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A094665
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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.
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16
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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, 191100, 4114110, 4579575, 2837835, 945945, 135135, 0, 14873104, 85389132, 197722980, 244909665
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,6
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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 . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jun 20 2004
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), 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, 2, 3, .. . 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, 2, .. and together with the recurrence relation this leads to ES1[-1,n] = 0.5 for n = 1, 2, .. .
We discovered that the n-th term of the row coefficients ES1[1-2*m,n] for m = 1, 2, 3, .., 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)
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}binomial(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).
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LINKS
| H.J.H. Tuenter, Walking into an absolute sum
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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.
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)
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, 3, .., with RES1(-1,n) = 0.5 for n = 1, 2, .. .
(End)
G.f. Sum_{n,k>=0}T(n,k)x^n/n!*y^k = sec(x/sqrt(2))^(2y)
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EXAMPLE
| 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, 191100, 4114110, 4579575, 2837835, 945945, 135135;
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), 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)
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MAPLE
| Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 27 2009: (Start)
restart; nmax:=8; imax := nmax: i:=0: 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(j)*(x+1)^(j), j=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: t:=0: for n from 0 to nmax do for k from 0 to n do a(t):= T(n, k): t:= t+1: od: od: seq(a(n), n=0..t-1);
(End)
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CROSSREFS
| Cf. A000364 A084938 A083061.
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)
Cf. A160464, A083061 and A160468.
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
(End)
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 27 2009: (Start)
Appears in A162005, A162006 and A162007.
(End)
Sequence in context: A068627 A074171 A180657 * A052439 A143073 A154725
Adjacent sequences: A094662 A094663 A094664 * A094666 A094667 A094668
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KEYWORD
| nonn,tabl
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AUTHOR
| DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jun 07 2004, Jun 12 2007
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