A162007 -id:A162007 - cp's OEIS frontend

A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

1, 2, 1, 4, 10, 1, 8, 60, 36, 1, 16, 296, 516, 116, 1, 32, 1328, 5168, 3508, 358, 1, 64, 5664, 42960, 64240, 21120, 1086, 1, 128, 23488, 320064, 900560, 660880, 118632, 3272, 1, 256, 95872, 2225728

Offset: 0

Johannes W. Meijer, Feb 20 2009, Jun 24 2009

Essentially the same as A185411. Row reverse of A185410. - Peter Bala, Jul 24 2012
The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.
From Peter Bala, Apr 03 2011: (Start)
Let D be the differential operator 2*x*d/dx. The row polynomials of this table come from repeated application of the operator D to the function g(x) = 1/sqrt(1 - x). For example,
  D(g) = x*g^3
  D^2(g) = x*(2 + x)*g^5
  D^3(g) = x*(4 + 10*x + x^2)*g^7
  D^4(g) = x*(8 + 60*x + 36*x^2 + x^3)*g^9.
Thus this triangle is analogous to the triangle of Eulerian numbers A008292, whose row polynomials come from the  repeated application of the operator x*d/dx to the function 1/(1 - x). (End)

			The first few rows of the triangle are:
  [1]
  [2, 1]
  [4, 10, 1]
  [8, 60, 36, 1]
  [16, 296, 516, 116, 1]
The first few P(z;n) are:
  P(z; n=0) = 1
  P(z; n=1) = 2 + z
  P(z; n=2) = 4 + 10*z + z^2
  P(z; n=3) = 8 + 60*z + 36*z^2 + z^3
The first few SF(z;n) are:
  SF(z; n=0) = (1/2)*(1)/(1-z)^(3/2);
  SF(z; n=1) = (1/4)*(2+z)/(1-z)^(5/2);
  SF(z; n=2) = (1/8)*(4+10*z+z^2)/(1-z)^(7/2);
  SF(z; n=3) = (1/16)*(8+60*z+36*z^2+z^3)/(1-z)^(9/2);
In the Savage-Viswanathan paper, the coefficients appear as
  1;
  1,    2;
  1,   10,     4;
  1,   36,    60,     8;
  1,  116,   516,   296,    16;
  1,  358,  3508,  5168,  1328,   32;
  1, 1086, 21120, 64240, 42960, 5664, 64;
  ...

D. H. Lehmer, Interesting Series Involving the Central Binomial Coefficient, Am. Math. Monthly 92 (1985) 449-457, Polynomial V in eq (17). [R. J. Mathar, Feb 24 2009]
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, arXiv: 1204.4963v3 [math.CO], 2012.
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11
S.-M. Ma and T. Mansour, The 1/k-Eulerian polynomials and k-Stirling permutations, arXiv preprint arXiv:1409.6525 [math.CO], 2014.
S.-M. Ma and Y.-N. Yeh, Stirling permutations, cycle structures of permutations and perfect matchings, arXiv preprint arXiv:1503.06601 [math.CO], 2015.
Carla D. Savage and Gopal Viswanathan, The 1/k-Eulerian polynomials, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012).
Eric Weisstein's World of Mathematics, Dirichlet Eta Function

A142963 and this sequence can be mapped onto the A156920 triangle.
FP1 sequences A000340, A156922, A156923, A156924.
FP2 sequences A050488, A142965, A142966, A142968.
Appears in A162005, A000182, A162006 and A162007.
Cf. A186695, A187075.
Cf. A185410 (row reverse), A185411.

A156919 := proc(n,m) if n=m then 1; elif m=0 then 2^n ; elif m<0 or m>n then 0; else 2*(m+1)*procname(n-1,m)+(2*n-2*m+1)*procname(n-1,m-1) ; end if; end proc: seq(seq(A156919(n,m), m=0..n), n=0..7); # R. J. Mathar, Feb 03 2011

g[0] = 1/Sqrt[1-x]; g[n_] := g[n] = 2x*D[g[n-1], x]; p[n_] := g[n] / g[0]^(2n+1) // Cancel; row[n_] := CoefficientList[p[n], x] // Rest; Table[row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Aug 09 2012, after Peter Bala *)
Flatten[Table[Rest[CoefficientList[Nest[2 x D[#, x] &, (1 - x)^(-1/2), k] (1 - x)^(k + 1/2), x]], {k, 10}]] (* Jan Mangaldan, Mar 15 2013 *)

SF(z; n) = Sum_{m >= 1} m^(n-1)*4^(-m)*z^(m-1)*Gamma(2*m+1)/(Gamma(m)^2) = P(z;n) / (2^(n+1)*(1-z)^((2*n+3)/2)) for n >= 0. The polynomials P(z;n) = Sum_{k = 0..n} a(k)*z^k generate the a(n) sequence.
If we write the sequence as a triangle the following relation holds: T(n,m) = (2*m+2)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 2^n and T(n,n) = 1, n >= 0 and 0 <= m <= n.
G.f.: 1/(1-xy-2x/(1-3xy/(1-4x/(1-5xy/(1-6x/(1-7xy/(1-8x/(1-... (continued fraction). - Paul Barry, Jan 26 2011
From Peter Bala, Apr 03 2011 (Start)
E.g.f.: exp(z*(x + 2)) * (1 - x)/(exp(2*x*z) - x*exp(2*z))^(3/2) = Sum_{n >= 0} P(x,n)*z^n/n! = 1 + (2 + x)*z + (4 + 10*x + x^2)*z^2/2! + (8 + 60*x + 36*x^2 + x^3)*z^3/3! + ... .
Explicit formula for the row polynomials:
P(x,n-1) = Sum_{k = 1..n} 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k)*x^(k-1)*(1 - x)^(n-k).
The polynomials x*(1 + x)^n * P(x/(x + 1),n) are the row polynomials of A187075.
The polynomials x^(n+1) * P((x + 1)/x,n) are the row polynomials of A186695.
Row sums are A001147(n+1). (End)
Sum_{k = 0..n} (-1)^k*T(n,k) = (-1)^binomial(n,2)*A012259(n+1). - Johannes W. Meijer, Sep 27 2011

Minor edits from Johannes W. Meijer, Sep 27 2011

A162005 The EG1 triangle.

Original entry on oeis.org

1, 2, 1, 16, 28, 1, 272, 1032, 270, 1, 7936, 52736, 36096, 2456, 1, 353792, 3646208, 4766048, 1035088, 22138, 1, 22368256, 330545664, 704357760, 319830400, 27426960, 199284, 1, 1903757312, 38188155904, 120536980224, 93989648000

Offset: 1

Johannes W. Meijer, Jun 27 2009, Jul 02 2009, Aug 31 2009

We define the EG1 matrix by EG1[2m-1,1] = 2*eta(2m-1) and the recurrence relation EG1[2m-1,n] = EG1[2m-1,n-1] - EG1[2m-3,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .., with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. For the EG2[2m,n] coefficients see A008955.
The n-th term of the row coefficients EG1[1-2*m,n] for m = 1, 2, .., can be generated with REG1(1-2*m,n) = (-1)^(m+1)*2^(1-m)*ECGP(1-2*m, n)*(1/n)*4^(-n)*(2*n)!/((n-1)!)^2 . For information about the ECGP polynomials see A094665 and the examples below.
We define the o.g.f.s. of the REG1(1-2*m,n) by GFREG1(z,1-2*m) = sum(REG1(1-2*m,n)* z^(n-1), n=1..infinity) for m = 1, 2, .., with GFREG1(z,1-2*m) = (-1)^(m+1)* RG(z,1-2*m)/ (2^(2*m-1)*(1-z)^((2*m+1)/2)). The RG(z,1-2m) polynomials led to the EG1 triangle.
We used the coefficients of the A156919 and A094665 triangles to determine those of the EG1 triangle, see the Maple program. The A156919 triangle gives information about the sums SF(p) = sum(n^(p-1)*4^(-n)*z^(n-1)*(2*n)!/((n-1)!)^2, n=1..infinity) for p= 0, 1, 2, .. .
Contribution from Johannes W. Meijer, Nov 23 2009: (Start)
The EG1 matrix is related to the ED2 array A167560 because sum(EG1(2*m-1,n)*z^(2*m-1), m=1..infinity) = ((2*n-1)!/(4^(n-1)*(n-1)!^2))*int(sinh(y*(2*z))/cosh(y)^(2*n),y=0..infinity).
(End)
Appears to equal triangle A322230 with rows read in reverse order. Triangle A322230 describes the e.g.f. S(x,k) = Integral C(x,k)*D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2*S(x,k)^2 = 1. - Paul D. Hanna, Dec 22 2018
Appears to equal triangle A325220, which has e.g.f. S(x,k) = -i * sn( i * Integral C(x,k) dx, k) such that C(x,k) = cn( i * Integral C(x,k) dx, k), where sn(x,k) and cn(x,k) are Jacobi Elliptic functions. - Paul D. Hanna, Apr 13 2019

			The first few rows of the EG1 triangle are :
[1]
[2, 1]
[16, 28, 1]
[272, 1032, 270, 1]
The first few RG(z,1-2*m) polynomials are:
RG(z,-1) = 1
RG(z,-3) = 2+z
RG(z,-5) = 16+28*z+z^2
RG(z,-7) = 272+1032*z+270*z^2+z^3
The first few GFREG1(z,1-2*m) are:
GFREG1(z,-1) = (1)*(1)/(2*(1-z)^(3/2))
GFREG1(z,-3) = (-1)*(2+z)/(2^3*(1-z)^(5/2))
GFREG1(z,-5) = (1)*(16+28*z+z^2)/( 2^5*(1-z)^(7/2))
GFREG1(z,-7) = (-1)*(272+1032*z+270*z^2+z^3)/(2^7*(1-z)^(9/2))
The first few REG1(1-2*m,n) are:
REG1(-1,n) = (1/1)*(1)*(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-3,n) = (-1/2)*(n) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-5,n) = (1/4) *(n+3*n^2) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-7,n) = (-1/8)*(4*n+15*n^2+15*n^3) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
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

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.

A079484 equals the row sums.
A000182 (ZAG numbers), A162006 and A162007 equal the first three left hand columns.
A000012, A004004 (2x), A162008, A162009 and A162010 equal the first five right hand columns.
Related to A094665, A083061 and A156919 (DEF triangle).
Cf. A161198 [(1-x)^((-1-2*n)/2)], A008955 (EG2[2m, n])
Cf. A167560 (ED2 array).
Cf. A322230 (reversed rows), A325220.

nmax:=7; mmax := nmax: 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 A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1)*A156919(n-1, m-1) end do end do: for n from 0 to nmax do SF(n) := sum(A156919(n, k1)*z^k1, k1=0..n)/(2^(n+1)*(1-z)^((2*n+3)/2)) od: GFREG1(z, -1) := A156919(0, 0)*A094665 (0, 0) / (2*(1-z)^(3/2)): for m from 2 to nmax do GFREG1(z, 1-2*m) := simplify((-1)^(m+1)*2^(1-m)* sum(A094665(m-1, k2)*SF(k2), k2=1..m-1)) od: for m from 1 to mmax do g(m) := sort((numer ((-1)^(m+1)* GFREG1(z, 1-2*m))), ascending) od: for n from 1 to nmax do for m from 1 to n do a(n, m) := abs(coeff(g(n), z, m-1)) od: od: seq(seq(a(n, m), m=1..n), n=1..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

A different form of the recurrence relation is EG1[1-2*m,n] = (EG1[3-2*m,n]-EG1[3-2*m,n+1])* (n^2) for m = 2, 3, .., with EG1[ -1,n] = (1/n)*4^(-n)*((2*n)!/(n-1)!^2).

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.

Original entry on oeis.org

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

Philippe Deléham, Jun 07 2004, Jun 12 2007

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)

			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)

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.

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)

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

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 *)

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).

Term corrected by Johannes W. Meijer, Sep 23 2012

A083061 Triangle of coefficients of a companion polynomial to the Gandhi polynomial.

Original entry on oeis.org

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

Offset: 0

Hans J. H. Tuenter, Apr 19 2003

This polynomial arises in the setting of a symmetric Bernoulli random walk and occurs in an expression for the even moments of the absolute distance from the origin after an even number of timesteps. The Gandhi polynomial, sequence A036970, occurs in an expression for the odd moments.
When formatted as a square array, first row is A002105, first column is A001147, second column is A001880.
Another version of the 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, 9, ...] = 1; 0, 1; 0, 1, 3; 0, 4, 15, 15; 0, 34, 147, 210, 105; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 07 2004
In A160464 we defined the coefficients of the ES1 matrix. Our discovery that the n-th term of the row coefficients ES1[1-2*m,n] for m>=1, can be generated with rather simple polynomials led to triangle A094665 and subsequently to this one. - Johannes W. Meijer, May 24 2009
Related to polynomials defined in A160485 by a shift of +-1/2 and scaling by a power of 2. - Richard P. Brent, Jul 15 2014

			Triangle starts (with an additional first column 1,0,0,...):
[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]

R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
Marc Joye, Pascal Paillier and Berry Schoenmakers, On Second-Order Differential Power Analysis, in Cryptographic Hardware and Embedded Systems-CHES 2005, editors: Josyula R. Rao and Berk Sunar, Lecture Notes in Computer Science 3659 (2005) 293-308, Springer-Verlag.
H. J. H. Tuenter, Walking into an absolute sum, The Fibonacci Quarterly, 40 (2002), 175-180.

Cf. A036970, A160485.
From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
Cf. A160464, A094665 and A160468.
A002105 equals the row sums (n>=2) and the first left hand column (n>=1).
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

imax := 6;
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
        a(i, j) := coeff(T1(i, x), x, j)
    od:
od:
seq(seq(a(i, j), j = 0..i), i = 0..imax);
# Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012

b[0, 0] = 1;
b[n_, k_] := b[n, k] = Sum[2^j*(Binomial[k + j, 1 + j] + Binomial[k + j + 1, 1 + j])*b[n - 1, k - 1 + j], {j, Max[0, 1 - k], n - k}];
a[0, 0] = 1;
a[n_, k_] := b[n, k]/2^(n - k);
Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 19 2018, after Philippe Deléham *)

# uses[fr2_row from A088874]
A083061_row = lambda n: [(-1)^(n-k)*m*2^(-n+k) for k,m in enumerate(fr2_row(n))]
for n in (0..7): print(A083061_row(n)) # Peter Luschny, Sep 19 2017

Let T(i, x)=(2x+1)(x+1)T(i-1, x+1)-2x^2T(i-1, x), T(0, x)=1; so that T(1, x)=1+3x; T(2, x)=4+15x+15x^2; T(3, x)=34+147x+210x^2+105x^3, etc. Then the (i, j)-th entry in the table is the coefficient of x^j in T(i, x).

a(n, k)*2^(n-k) = A085734(n, k). - Philippe Deléham, Feb 27 2005

A162006 Second left hand column of the EG1 triangle A162005.

Original entry on oeis.org

1, 28, 1032, 52736, 3646208, 330545664, 38188155904, 5488365862912, 961530104709120, 201865242068910080, 50052995352723193856, 14476381898608390176768, 4831399425299156001882112

Offset: 2

Johannes W. Meijer, Jun 27 2009

Second left hand column of the EG1 triangle A162005.
Other left hand columns are A000182 and A162007.
Related to A094665, A083061 and A156919.
A000079 and A036289 appear in the Maple program.

nmax := 14; mmax := nmax: 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 A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1) * A156919(n-1, m-1) end do end do: m:=2; for n from m to nmax do a(n, m) := sum((-1)^(m-p1-1)*sum(2^(n-q-1)*binomial(n-q-1, m-p1-1) * A094665(n-1, q) * A156919(q, p1), q=1..n-m+p1), p1=0..m-1) od: seq(a(n, m), n = m..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

a(n) = sum((-1)^(m-p-1)*sum(2^(n-q-1)*binomial(n-q-1,m-p-1)*A094665(n-1,q)*A156919(q,p),q=1..n-m+p), p=0..m-1) with m = 2.

A100381 a(n) = 2^n*binomial(n,2).

Original entry on oeis.org

0, 0, 4, 24, 96, 320, 960, 2688, 7168, 18432, 46080, 112640, 270336, 638976, 1490944, 3440640, 7864320, 17825792, 40108032, 89653248, 199229440, 440401920, 968884224, 2122317824, 4630511616, 10066329600, 21810380800, 47110422528

Offset: 0

Jorge Coveiro, Dec 30 2004

From Enrique Navarrete, Jun 13 2023: (Start)
a(n) is the number of ways to partition the set [n]={1,2,...,n} into two sets S,T and select 2 elements in total (from either S or T or both).
Example. For n=4, sample partitions are given (where S(i),T(j) means i elements are selected from S, j elements are selected from T):
S={ }, T={1,2,3,4}: partition [4] in 1 way, S(0),T(2) (6 ways);
S={1}, T={2,3,4}: partition [4] in 4 such ways, S(1),T(1) or S(0),T(2) (24 ways);
S={1,2}, T={3,4}: partition [4], in such 6 ways, S(1),T(1) or S(0),T(2) or S(2),T(0) (36 ways);
S={1,2,3}, T={4}: partition [4] in 4 such ways, S(1),T(1) or S(2),T(0) (24 ways);
S={1,2,3,4}, T={ }: partition [4] in 1 way, S(2),T(0) (6 ways). (End)

Jolley, Summation of Series, Dover (1961), eq (214) page 40.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
The Ramanujan Machine, Using algorithms to discover new mathematics.
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A001788, A001815, A162007, A244009.

Maple
```
seq(2^n*binomial(n,2),n=0..20);
```

Range[0,20]! CoefficientList[Series[2x^2 Exp[2x],{x,0,20}],x]
Table[2^n Binomial[n,2],{n,0,30}] (* or *) LinearRecurrence[{6,-12,8},{0,0,4},30] (* Harvey P. Dale, Aug 15 2020 *)

a(n)=binomial(n,2)<Charles R Greathouse IV, Oct 16 2015

Sum_{n>=2} 1/a(n) = 1 - log(2) = 0.3068528.... - Graeme McRae, Jul 28 2006
a(n) = Sum_{k=0..n} k*2^k = 2*A001815(n). - Zerinvary Lajos, Oct 09 2006
E.g.f.: 2*x^2*exp(2x).
a(n) = 4*A001788(n-1). - Johannes W. Meijer, Jun 27 2009
Sum_{j=1..k} (j+2)/a(j+1) = 1 - 1/((k+1)*2^k). [Jolley]
G.f.: -4*x^2 / (2*x-1)^3. - R. J. Mathar, Oct 05 2011
Sum_{n>=2} (-1)^n/a(n) = 3*log(3/2) - 1. - Amiram Eldar, Jul 20 2020
From Peter Bala Mar 04 2024: (Start)
Sum_{k = 2..n+2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/(3*n + 4)))).
Sum_{k = 2..n+2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/(n + 4)))).
Letting n -> oo in the above gives the continued fraction representations
1 - log(2) = Sum_{k >= 2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/((3*n + 4) - ... )))) (an equivalent continued fraction for 1 - log(2) was conjectured by the Ramanujan machine) and
3*log(3/2) - 1 = Sum_{k >= 2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/((n + 4) + ... )))). (End)

A322230 E.g.f.: S(x,k) = Integral C(x,k)D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2S(x,k)^2 = 1, as a triangle of coefficients read by rows.

Original entry on oeis.org

1, 1, 2, 1, 28, 16, 1, 270, 1032, 272, 1, 2456, 36096, 52736, 7936, 1, 22138, 1035088, 4766048, 3646208, 353792, 1, 199284, 27426960, 319830400, 704357760, 330545664, 22368256, 1, 1793606, 702812568, 18598875760, 93989648000, 120536980224, 38188155904, 1903757312, 1, 16142512, 17753262208, 1002968825344, 10324483102720, 28745874079744, 24060789342208, 5488365862912, 209865342976, 1, 145282674, 445736371872, 51882638754240, 1013356176688128, 5416305638467584, 9498855414644736, 5590122715250688, 961530104709120, 29088885112832

Offset: 0

Paul D. Hanna, Dec 14 2018

Equals a row reversal of triangle A325220.
Appears to be a row reversal of EG1 triangle A162005, which has other formulas.
Compare to sn(x,k) = Integral cn(x,k)*dn(x,k) dx, where sn(x,k), cn(x,k), and dn(x,k) are Jacobi elliptic functions (see triangle A060628).
Compare also to Michael Pawellek's generalized elliptic functions.

			E.g.f.: S(x,k) = x + (2*k^2 + 1)*x^3/3! + (16*k^4 + 28*k^2 + 1)*x^5/5! + (272*k^6 + 1032*k^4 + 270*k^2 + 1)*x^7/7! + (7936*k^8 + 52736*k^6 + 36096*k^4 + 2456*k^2 + 1)*x^9/9! + (353792*k^10 + 3646208*k^8 + 4766048*k^6 + 1035088*k^4 + 22138*k^2 + 1)*x^11/11! + (22368256*k^12 + 330545664*k^10 + 704357760*k^8 + 319830400*k^6 + 27426960*k^4 + 199284*k^2 + 1)*x^13/13! + ...
such that C(x,k)^2 - S(x,k)^2 = 1.
This triangle of coefficients T(n,j) of x^(2*n+1)*k^(2*j)/(2*n+1)! in e.g.f. S(x,k) begins:
1;
1, 2;
1, 28, 16;
1, 270, 1032, 272;
1, 2456, 36096, 52736, 7936;
1, 22138, 1035088, 4766048, 3646208, 353792;
1, 199284, 27426960, 319830400, 704357760, 330545664, 22368256;
1, 1793606, 702812568, 18598875760, 93989648000, 120536980224, 38188155904, 1903757312;
1, 16142512, 17753262208, 1002968825344, 10324483102720, 28745874079744, 24060789342208, 5488365862912, 209865342976; ...
RELATED SERIES.
The related series C(x,k), where C(x,k)^2 - S(x,k)^2 = 1, starts
C(x,k) = 1 + x^2/2! + (8*k^2 + 1)*x^4/4! + (136*k^4 + 88*k^2 + 1)*x^6/6! + (3968*k^6 + 6240*k^4 + 816*k^2 + 1)*x^8/8! + (176896*k^8 + 513536*k^6 + 195216*k^4 + 7376*k^2 + 1)*x^10/10! + (11184128*k^10 + 51880064*k^8 + 39572864*k^6 + 5352544*k^4 + 66424*k^2 + 1)*x^12/12! + (951878656*k^12 + 6453433344*k^10 + 8258202240*k^8 + 2458228480*k^6 + 139127640*k^4 + 597864*k^2 + 1)*x^14/14! + ...
The related series D(x,k), where D(x,k)^2 - k^2*S(x,k)^2 = 1, starts
D(x,k) = 1 + k^2*x^2/2! + (5*k^4 + 4*k^2)*x^4/4! + (61*k^6 + 148*k^4 + 16*k^2)*x^6/6! + (1385*k^8 + 6744*k^6 + 2832*k^4 + 64*k^2)*x^8/8! + (50521*k^10 + 410456*k^8 + 383856*k^6 + 47936*k^4 + 256*k^2)*x^10/10! + (2702765*k^12 + 32947964*k^10 + 54480944*k^8 + 17142784*k^6 + 780544*k^4 + 1024*k^2)*x^12/12! + (199360981*k^14 + 3402510924*k^12 + 8760740640*k^10 + 5199585280*k^8 + 686711040*k^6 + 12555264*k^4 + 4096*k^2)*x^14/14! + ...

Paul D. Hanna, Table of n, a(n) for n = 0..860 terms of this triangle read by rows 0..40.

Cf. A000182 (diagonal), A162006, A162007, A162008, A162009, A162010.
Cf. A322231 (C), A322232 (D).
Cf. A325220 (row reversal), A162005.

N=10;
{S=x;C=1;D=1; for(i=1,2*N, S = intformal(C*D^2 +O(x^(2*N+1))); C = 1 + intformal(S*D^2); D = 1 + k^2*intformal(S*C*D));}
for(n=0,N, for(j=0,n, print1( (2*n+1)!*polcoeff(polcoeff(S,2*n+1,x),2*j,k),", ")) ;print(""))

E.g.f. S = S(x,k) = Sum_{n>=0} Sum_{j=0..n} T(n,j) * x^(2*n+1) * k^(2*j) / (2*n+1)!, along with related series C = C(x,k) and D = D(x,k), satisfies:
(1a) S = Integral C*D^2 dx.
(1b) C = 1 + Integral S*D^2 dx.
(1c) D = 1 + k^2 * Integral S*C*D dx.
(2a) C^2 - S^2 = 1.
(2b) D^2 - k^2*S^2 = 1.
(3a) C + S = exp( Integral D^2 dx ).
(3b) D + k*S = exp( k * Integral C*D dx ).
(4a) S = sinh( Integral D^2 dx ).
(4b) S = sinh( k * Integral C*D dx ) / k.
(4c) C = cosh( Integral D^2 dx ).
(4d) D = cosh( k * Integral C*D dx ).
(5a) d/dx S = C*D^2.
(5b) d/dx C = S*D^2.
(5c) d/dx D = k^2 * S*C*D.
From Paul D. Hanna, Mar 31 2019, Apr 20 2019 (Start):
Given sn(x,k), cn(x,k), and dn(x,k) are Jacobi elliptic functions, with i^2 = -1, k' = sqrt(1-k^2), then
(6a) S = -i * sn( i * Integral D dx, k),
(6b) C = cn( i * Integral D dx, k),
(6c) D = dn( i * Integral D dx, k).
(7a) S = sc( Integral D dx, k') = sn(Integral D dx, k')/cn(Integral D dx, k'),
(7b) C = nc( Integral D dx, k') = 1/cn(Integral D dx, k'),
(7c) D = dc( Integral D dx, k') = dn(Integral D dx, k')/cn(Integral D dx, k'). (End)
Row sums equal (2*n+1)!*(2*n)!/(n!^2*4^n) = A079484(n), the product of two consecutive odd double factorials.
Main diagonal equals A000182, the tangent numbers.

cp's OEIS Frontend

A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

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A162005 The EG1 triangle.

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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.

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A083061 Triangle of coefficients of a companion polynomial to the Gandhi polynomial.

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A162006 Second left hand column of the EG1 triangle A162005.

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A100381 a(n) = 2^n*binomial(n,2).

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A322230 E.g.f.: S(x,k) = Integral C(x,k)*D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2*S(x,k)^2 = 1, as a triangle of coefficients read by rows.

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A322230 E.g.f.: S(x,k) = Integral C(x,k)D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2S(x,k)^2 = 1, as a triangle of coefficients read by rows.