A009843 -id:A009843 - cp's OEIS frontend

A002111 Glaisher's G numbers.

1, 5, 49, 809, 20317, 722813, 34607305, 2145998417, 167317266613, 16020403322021, 1848020950359841, 252778977216700025, 40453941942593304589, 7488583061542051450829, 1587688770629724715374457, 382218817191632327375004833

Offset: 1

N. J. A. Sloane

Related to the formula Sum_{k>0} sin(kx)/k^(2n+1)=(-1)^(n+1)/2*x^(2n+1)/(2n+1)! * Sum_{i=0..2n} (2Pi/x)^i*B(i)*C(2n+1,i). - Benoit Cloitre, May 01 2002

Named after the English mathematician and astronomer James Whitbread Lee Glaisher (1848-1928). - Amiram Eldar, Jun 16 2021

			G.f. = x + 5*x^2 + 49*x^3 + 809*x^4 + 20317*x^5 + 722813*x^6 + 34607305*x^7 + ...

A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 76.
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).

Seiichi Manyama, Table of n, a(n) for n = 1..254 (first 50 terms from T. D. Noe)
Shaun Cooper, Cubic elliptic functions, Res. Lett. Inf. Math. Sci., Vol. 5 (2003), pp. 23-59, see page 30.
Ira M. Gessel, On the Almkvist-Meurman Theorem for Bernoulli Polynomials, Integers (2023) Vol. 23, #A14.
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., Vol. 31 (1899), pp. 216-235.
René Gy, Bernoulli-Stirling Numbers, Integers, Vol. 20, (2020), #A67.
J. C. P. Miller, Letter to N. J. A. Sloane, Mar 26 1971.
N. J. A. Sloane, Transforms.
Wikipedia, Bernoulli Polynomials.
Index entries for sequences related to Glaisher's numbers

Cf. A083007, A033470.

read transforms; t1 := (3/2)/(1+exp(x)+exp(-x)); series(t1,x,50): t2 := SERIESTOLISTMULT(t1); [seq(n*t2[n],n=1..nops(t5))];

s[n_] := CoefficientList[Series[(1/2)*(Sin[t/2]/Sin[3*(t/2)]), {t, 0, 32}], t][[n + 1]]*n!*(-1)^Floor[n/2]; a[n_] := (-1)^n*(6*n + 3)*s[2*n]; Table[a[n], {n, 1, 16}] (* Jean-François Alcover, Mar 22 2011, after Michael Somos' formula *)
a[ n_] := If[ n < 1, 0, (2 n + 1)! SeriesCoefficient[ 3 / (2 + 4 Cos[x]), {x, 0, 2 n}]]; (* Michael Somos, Jun 01 2012 *)

{a(n) = if( n<1, 0, n*=2; (n+1)! * polcoeff( 3 / (2 + 4 * cos( x + O(x^n))), n))}; /* Michael Somos, Feb 26 2004 */

a(n)=if(n<1,0,-(-1)^n*sum(i=0,2*n,binomial(2*n+1,i)*bernfrac(i)*3^i)) \\ Benoit Cloitre, May 01 2002

def A002111(n):
    return add(add(add(((-1)^(n+1-v)/(j+1))*binomial(2*n+1,k)*binomial(j,v)*(3*v)^k for v in (0..j)) for j in (0..k)) for k in (0..2*n+1))
[A002111(n) for n in (1..16)]  # Peter Luschny, Jun 03 2013

To get these numbers, expand the e.g.f. (3/2)/(1+exp(x)+exp(-x)), multiply coefficient of x^n by (n+1)! and take absolute values.
Or expand the e.g.f. (3/2)/(1+2*cos(x)) and multiply coefficient of x^n by (n+1)!. - Herb Conn, Feb 25 2002
a(n) = (2n+1)*I(n), where I(n) is given by A047788/A047789.
a(n) = Sum_{i=0, 2n} B(i)*C(2n+1, i)*3^i where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers. - Benoit Cloitre, May 01 2002
a(n) = (-1)^n * (6*n + 3) * s(2*n), if n>0, where s(n) are the cubic Bernoulli numbers. - Michael Somos, Feb 26 2004
E.g.f.: 3*x / (2 + 4*cos(x)) = Sum_{n>=0} a(n) * x^(2*n+1) / (2*n+1)!. - Michael Somos, Feb 26 2004
E.g.f.: E(x) = (3/2)/(1+2*cos(x)) - 1/2 = x^2/(3*G(0)+x^2); G(k) = 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step). Let f[n]:=coeftayl(E(x), x=0, n) then: A002111[n]=f[2*n+2]*((2*n+3)!). - Sergei N. Gladkovskii, Jan 14 2012
a(n) = Sum_{k=0..2n+1} Sum_{j=0..k} Sum_{v=0..j} ((-1)^(n-v+1)/(j+1))* binomial(2*n+1,k)*binomial(j,v)*(3*v)^k. - Peter Luschny, Jun 03 2013
a(n) ~ (2*n+1)! * sqrt(3) * (3/(2*Pi))^(2*n+1). - Vaclav Kotesovec, Jul 30 2013
From Peter Bala, Mar 02 2015: (Start)
a(n) = (-1)^(n+1)*3^(2*n+1)*B(2*n+1,1/3), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A009843, A069852, A069994.
Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/3). Cf. A033470.
Essentially a bisection of |A083007|.
G.f. for signed version of sequence: 1/2 + 1/2*Sum_{n >= 0} { 1/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (3*k + 1)*x)*(1 - (3*k + 2)*x) ) } = x^2 - 5*x^4 + 49*x^6 - .... (End)

A083008 a(n) = Sum_{k=0..n-1} 4^kB(k)C(n,k) where B(k) is the k-th Bernoulli number and C(n,k) = binomial(n,k).

Original entry on oeis.org

0, 1, -3, 3, 9, -25, -99, 427, 2193, -12465, -79515, 555731, 4247577, -35135945, -313193811, 2990414715, 30461046561, -329655706465, -3777604994187, 45692713833379, 581778811909545, -7777794952988025, -108933009112011843, 1595024111042171723, 24370176181315498929

Offset: 0

Benoit Cloitre, May 31 2003

sign

Seiichi Manyama, Table of n, a(n) for n = 0..485

Cf. A001469.
Bisection is A009843.
Cf. A036968, A083007, A083009, A083010, A083011, A083012, A083013, A083014.

Range[0, 15]! CoefficientList[ Series[ 4x/(1 + Exp[x] + Exp[ 2x] + Exp[ 3x]), {x, 0, 15}], x] (* Robert G. Wilson v, Oct 26 2012 *)
Table[Sum[4^k*BernoulliB[k] Binomial[n, k], {k, 0, n - 1}], {n, 0, 24}] (* Michael De Vlieger, Sep 28 2016 *)

a(n)=sum(k=0,n-1,4^k*binomial(n,k)*bernfrac(k))

E.g.f.: 4*x/(1+exp(x)+exp(2*x)+exp(3*x)). - Ira M. Gessel, Feb 23 2012

a(n) ~ n! * (cos(n*Pi/2)-sin(n*Pi/2)) * 2^(n+1) / Pi^n. - Vaclav Kotesovec, Mar 02 2014

Offset changed to 0 by Seiichi Manyama, Sep 28 2016

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

Richard P. Brent, Jul 14 2014

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.

			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)

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

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. P. Brent, Generalizing Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
L. Carlitz, Explicit formulas for the Dumont-Foata polynomials, Discrete Mathematics, 30 (1980), 211-255.
H. J. H. Tuenter, Walking into an absolute sum, The Fibonacci Quarterly, 40 (2002), 175-180.

(-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.

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

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

More terms from Joerg Arndt, Jan 27 2015

A104033 Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2n+1,2k+1).

Original entry on oeis.org

1, -3, 1, 25, -10, 1, -427, 175, -21, 1, 12465, -5124, 630, -36, 1, -555731, 228525, -28182, 1650, -55, 1, 35135945, -14449006, 1782495, -104676, 3575, -78, 1, -2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1, 329655706465, -135565467080, 16724709820, -982532408

Offset: 0

Paul D. Hanna, Feb 28 2005

Column 0 equals signed A009843 (expansion of x/cosh(x)). Row sums form signed A000182 (expansion of tanh(x)).
The matrix logarithm is L(n,k) = -(-1)^(n-k)*A000182(n-k)*A103327(n,k), where A000182 = tangent numbers.
Let E(y) = cosh(sqrt(y)) = 1 + 3*y/3! + 5*y^2/5! + 7*y^3/7! + .... so that 1/E(y) = 1 - 3*y/3! + 25*y^2/5! - 427*y^3/7! + .... Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence (2*n+1)! as defined in Wang and Wang. - Peter Bala, Aug 06 2013

			Rows begin:
1;
-3, 1;
25, -10, 1;
-427, 175, -21, 1;
12465, -5124, 630, -36, 1;
-555731 ,228525, -28182, 1650, -55, 1;
35135945, -14449006, 1782495, -104676, 3575, -78, 1;
-2990414715, 1229758075, -151714563, 8912475, -305305, 6825, -105, 1;
329655706465, -135565467080, 16724709820, -982532408, 33669350, -754936, 11900, -136, 1; ...
From _Peter Bala_, Aug 06 2013: (Start)
The real zeros of the row polynomials R(n,x) seem to converge to the even squares as n increases.
Polynomial |        Real zeros to 6 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 3.999986
R(10,x)    | 4.000000, 15.999978
R(15,x)    | 4.000000, 16.000000, 35.999992, 64.414273, 76.998346
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
(End)

W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A000364, A103327, A009843, A000182 (unsigned row sums), A055133, A086645, A086646, A096053, A103364.

{T(n,k) = if(n=j, binomial(2*m-1,2*j-1))))^-1)[n+1,k+1])}
for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))

{T(n,k) = binomial(2*n+1,2*k+1) * polcoeff(1/cosh(x+x*O(x^(2*n))), 2*n-2*k) * (2*n-2*k)!}
for(n=0,10,for(k=0,n, print1(T(n,k),", "));print(""))

Column k: Sum_{j=0..n} C(2*n+1, 2*j+1) * T(j, k) = 0 (n>k), or 1 (n=k).

Row n: Sum_{j=0..n} T(n, j) * C(2*j+1, 2*k+1) = 0 (k

Sum_{k=0..n} T(n, k) * 4^k = 1 for n >= 0.

T(n, k) = (-1)^(n-k)*A000364(n-k)*A103327(n, k), where A000364 = Euler numbers.

Sum_{k=0..n} (-1)^(n-k)*T(n, k) = A002084(n). - Philippe Deléham, Aug 27 2005

From Peter Bala, Aug 06 2013: (Start)

Generating function: 1/sqrt(x)*sinh(sqrt(x)*t)/cosh(t) = t + (-3 + x)*t^3/3! + (25 - 10*x + x^2)*t^5/5! + ....

Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} binomial(2*n+1,2*k+1)*R(k,x) with initial value R(0,x) = 1.

It appears that for arbitrary nonzero complex x we have

lim_{n -> inf} R(n,x^2)/R(n,0) = (1/(Pi/2*x))*sin(Pi/2*x).

A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,x) seem to converge to the even squares 4, 16, 36, ... as n increases. Some numerical examples are given below. Cf. A055133, A086646 and A103364.

If p = 2*n + 1 is a prime then all the entries in row n are divisible by p, apart from T(n,n) = 1. Thus the row sum is congruent to 1 modulo p.

Row sums R(n,1) = (-1)^n*A000182(n+1).

R(n,4) = 1; R(n,16) = (1/2)*( 3^(2*n+1) - 1 ) = A096053(n);

R(n,36) = (1/3)*( 5^(2*n+1) - 3^(2*n+1) + 1 );

R(n,64) = (1/4)*( 7^(2*n+1) - 5^(2*n+1) + 3^(2*n+1) - 1 ). (End)

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

A160144 Numerator of (2n+1)/(2^(2n+1)-1).

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 3, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 9, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 15, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127

Offset: 0

Peter Luschny, May 03 2009

This first differs from A005408 (the odd numbers 2n+1) at a(10). The sequence of differences is A160145. This explains the similarity of A009843 (expansion of x/cos(x)) and A160143. A156769 describes a similar companion to A036279 (expansion of tan(x)).

Altug Alkan, Table of n, a(n) for n = 0..10000

Cf. A005408, A009843, A036279, A156769, A160143, A160145, A303449 (denominators).

[Numerator((2*n+1)/(2^(2*n+1)-1)): n in [0..70]]; // Vincenzo Librandi, Apr 25 2018

seq(numer((2*n+1)/(4^(2*n+1)-2^(2*n+1))),n=0..32);
seq(numer((2*n+1)/(2^(2*n+1)-1)),n=0..50); # Altug Alkan, Apr 21 2018

Array[Numerator[(2 # + 1)/(2^(2 # + 1) - 1)] &, 64, 0] (* Michael De Vlieger, Apr 21 2018 *)

vector(80,n, n--; numerator((2*n+1)/(4^(2*n+1)-2^(2*n+1)))) \\ Michel Marcus, Jan 31 2015

forstep(k=1, 1e3, 2, print1(numerator(k/(2^k-1)), ", ")); \\ Altug Alkan, Apr 21 2018

More terms from Michel Marcus, Jan 31 2015
Name simplified by Altug Alkan, Apr 21 2018
Further edited by N. J. A. Sloane, Apr 24 2018

A069852 a(n) = Sum_{i=0..2n} B(i)C(2n+1,i)5^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial numbers.

Original entry on oeis.org

6, -74, 1946, -88434, 6154786, -607884394, 80834386026, -13923204233954, 3015393801263666, -801997872697905114, 256982712667627683706, -97641716941862894337874, 43406301788286350509870146, -22319737637152541506923644234

Offset: 1

Benoit Cloitre, May 01 2002

Related to those formulas derived from Bernoulli polynomials: Sum_{k>0} sin(k*x)/k^(2n+1) = (-1)^(n+1)/2*x^(2n+1)/(2n+1)!*Sum_{i=0..2n}(2Pi/x)^i*B(i)*C(2n+1,i).

Robert Israel, Table of n, a(n) for n = 1..232
Wikipedia, Bernoulli Polynomials

Cf. A002111, A009843, A069994.

seq(5^(2*n+1)*bernoulli(2*n+1,1/5),n=1..14); # (after Peter Bala) Peter Luschny, Mar 08 2015

Table[5^(2n+1) BernoulliB[2n+1, 1/5], {n, 1, 14}] (* Jean-François Alcover, Jun 03 2019, from Maple *)

for(n=1,25,print1(sum(i=0,2*n,binomial(2*n+1,i)*bernfrac(i)*5^i),","))

From Peter Bala, Mar 02 2015: (Start)
a(n) = 5^(2*n + 1)*B(2*n + 1,1/5), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A009843 and A069994.
Conjecturally, a(n) = the signed numerator of B(2*n + 1,1/5).
G.f.: t/2*( 3 - 5*sinh(3*t/2)/sinh(5*t/2) ) = 6*t^3/3! - 74*t^5/5! + 1946*t^7/7! - ....
G.f. for signed version of sequence: 3/2 + 3/2*Sum_{n >= 0} { 1/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (5*k + 1)*x)*(1 - (5*k + 4)*x) ) } = 6*x^2 - 74*x^4 + 1946*x^6 + .... (End)

A099028 Euler-Seidel matrix T(k,n) with start sequence e.g.f. 2x/(1+e^(2x)), read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 0, -1, -2, -3, -3, -2, 0, 0, 3, 6, 8, 8, 25, 25, 22, 16, 8, 0, 0, -25, -50, -72, -88, -96, -96, -427, -427, -402, -352, -280, -192, -96, 0, 0, 427, 854, 1256, 1608, 1888, 2080, 2176, 2176, 12465, 12465, 12038, 11184, 9928, 8320, 6432, 4352, 2176, 0

Offset: 0

Ralf Stephan, Sep 27 2004

In an Euler-Seidel matrix, the rows are consecutive pairwise sums and the columns consecutive differences.

			Seidel matrix:
[    0     1    -2     0     8     0   -96     0  2176     0]
[    1    -1    -2     8     8   -96   -96  2176  2176     .]
[    0    -3     6    16   -88  -192  2080  4352     .     .]
[   -3     3    22   -72  -280  1888  6432     .     .     .]
[    0    25   -50  -352  1608  8320     .     .     .     .]
[   25   -25  -402  1256  9928     .     .     .     .     .]
[    0  -427   854 11184     .     .     .     .     .     .]
[ -427   427 12038     .     .     .     .     .     .     .]
[    0 12465     .     .     .     .     .     .     .     .]
[12465     .     .     .     .     .     .     .     .     .]

D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.

First column (odd part) is A009843, main diagonal is in A099029. Antidiagonal sums are in A065619. Cf. A009752.

T[k_, n_] := T[k, n] = If[k == 0, SeriesCoefficient[2x/(1 + E^(2x)), {x, 0, n}] n!, T[k-1, n] + T[k-1, n+1]];
Table[T[k-n, n], {k, 0, 9}, {n, 0, k}] (* Jean-François Alcover, Jun 11 2019 *)

def SeidelMatrixA099028(dim):
    E = matrix(ZZ, dim)
    t = taylor(2*x/(1+exp(2*x)), x, 0, dim + 1)
    for k in (0..dim-1):
        E[0, k] = factorial(k) * t.coefficient(x, k)
    R = [0]
    for n in (1..dim-1):
        for k in (0..dim-n-1):
            E[n, k] = E[n-1, k] + E[n-1, k+1]
        R.extend([E[n-k,k] for k in (0..n)])
    return R
print(SeidelMatrixA099028(10)) # Peter Luschny, Jul 02 2016

Recurrence: T(k, n) = T(k-1, n) + T(k-1, n+1).

A160145 a(n) = the odd number 2n+1 minus the numerator of (2n+1)/(2^(2n+1)-1).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 54, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 90, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 144, 0, 0, 0, 150

Offset: 0

Peter Luschny, May 03 2009

nonn

Explains the similarity of the sequences A009843 and A160143. (Cf. also the pair A036279 and A156769.) The first nonzero values occur at n = 10, 31, 52 and 73.

Previous name was: Odd numbers 2n+1 minus the numerators of (2n+1)/(4^(2n+1)-2^(2n+1)), (A005408 - A160144). - Altug Alkan, Apr 21 2018

Altug Alkan, Table of n, a(n) for n = 0..10000

Cf. A005408, A009843, A160143, A160144.

seq((2*n+1)-numer((2*n+1)/(4^(2*n+1)-2^(2*n+1))),n=0..77);
seq((2*n+1)-numer((2*n+1)/(2^(2*n+1)-1)),n=0..100); # Altug Alkan, Apr 21 2018

Array[# - Numerator[#/(2^# - 1)] &[2 # + 1] &, 78, 0] (* Michael De Vlieger, Apr 21 2018 *)

forstep(k=1, 1e2, 2, print1(k - numerator(k/(2^k-1)), ", ")); \\ Altug Alkan, Apr 21 2018

a(n) = A005408(n) - A160144(n).

Name simplified by Altug Alkan, Apr 21 2018

A069994 a(n) = Sum_{i=0..2n} B(i)C(2n+1,i)6^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial coefficients.

Original entry on oeis.org

-2, 10, -170, 6370, -415826, 41649850, -5922729722, 1134081384850, -281284596509858, 87722769712529770, -33597252908389628234, 15502327024398065811010, -8481855507605264686660850, 5429636257086663655134162970

Offset: 1

Benoit Cloitre, May 01 2002

Related to those formulas derived from Bernoulli polynomials: Sum_{k>0} sin(k*x)/k^(2n+1) = (-1)^(n+1)/2*x^(2n+1)/(2n+1)!*Sum_{i=0..2n} (2Pi/x)^i*B(i)*C(2n+1,i).

Wikipedia, Bernoulli Polynomials

Cf. A002111, A009843, A069852, A158073.

seq(6^(2*n-1)*bernoulli(2*n-1,1/6),n=1..14); # (after Peter Bala) Peter Luschny, Mar 08 2015

for(n=1,25,print1(sum(i=0,2*n,binomial(2*n+1,i)*bernfrac(i)*6^i),","))

From Peter Bala, Mar 02 2015: (Start)
a(n) = 6^(2*n - 1)*B(2*n - 1,1/6), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A009843, A069852.
Conjecturally, a(n) = 2 * the unsigned numerator of B(2*n - 1,1/6). If true then this sequence is a bisection of 2*A158073.
G.f.: -3*t*sinh(2*t)/sinh(3*t) = -2*t + 10*t^3/3! - 170*t^5/5! + ....
G.f.: Sum_{n >= 0} { 2/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (6*k + 1)*x)*(1 - (6*k + 5)*x) ) } = -2 + 10*x^2 - 170*x^4 + 6370*x^6 - ....
(End)

A160143 a(n) = Numerator((-1)^nEuler(2n)(2n+1)/(4^(2n+1)-2^(2n+1))), where Euler(n) = A122045(n).

Original entry on oeis.org

1, 3, 25, 427, 12465, 555731, 35135945, 2990414715, 329655706465, 45692713833379, 1111113564712575, 1595024111042171723, 387863354088927172625, 110350957750914345093747

Offset: 0

Peter Luschny, May 03 2009

Resembles the coefficients of the series for x/cos(x).
The first difference with sequence A009843 (expansion of x/cos(x)) occurs at a(10). An explanation can be found in the similarity of the numerators of (2*n+1)/(2^(2*n+1)-1) and the odd numbers 2n+1 (cf. A160144).
Similarly, A156769 resembles A036279 (from the expansion of tan(x)).

Cf. A009843, A122045, A160144, A160145, A036279, A156769.

a := n -> (-1)^iquo(n,2)*euler(n)*(n+1)/(4^(n+1)-2^(n+1));
seq(numer(a(2*n)),n=0..13);

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A002111 Glaisher's G numbers.

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A083008 a(n) = Sum_{k=0..n-1} 4^k*B(k)*C(n,k) where B(k) is the k-th Bernoulli number and C(n,k) = binomial(n,k).

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

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A104033 Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2*n+1,2*k+1).

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A160144 Numerator of (2*n+1)/(2^(2*n+1)-1).

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A069852 a(n) = Sum_{i=0..2n} B(i)*C(2n+1,i)*5^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial numbers.

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A083008 a(n) = Sum_{k=0..n-1} 4^kB(k)C(n,k) where B(k) is the k-th Bernoulli number and C(n,k) = binomial(n,k).

A104033 Triangle, read by rows, equal to the matrix inverse of triangle A103327, where A103327(n,k) = binomial(2n+1,2k+1).

A160144 Numerator of (2n+1)/(2^(2n+1)-1).

A069852 a(n) = Sum_{i=0..2n} B(i)C(2n+1,i)5^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial numbers.

A069994 a(n) = Sum_{i=0..2n} B(i)C(2n+1,i)6^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial coefficients.

A160143 a(n) = Numerator((-1)^nEuler(2n)(2n+1)/(4^(2n+1)-2^(2n+1))), where Euler(n) = A122045(n).