cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 14 results. Next

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

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Author

Philippe Deléham, Jun 07 2004, Jun 12 2007

Keywords

Comments

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)
		

Crossrefs

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

A160470 Third right hand column of triangles A094665 and A083061.

Original entry on oeis.org

4, 147, 4095, 107415, 2837835, 77567490, 2219186970, 66782365650, 2117393828550, 70731036701325, 2487181536464625, 91943495971952625, 3567666639475063125, 145075480675658032500, 6172232487370191742500
Offset: 0

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Author

Johannes W. Meijer, May 24 2009

Keywords

Crossrefs

Equals third right hand column of the triangles A094665 and A083061. Other right hand columns are A001147, A001880, A160471 and A160472.

Formula

a(n+1) = (10*n^3+117*n^2+455*n+588)*a(n)/(5*n^2+21*n+16) with a(0) = 4.

A160471 Fourth right hand column of triangles A094665 and A083061.

Original entry on oeis.org

34, 2370, 111705, 4579575, 178378200, 6873506640, 267565948650, 10644585301350, 435650179214250, 18411603581621250, 805168495577320875, 36472461710322763125, 1711907729642450340000, 83253368434295609550000
Offset: 0

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Author

Johannes W. Meijer, May 24 2009

Keywords

Crossrefs

Equals fourth right hand column of the triangles A094665 and A083061. Other right hand columns are A001147, A001880, A160470 and A160472.

Formula

a(n+1) = ((70*n^4+1407*n^3+10520*n^2+34707*n+42660)*a(n))/(35*n^3+336*n^2+913*n +612) with a(0) = 34.

A160472 Fifth right hand column of triangles A094665 and A083061.

Original entry on oeis.org

496, 56958, 4114110, 244909665, 13285256985, 689604310395, 35192429807535, 1794448057577175, 92348727438342375, 4828071577102648875, 257517575815061937375, 14052167988805344147750, 785907013343403755553750
Offset: 0

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Author

Johannes W. Meijer, May 24 2009

Keywords

Crossrefs

Equals fifth right hand column of the triangles A094665 and A083061. Other right hand columns are A001147, A001880, A160470 and A160471.

Formula

a(n+1) = ((350*n^5+10745*n^4+130432*n^3+783031*n^2+2325618*n+2733984)* a(n))/(175*n^4+3010*n^3+17441*n^2+38414*n+23808) with a(0) = 496.

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

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Author

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

Keywords

Comments

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

Examples

			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
		

Crossrefs

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.

Programs

  • Maple
    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

Formula

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).

A036970 Triangle of coefficients of Gandhi polynomials.

Original entry on oeis.org

1, 1, 2, 3, 8, 6, 17, 54, 60, 24, 155, 556, 762, 480, 120, 2073, 8146, 12840, 10248, 4200, 720, 38227, 161424, 282078, 263040, 139440, 40320, 5040, 929569, 4163438, 7886580, 8240952, 5170800, 1965600, 423360, 40320, 28820619, 135634292
Offset: 1

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Keywords

Comments

Another version of triangle T(n,k), 0 <= k <= n, read by rows; given by [0, 1, 2, 4, 6, 9, 12, 16, 20, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, ...] = 1; 0, 1; 0, 1, 2; 0, 3, 8, 6; 0, 17, 54, 60, 24; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 07 2004

Examples

			Triangle begins:
    1;
    1,   2;
    3,   8,   6;
   17,  54,  60,  24;
  155, 556, 762, 480, 120;
  ...
		

Crossrefs

First 2 columns are Genocchi numbers A001469, A005440, row sums are also A001469.

Programs

  • Maple
    B[1]:= X -> X^2:
    for n from 2 to 12 do B[n]:= unapply(expand(X^2*(B[n-1](X+1)-B[n-1](X))),X) od:
    seq(seq(coeff(B[i](X),X,1+j),j=1..i),i=1..12); # Robert Israel, Apr 21 2016
  • Mathematica
    B[1][X_] = X^2;
    B[n_][X_] := B[n][X] = X^2*(B[n-1][X+1] - B[n-1][X]) // Simplify;
    Table[Coefficient[B[i][X], X, j+1], {i, 1, 12}, {j, 1, i}] // Flatten (* Jean-François Alcover, Sep 19 2018, from Maple *)

Formula

Let B(X, n) = X^2 (B(X+1, n-1) - B(X, n-1)), B(X, 1) = X^2; then the (i, j)-th entry in the table is the coefficient of X^(1+j) in B(X, i). - Mike Domaratzki (mdomaratzki(AT)alumni.uwaterloo.ca), Nov 17 2001
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row = top row of M^(n-1), M = an infinite square matrix in which the first "1" and right border of 1's of Pascal's triangle are deleted, as follows:
1, 2, 0, 0, 0, 0, ...
1, 3, 3, 0, 0, 0, ...
1, 4, 6, 4, 0, 0, ...
1, 5, 10, 10, 5, 0, ...
1, 6, 15, 20, 15, 6, ...
...
(End)
Let G(n,x) = (-1)^(n+1)*B(-x,n). Then G(n,x) = (2*x/(x+1))*( 1 + 2^(2*n+1)*(x-1)/(x+2) + 3^(2*n+1)*(x-1)*(x-2)/((x+2)*(x+3)) + ... ). Cf. A083061. - Peter Bala, Feb 04 2019

Extensions

More terms from David W. Wilson, Jan 12 2001

A001880 Coefficients of Bessel polynomials y_n (x).

Original entry on oeis.org

1, 15, 210, 3150, 51975, 945945, 18918900, 413513100, 9820936125, 252070693875, 6957151150950, 205552193096250, 6474894082531875, 216659917377028125, 7675951358500425000, 287080580807915895000, 11303797869311688365625, 467445288360359818884375
Offset: 4

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Keywords

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A001518.
Column 4 of triangle A001497.
Equals the second right hand column of the triangles A094665 and A083061.
Other right hand columns are A001147, A160470, A160471 and A160472.

Programs

  • Mathematica
    nn = 25; t = Range[0, nn]! CoefficientList[Series[x (1 + x/2)/(1 - 2 x)^(7/2), {x, 0, nn}], x]; Drop[t, 1] (* T. D. Noe, Aug 10 2012 *)
  • PARI
    x='x+O('x^50); Vec(serlaplace(x*(1 + x/2)/(1 - 2*x)^(7/2))) \\ G. C. Greubel, Aug 13 2017

Formula

E.g.f.: x*(1 + x/2)/(1 - 2*x)^(7/2); or, if shifted, (1+ 6x+ 3x^2/2!) / (1-2x)^(9/2).
a(n) = (2*n-4)!/(4!*(n-4)!*2^(n-4)).
(n-4)*a(n) = (n-2)*(2*n-5)*a(n-1) for n = 5, 6, .. , with a(4) = 1. - Johannes W. Meijer, May 24 2009
G.f.: x^4*2F0(5/2,3;;2x). - R. J. Mathar, Aug 08 2015

A160485 Triangle of the RBS1 polynomial coefficients.

Original entry on oeis.org

1, 1, -2, 1, -8, 12, 1, -2, 60, -120, 1, -128, -168, 0, 1680, 1, 2638, 7320, -5040, -25200, -30240, 1, -98408, -300828, 52800, 1053360, 1330560, 665280, 1, 5307118, 17914260, 2522520, -56456400, -90810720, -60540480, -17297280
Offset: 1

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Author

Johannes W. Meijer, May 24 2009, Jul 06 2009, Sep 19 2012

Keywords

Comments

In A160480 we defined the BS1 matrix by BS1[2*m-1,n=1] = 2*beta(2*m) and the recurrence relation BS1 [2*m-1,n] = (2*n-3)/(2*n-2)*(BS1[2*m-1,n-1]- BS1[2*m-3,n-1]/(2*n-3)^2), for positive and negative values of m and n= 1, 2, .. . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). It is well-known that BS1[1-2*m,n=1] = euler(2*m-2) for m = 1, 2, .., with euler(2*m-2) the Euler numbers A000364. These values together with the recurrence relation lead to BS1[ -1,n] = 1 for n = 1, 2, .. .
We discovered that the n-th term of the row coefficients BS1[1-2*m,n] for m = 1, 2, .., can be generated with the rather simple polynomials RBS1(1-2*m,n). Our discovery was enabled by the recurrence relation for the RBS1(1-2*m,n) polynomials which we derived from the recurrence relation for the BS[2*m-1,n] coefficients and the fact that RBS1(-1,n) = 1.
The RBS1 polynomials and the polynomials defined by sequence A083061 are related by a shift of +-1/2 and scaling by a power of 2 (see arXiv link). - Richard P. Brent, Jul 15 2014

Examples

			The first few rows of the triangle are:
[1]
[1, -2]
[1, -8, 12]
[1, -2, 60, -120]
[1, -128, -168, 0, 1680]
The first few RBS1(1-2*m,n) polynomials are:
RBS1(-1,n) = 1
RBS1(-3,n) = 1 - 2*n
RBS1(-5,n) = 1 - 8*n + 12*n^2
RBS1(-7,n) = 1 - 2*n + 60*n^2 - 120*n^3
From _Peter Bala_, Jan 22 2019: (Start)
Qbar(r,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n+k+1,k)*(2*k + 1)^(2*r):
Case r = 2: Qbar(2,n) = binomial(2*n+2,n+1)/2^(2*n+1) * ( 1 + 3^4*n/(n+2) + 5^4*n*(n-1)/((n+2)*(n+3)) + 7^4*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 12*n^2 + 8*n + 1, valid for n a nonnegative integer (when the series terminates). The identity is also valid for complex n with real part greater than 1 (provided the factor binomial(2*n,n) is replaced with the appropriate expression involving the gamma function).
Case r = 3: Qbar(3,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * ( 1 + 3^6*n/(n+2) + 5^6*n*(n-1)/((n+2)*(n+3)) + 7^6*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 120*n^3 + 60*n^2 + 2*n + 1, valid for n a nonnegative integer. The identity is also valid for complex n with real part greater than 2.
Note, the case r = 0 is equivalent to the identity 1 = binomial(2*n,n)/2^(2*n-1) * ( 1 + (n-1)/(n+1) + (n-1)*(n-2)/((n+1)*(n+2)) + (n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... ), which is valid for complex n with real part greater than 0. This identity was found by Ramanujan. See Example 6, Chapter 10 in Berndt. (End)
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 21.

Crossrefs

A160480 is the Beta triangle.
A009389(2*n) equals the second left hand column divided by 2.
A001813 equals the first right hand column.
The absolute values of the row sums equal the Euler numbers A000364.

Programs

  • Maple
    nmax := 8; mmax := nmax: A(1, 1) := 1: RBS1(n, 2) := (2*n-1)^2*1-(2*n)*(2*n-1)*1: for m from 3 to mmax do for k from 0 to m-1 do A(m-1, k+1) := coeff(RBS1(n, m-1), n, k) od; RBS1(n+1, m-1) := 0: for k from 0 to m-1 do RBS1(n+1, m-1) := RBS1(n+1, m-1) + A(m-1, k+1)*(n+1)^k od: RBS1(n, m) := (2*n-1)^2*RBS1(n, m-1)-(2*n)*(2*n-1) * RBS1(n+1, m-1) od: for k from 0 to nmax-1 do A(nmax, k+1) := coeff(RBS1(n, nmax), n, k) od: seq(seq(A(n, m), m=1..n), n=1..nmax);

Formula

RBS1(1-2*m,n) = (2*n-1)^2*RBS1(3-2*m,n)-(2*n)*(2*n-1)*RBS1(3-2*m,n+1) for m = 2, 3, .., with RBS1(-1,n) =1 for n = 1, 2, .. .
From Peter Bala, Jan 22 2019: (Start)
The row polynomials RBS1 of the triangle are related to the polynomials Qbar(r,n), r = 0,1,2,..., introduced by Brent by Qbar(r,n) = RBS1(-2*r-1,-n).
Recurrence: Qbar(r+1,n) = (2*n + 1)^2*Qbar(r,n) - 2*n(2*n + 1)*Qbar(r,n-1) with Qbar(0,n) = 1 (Brent, equation 19, p.7).
Qbar(r,n) = binomial(2*n + 2,n + 1)/(2^(2*n + 1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n + k + 1,k)*(2*k + 1)^(2*r); this follows easily from the above recurrence. Two examples are given below.
Qbar(r,n) = 1/4^n * Sum_{k = 0..n} binomial(2*n + 1,n - k)*(2*k + 1)^(2*r) For related polynomial sequences see A036970, A083061 and A245244. (End)

A245244 Triangle of coefficients of the Pbar polynomials, read by rows.

Original entry on oeis.org

1, -3, 4, 25, -56, 32, -427, 1228, -1184, 384, 12465, -41840, 52416, -29184, 6144, -555731, 2079892, -3076288, 2258688, -829440, 122880, 35135945, -142843304, 237829600, -208562688, 102279168, -26787840, 2949120, -2990414715, 12987478876, -23672564832, 23581133952, -13947525120, 4929576960, -970260480
Offset: 0

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Author

Richard P. Brent, Jul 14 2014

Keywords

Comments

Pbar(r,n) is a polynomial of degree r defined by the recurrence
Pbar(r+1,n) = (2*n-1)^2 * Pbar(r,n) - 4*(n-1)^2 * Pbar(r,n-1)
with initial condition Pbar(0,n) = 1.

Examples

			Pbar(1,n) = 4*n-3, Pbar(2,n) = 32*n^2 - 56*n + 25.
Triangle begins:
1,
-3, 4,
25, -56, 32,
-427, 1228, -1184, 384,
12465, -41840, 52416, -29184, 6144,
...
From _Peter Bala_, Jan 22 2018: (Start)
The polynomials Pbar(r,n) as hypergeometric series:
r = 0: n*Pbar(0,n) = n = 1 + 3*(n-1)/(n+1) + 5*(n-1)*(n-2)/((n+1)*(n+2)) + 7*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 1/2.
r = 1: n*Pbar(1,n) = n*(4*n - 3) = 1 + 3^3*(n-1)/(n+1) + 5^3*(n-1)*(n-2)/((n+1)*(n+2)) + 7^3*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 3/2.
r = 2: n*Pbar(2,n) = n*(32*n^2 - 56*n + 25) = 1 + 3^5*(n-1)/(n+1) + 5^5*(n-1)*(n-2)/((n+1)*(n+2)) + 7^5*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ..., for n a positive integer (when the series terminates). The identity is also valid for complex n with real part greater than 5/2.
The above identities when r = 0 and r = 1 were found by Ramanujan. See Example 5 and Example 13 in Chapter 10 of Berndt. (End)
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 20 and p. 23.

Crossrefs

(-1)^r Pbar(r,0) is sequence A009843. The leading coefficient of Pbar(r,n) is sequence A047053. Cf. also A036970, A083061, A160485 for analogous moments of Bernoulli random walks.

Programs

  • PARI
    N=10; P=vector(N+2); P[1]=1;
    Pbar(r)=P[r+1];
    for (r=0, N, P[r+2] = (2*n-1)^2 * Pbar(r) - 4*(n-1)^2 * subst(Pbar(r),n,n-1) );
    seq=[];  for(r=1,N, seq=concat(seq, Vecrev(P[r])); );  seq
    \\ Joerg Arndt, Jan 27 2015

Formula

In terms of Dumont-Foata polynomials F(r,x,y,z),
Pbar(r,n) = (-4)^r F(r+1,1/2-n,1/2,1/2).
In terms of odd absolute moments of a symmetric Bernoulli random walk with an odd number of steps,
n*C(2*n,n)*Pbar(r,n) = Sum_{k} C(2*n-1,k) * |2*n-1-2*k|^(2*r+1).
In terms of the Pochhammer symbol or ascending factorial (x)_k,
Pbar(r,n) = Sum_{1 <= j <= k <= r+1} (-1)^(j+1)*(1-n)_{k-1}*(2j-1)^(2r+1)/((k-j)!(k)_j).
n*Pbar(r,n) = 1 + 3^(2*r+1)*(n-1)/(n+1) + 5^(2*r+1)*(n-1)*(n-2)/((n+1)*(n+2)) + 7^(2*r+1)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... = Sum_{k = 0..n-1} binomial(n-1,k)/binomial(n+k,k)*(2*k + 1)^(2*r+1); this follows easily from the above recurrence. Examples are given below. - Peter Bala, Jan 22 2018

Extensions

More terms from Joerg Arndt, Jan 27 2015

A160468 Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.

Original entry on oeis.org

1, 1, 2, 1, 17, 26, 2, 62, 192, 60, 1, 1382, 7192, 5097, 502, 2, 21844, 171511, 217186, 55196, 2036, 2, 929569, 10262046, 20376780, 9893440, 1089330, 16356, 4, 6404582, 94582204, 271154544, 215114420, 48673180, 2567568, 16376, 1
Offset: 1

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Author

Johannes W. Meijer, May 24 2009

Keywords

Comments

In A160464 we defined the ES1 matrix by ES1[2*m-1,n=1] and in A094665 it was shown that the n-th term of the coefficients of matrix row ES1[1-2*m,n] for m >= 1 can be generated with the RES1(1-2*m,n) polynomials.
We define the o.g.f.s. of these polynomials by GFRES1(z,1-2*m) = sum(RES1(1-2*m,n)*z^(n-1), n=1..infinity) for m >= 1. The general expression of the o.g.f.s. is GFRES1(z,1-2*m) = (-1)*RE(z,1-2*m)/(2*p(m-1)*(z-1)^(m)). The p(m-1), m >= 1, sequence is Gould's sequence A001316.
The coefficients of the RE(z,1-2*m) polynomials lead to the triangle given above.
The E(z,n) = numer(sum((-1)^(n+1)*k^n*z^(k-1), k=1..infinity)) polynomials with n >= 1, see the Maple algorithm, lead to the Eulerian numbers A008292.
Some of our results are conjectures based on numerical evidence.

Examples

			The first few rows are:
[1]
[1]
[2, 1]
[17, 26, 2]
[62, 192, 60, 1]
The first few polynomials RE(z,m) are:
RE(z,-1) = 1
RE(z,-3) = 1
RE(z,-5) = 2+z
RE(z,-7) = 17+26*z+2*z^2
The first few GFRES1(z,m) are:
GFRES1(z,-1) = -(1/1)*(1)/(2*(z-1)^1)
GFRES1(z,-3) = -(1/2)*(1)/(2*(z-1)^2)
GFRES1(z,-5) = -(1/2)*(2+z)/(2*(z-1)^3)
GFRES1(z,-7) = -(1/4)*(17+26*z+2*z^2)/(2*(z-1)^4)
		

Crossrefs

For the Eulerian numbers E(n, k) see A008292.
The p(n) sequence equals Gould's sequence A001316.
The first right hand column of the triangle equals A048896.
The first left hand column equals A160469.
The row sums equal the absolute values of A117972.

Programs

  • Maple
    nmax := 8; mmax := nmax: T(0, x) := 1: for i from 1 to nmax do dgr := degree(T(i-1, x), x): for na from 0 to dgr do c(na) := coeff(T(i-1, x), x, na) od: T(i-1, x+1) := 0: for nb from 0 to dgr do T(i-1, x+1) := T(i-1, x+1) + c(nb)*(x+1)^nb od: for nc from 0 to dgr do ECGP(i-1, nc+1) := coeff(T(i-1, x), x, nc) od: T(i, x) := expand((2*x+1)*(x+1)*T(i-1, x+1) - 2*x^2*T(i-1, x)) od: dgr := degree (T(nmax, x), x): kmax := nmax: for k from 1 to kmax do p := k: for m from 1 to k do E(m, k) := sum((-1)^(m-q)*(q^k)*binomial(k+1, m-q), q=1..m) od: fx(p) := (-1)^(p+1) * (sum(E(r, k)*z^(k-r), r=1..k))/(z-1)^(p+1): GF(-(2*p+1)) := sort(simplify(((-1)^p* 1/2^(p+1)) * sum(ECGP(k-1, k-s)*fx(k-s), s=0..k-1)), ascending): NUMGF(-(2*p+1)) := -numer(GF(-(2*p+1))): for n from 1 to mmax+1 do A(k+1, n) := coeff(NUMGF(-(2*p+1)), z, n-1) od: od: for m from 2 to mmax do A(1, m) := 0 od: A(1, 1) := 1: FT(1) := 1: for n from 1 to nmax do for m from 1 to n do FT((n)*(n-1)/2+m+1) := A(n+1, m) end do end do: a := n-> FT(n): seq(a(n), n = 1..(nmax+1)*(nmax)/2+1);
  • Mathematica
    T[ n_, k_] := Coefficient[a[2 n]/2^IntegerExponent[(2 n)!, 2], x, n + k];
    a[0] = a[1] = 1; a[ m_] := a[m] = With[{n = m - 1}, x Sum[ a[k] a[n - k] Binomial[n, k], {k, 0, n}]]; Join[{1}, Flatten@Table[T[n, k], {n, 1, 8}, {k, 0, n - 1}]] (* Michael Somos, Apr 22 2020 *)

Extensions

Edited by Johannes W. Meijer, Sep 23 2012
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