]]^(2y), {x, 0, 2n}, {y, 0, k}]* (2n)!;
%t Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 10 2018 *)
%Y Cf. A000364 A084938 A083061.
%Y From _Johannes W. Meijer_, May 24 2009 and Jun 27 2009: (Start)
%Y Cf. A160464, A083061 and A160468.
%Y A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
%Y Appears in A162005, A162006 and A162007.
%Y (End)
%K nonn,tabl
%O 0,6
%A _Philippe Deléham_, Jun 07 2004, Jun 12 2007
%E Term corrected by _Johannes W. Meijer_, Sep 23 2012