A160485 Triangle of the RBS1 polynomial coefficients.

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

1,3

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

References

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

Links

Table of n, a(n) for n=1..36.

P. Bala, A245244 and A160485 and some hypergeometric series evaluations of Ramanujan

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.

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)

Example

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)

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);

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.

Cf. A083061, A036970, A245244.

Sequence in context: A307049 A164662 A007026 * A328821 A367382 A118708

Adjacent sequences: A160482 A160483 A160484 * A160486 A160487 A160488

Keyword

easy,sign,tabl

Author

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

Status

approved