A160485 -id:A160485 - cp's OEIS frontend

A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

1, 1, 5, 61, 1385, 50521, 2702765, 199360981, 19391512145, 2404879675441, 370371188237525, 69348874393137901, 15514534163557086905, 4087072509293123892361, 1252259641403629865468285, 441543893249023104553682821, 177519391579539289436664789665

Offset: 0

N. J. A. Sloane

Inverse Gudermannian gd^(-1)(x) = log(sec(x) + tan(x)) = log(tan(Pi/4 + x/2)) = arctanh(sin(x)) = 2 * arctanh(tan(x/2)) = 2 * arctanh(csc(x) - cot(x)). - Michael Somos, Mar 19 2011
a(n) is the number of downup permutations of [2n]. Example: a(2)=5 counts 4231, 4132, 3241, 3142, 2143. - David Callan, Nov 21 2011
a(n) is the number of increasing full binary trees on vertices {0,1,2,...,2n} for which the leftmost leaf is labeled 2n. - David Callan, Nov 21 2011
a(n) is the number of unordered increasing trees of size 2n+1 with only even degrees allowed and degree-weight generating function given by cosh(t). - Markus Kuba, Sep 13 2014
a(n) is the number of standard Young tableaux of skew shape (n+1,n,n-1,...,3,2)/(n-1,n-2,...2,1). - Ran Pan, Apr 10 2015
Since cos(z) has a root at z = Pi/2 and no other root in C with a smaller |z|, the radius of convergence of the e.g.f. (intended complex-valued) is Pi/2 = A019669 (see also A028296). - Stanislav Sykora, Oct 07 2016
All terms are odd. - Alois P. Heinz, Jul 22 2018
The sequence starting with a(1) is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 2]. - Allen Stenger, Aug 03 2020
Conjecture: taking the sequence [a(n) : n >= 1] modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, ...] with an apparent period of 6 = phi(21)/2. - Peter Bala, May 08 2023

			G.f. = 1 + x + 5*x^2 + 61*x^3 + 1385*x^4 + 50521*x^5 + 2702765*x^6 + 199360981*x^7 + ...
sec(x) = 1 + 1/2*x^2 + 5/24*x^4 + 61/720*x^6 + ...
From _Gary W. Adamson_, Jul 18 2011: (Start)
The first few rows of matrix M are:
   1,  1,  0,  0,  0, ...
   4,  4,  4,  0,  0, ...
   9,  9,  9,  9,  0, ...
  16, 16, 16, 16, 16, ... (End)

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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
J. M. Borwein and D. M. Bailey, Mathematics by Experiment, Peters, Boston, 2004; p. 49
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G. Chrystal, Algebra, Vol. II, p. 342.
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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).
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J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 269.

Seiichi Manyama, Table of n, a(n) for n = 0..242 (terms 0..99 from N. J. A. Sloane)
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Bishal Deb and Alan D. Sokal, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022.
Bishal Deb and Alan D. Sokal, Continued fractions for cycle-alternating permutations, arXiv:2304.06545 [math.CO], 2023.
K. Dilcher and C. Vignat, Euler and the Strong Law of Small Numbers, Amer. Math. Mnthly, 123 (May 2016), 486-490.
Filippo Disanto and Emanuele Munarini, Local height in weighted Dyck models of random walks and the variability of the number of coalescent histories for caterpillar-shaped gene trees and species trees, SN Applied Sciences (2019), 1:578.
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A. L. Edmonds and S, Klee, The combinatorics of hyperbolized manifolds, arXiv preprint arXiv:1210.7396 [math.CO], 2012. - From _N. J. A. Sloane_, Jan 02 2013
C. J. Fewster and D. Siemssen, Enumerating Permutations by their Run Structure, arXiv preprint arXiv:1403.1723 [math.CO], 2014.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 144.
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Dominique Foata and Guo-Niu Han, Seidel Triangle Sequences and Bi-Entringer Numbers, November 20, 2013.
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23. (Annotated scanned copy)
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Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, vol.6, no.1, #R21, (1999).
Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.
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Sam Wagstaff, Prime divisors of the Bernoulli and Euler numbers.
Eric Weisstein's World of Mathematics, Euler Number, Secant Number, Alternating Permutation.
Wolfram Research, Generating functions for E_n.
Index entries for "core" sequences.
Index entries for sequences related to boustrophedon transform.

Cf. A000111, A000182, A011248, A019669, A034947, A060075, A013525, A000816, A002436, A000464, A002105, A002439, A079144, A158690.
Essentially same as A028296 and A122045.
First column of triangle A060074.
Two main diagonals of triangle A060058 (as iterated sums of squares).
Absolute values of row sums of A160485. - Johannes W. Meijer, Jul 06 2009
Left edge of triangle A210108, see also A125053, A076552. Cf. A255881.
Bisection (even part) of A317139.
The sequences [(-k^2)^n*Euler(2*n, 1/k), n = 0, 1, ...] are: A000007 (k=1), A000364 (k=2), |A210657| (k=3), A000281 (k=4), A272158 (k=5), A002438 (k=6), A273031 (k=7).

series(sec(x),x,40): SERIESTOSERIESMULT(%): subs(x=sqrt(y),%): seriestolist(%);
# end of program
A000364_list := proc(n) local S,k,j; S[0] := 1;
for k from 1 to n do S[k] := k*S[k-1] od;
for k from  1 to n do
    for j from k to n do
        S[j] := (j-k)*S[j-1]+(j-k+1)*S[j] od od;
seq(S[j], j=1..n)  end:
A000364_list(16);  # Peter Luschny, Apr 02 2012
A000364 := proc(n)
    abs(euler(2*n)) ;
end proc: # R. J. Mathar, Mar 14 2013

Take[ Range[0, 32]! * CoefficientList[ Series[ Sec[x], {x, 0, 32}], x], {1, 32, 2}] (* Robert G. Wilson v, Apr 23 2006 *)
Table[Abs[EulerE[2n]], {n, 0, 30}] (* Ray Chandler, Mar 20 2007 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Sec[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n + 1}, m! SeriesCoefficient[ InverseGudermannian[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
a[n_] := Sum[Sum[Binomial[k, m] (-1)^(n+k)/(2^(m-1)) Sum[Binomial[m, j]* (2j-m)^(2n), {j, 0, m/2}] (-1)^(k-m), {m, 0, k}], {k, 1, 2n}]; Table[ a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 26 2019, after Vladimir Kruchinin *)
a[0] := 1; a[n_] := a[n] = -Sum[a[n - k]/(2 k)!, {k, 1, n}]; Map[(-1)^# (2 #)! a[#] &, Range[0, 16]] (* Oliver Seipel, May 18 2024 *)

a(n):=sum(sum(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*sum(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */

a[n]:=if n=0 then 1 else sum(sum((i-k)^(2*n)*binomial(2*k, i)*(-1)^(i+k+n), i, 0, k-1)/ (2^(k-1)), k, 1, 2*n); makelist(a[n], n, 0, 16); /* Vladimir Kruchinin, Oct 05 2012 */

{a(n)=local(CF=1+x*O(x^n));if(n<0,return(0), for(k=1,n,CF=1/(1-(n-k+1)^2*x*CF));return(Vec(CF)[n+1]))} \\ Paul D. Hanna Oct 07 2005

{a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */

{a(n) = my(A); if( n<0, 0, n = 2*n+1 ; A = x * O(x^n); n! * polcoeff( log(1 / cos(x + A) + tan(x + A)), n))}; /* Michael Somos, Aug 15 2007 */

{a(n)=polcoeff(sum(m=0, n, (2*m)!/2^m * x^m/prod(k=1, m, 1+k^2*x+x*O(x^n))), n)} \\ Paul D. Hanna, Sep 20 2012

list(n)=my(v=Vec(1/cos(x+O(x^(2*n+1)))));vector(n,i,v[2*i-1]*(2*i-2)!) \\ Charles R Greathouse IV, Oct 16 2012

a(n)=subst(bernpol(2*n+1),'x,1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1) \\ Charles R Greathouse IV, Dec 10 2014

a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022

\\ Based on an algorithm of Peter Bala, cf. link in A110501.
upto(n) = my(v1, v2, v3); v1 = vector(n+1, i, 0); v1[1] = 1; v2 = vector(n, i, i^2); v3 = v1; for(i=2, n+1, for(j=2, i-1, v1[j] += v2[i-j+1]*v1[j-1]); v1[i] = v1[i-1]; v3[i] = v1[i]); v3 \\ Mikhail Kurkov, Aug 30 2025

from functools import lru_cache
from math import comb
@lru_cache(maxsize=None)
def A000364(n): return 1 if n == 0 else (1 if n % 2 else -1)*sum((-1 if i % 2 else 1)*A000364(i)*comb(2*n,2*i) for i in range(n)) # Chai Wah Wu, Jan 14 2022

# after Mikhail Kurkov, based on an algorithm of Peter Bala, cf. link in A110501.
def euler_list(len: int) -> list[int]:
    if len == 0: return []
    v1 = [1] + [0] * (len - 1)
    v2 = [i**2 for i in range(1, len + 1)]
    result = [0] * len
    result[0] = 1
    for i in range(1, len):
        for j in range(1, i):
            v1[j] += v2[i - j] * v1[j - 1]
        v1[i] = v1[i - 1]
        result[i] = v1[i]
    return result
print(euler_list(1000))  # Peter Luschny, Aug 30 2025

# Algorithm of L. Seidel (1877)
# n -> [a(0), a(1), ..., a(n-1)] for n > 0.
def A000364_list(len) :
    R = []; A = {-1:0, 0:1}; k = 0; e = 1
    for i in (0..2*len-1) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        if e < 0 : R.append(A[-i//2])
    return R
A000364_list(17) # Peter Luschny, Mar 31 2012

E.g.f.: Sum_{n >= 0} a(n) * x^(2*n) / (2*n)! = sec(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n) * x^(2*n+1) / (2*n+1)! = gd^(-1)(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n+1)/(2*n+1)! = 2*arctanh(cosec(x)-cotan(x)). - Ralf Stephan, Dec 16 2004
Pi/4 - [Sum_{k=0..n-1} (-1)^k/(2*k+1)] ~ (1/2)*[Sum_{k>=0} (-1)^k*E(k)/(2*n)^(2k+1)] for positive even n. [Borwein, Borwein, and Dilcher]
Also, for positive odd n, log(2) - Sum_{k = 1..(n-1)/2} (-1)^(k-1)/k ~ (-1)^((n-1)/2) * Sum_{k >= 0} (-1)^k*E(k)/n^(2*k+1), where E(k) is the k-th Euler number, by Borwein, Borwein, and Dilcher, Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then replace x with (n-1)/2. - Peter Bala, Oct 29 2016
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i, 2*(j-1)) = A086645(i, j-1); then for n>0, a(n) = det(M_n); example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385. - Philippe Deléham, Sep 04 2005
This sequence is also (-1)^n*EulerE(2*n) or abs(EulerE(2*n)). - Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 14 2006
a(n) = 2^n * E_n(1/2), where E_n(x) is an Euler polynomial.
a(k) = a(j) (mod 2^n) if and only if k == j (mod 2^n) (k and j  are even). [Stern; see also Wagstaff and Sun]
E_k(3^(k+1)+1)/4 = (3^k/2)*Sum_{j=0..2^n-1} (-1)^(j-1)*(2j+1)^k*[(3j+1)/2^n] (mod 2^n) where k is even and [x] is the greatest integer function. [Sun]
a(n) ~ 2^(2*n+2)*(2*n)!/Pi^(2*n+1) as n -> infinity. [corrected by Vaclav Kotesovec, Jul 10 2021]
a(n) = Sum_{k=0..n} A094665(n, k)*2^(n-k). - Philippe Deléham, Jun 10 2004
Recurrence: a(n) = -(-1)^n*Sum_{i=0..n-1} (-1)^i*a(i)*binomial(2*n, 2*i). - Ralf Stephan, Feb 24 2005
O.g.f.: 1/(1-x/(1-4*x/(1-9*x/(1-16*x/(...-n^2*x/(1-...)))))) (continued fraction due to T. J. Stieltjes). - Paul D. Hanna, Oct 07 2005
a(n) = (Integral_{t=0..Pi} log(tan(t/2)^2)^(2n)dt)/Pi^(2n+1). - Logan Kleinwaks (kleinwaks(AT)alumni.princeton.edu), Mar 15 2007
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function: 2*exp(-t)*Sum {n >= 0} Product_{k = 1..n} (1-exp(-(4*k-2)*t))*exp(-2*n*t)/Product_{k = 1..n+1} (1+exp(-(4*k-2)*t)) = 1 + t + 5*t^2/2! + 61*t^3/3! + .... For other sequences with generating functions of a similar type see A000464, A002105, A002439, A079144 and A158690.
a(n) = 2*(-1)^n*L(-2*n), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s + 1/5^s - + .... (End)
Sum_{n>=0} a(n)*z^(2*n)/(4*n)!! = Beta(1/2-z/(2*Pi),1/2+z/(2*Pi))/Beta(1/2,1/2) with Beta(z,w) the Beta function. - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_(Sum_(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*Sum_(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
If n is prime, then a(n)==1 (mod 2*n). - Vladimir Shevelev, Sep 04 2010
From Peter Bala, Jan 21 2011: (Start)
(1)... a(n) = (-1/4)^n*B(2*n,-1),
where {B(n,x)}n>=1 = [1, 1+x, 1+6*x+x^2, 1+23*x+23*x^2+x^3, ...] is the sequence of Eulerian polynomials of type B - see A060187. Equivalently,
(2)... a(n) = Sum_{k = 0..2*n} Sum_{j = 0..k} (-1)^(n-j) *binomial(2*n+1,k-j)*(j+1/2)^(2*n).
We also have
(3)... a(n) = 2*A(2*n,i)/(1+i)^(2*n+1),
where i = sqrt(-1) and where {A(n,x)}n>=1 = [x, x + x^2, x + 4*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials - see A008292. Equivalently,
(4)... a(n) = i*Sum_{k = 1..2*n} (-1)^(n+k)*k!*Stirling2(2*n,k) *((1+i)/2)^(k-1)
= i*Sum_{k = 1..2*n} (-1)^(n+k)*((1+i)/2)^(k-1) Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^(2*n).
Either this explicit formula for a(n) or (2) above may be used to obtain congruence results for a(n). For example, for prime p
(5a)... a(p) = 1 (mod p)
(5b)... a(2*p) = 5 (mod p)
and for odd prime p
(6a)... a((p+1)/2) = (-1)^((p-1)/2) (mod p)
(6b)... a((p-1)/2) = -1 + (-1)^((p-1)/2) (mod p).
(End)
a(n) = (-1)^n*2^(4*n+1)*(zeta(-2*n,1/4) - zeta(-2*n,3/4)). - Gerry Martens, May 27 2011
a(n) may be expressed as a sum of multinomials taken over all compositions of 2*n into even parts (Vella 2008): a(n) = Sum_{compositions 2*i_1 + ... + 2*i_k = 2*n} (-1)^(n+k)* multinomial(2*n, 2*i_1, ..., 2*i_k). For example, there are 4 compositions of the number 6 into even parts, namely 6, 4+2, 2+4 and 2+2+2, and hence a(3) = 6!/6! - 6!/(4!*2!) - 6!/(2!*4!) + 6!/(2!*2!*2!) = 61. A companion formula expressing a(n) as a sum of multinomials taken over the compositions of 2*n-1 into odd parts has been given by Malenfant 2011. - Peter Bala, Jul 07 2011
a(n) = the upper left term in M^n, where M is an infinite square production matrix; M[i,j] = A000290(i) = i^2, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2 (see examples). - Gary W. Adamson, Jul 18 2011
E.g.f. A'(x) satisfies the differential equation A'(x)=cos(A(x)). - Vladimir Kruchinin, Nov 03 2011
From Peter Bala, Nov 28 2011: (Start)
a(n) = D^(2*n)(cosh(x)) evaluated at x = 0, where D is the operator cosh(x)*d/dx. a(n) = D^(2*n-1)(f(x)) evaluated at x = 0, where f(x) = 1+x+x^2/2! and D is the operator f(x)*d/dx.
Other generating functions: cosh(Integral_{t = 0..x} 1/cos(t)) dt = 1 + x^2/2! + 5*x^4/4! + 61*x^6/6! + 1385*x^8/8! + .... Cf. A012131.
A(x) := arcsinh(tan(x)) = log( sec(x) + tan(x) ) = x + x^3/3! + 5*x^5/5! + 61*x^7/7! + 1385*x^9/9! + .... A(x) satisfies A'(x) = cosh(A(x)).
B(x) := Series reversion( log(sec(x) + tan(x)) ) = x - x^3/3! + 5*x^5/5! - 61*x^7/7! + 1385*x^9/9! - ... = arctan(sinh(x)). B(x) satisfies B'(x) = cos(B(x)). (End)
HANKEL transform is A097476. PSUM transform is A173226. - Michael Somos, May 12 2012
a(n+1) - a(n) = A006212(2*n). - Michael Somos, May 12 2012
a(0) = 1 and, for n > 0, a(n) = (-1)^n*((4*n+1)/(2*n+1) - Sum_{k = 1..n} (4^(2*k)/2*k)*binomial(2*n,2*k-1)*A000367(k)/A002445(k)); see the Bucur et al. link. - L. Edson Jeffery, Sep 17 2012
O.g.f.: Sum_{n>=0} (2*n)!/2^n * x^n / Product_{k=1..n} (1 + k^2*x). - Paul D. Hanna, Sep 20 2012
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 11 2013: (Start)
Continued fractions:
E.g.f.: (sec(x)) = 1+x^2/T(0), T(k) = 2(k+1)(2k+1) - x^2 + x^2*(2k+1)(2k+2)/T(k+1).
E.g.f.: 2/Q(0) where Q(k) = 1 + 1/(1 - x^2/(x^2 - 2*(k+1)*(2*k+1)/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + x*k*(3*k-1) - x*(k+1)*(2*k+1)*(x*k^2+1)/Q(k+1).
E.g.f.: (2 + x^2 + 2*U(0))/(2 + (2 - x^2)*U(0)) where U(k)=  4*k + 4 + 1/( 1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
E.g.f.: 1/cos(x) = 8*(x^2+1)/(4*x^2 + 8 - x^4*U(0)) where U(k) = 1 + 4*(k+1)*(k+2)/(2*k+3 - x^2*(2*k+3)/(x^2 - 8*(k+1)*(k+2)*(k+3)/U(k+1))).
G.f.: 1/U(0) where U(k) = 1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 + x - x*(2*k+2)*(2*k+3)/(1 - x*(2*k+2)*(2*k+3)/G(k+1)).
Let F(x) = sec(x^(1/2)) = Sum_{n>=0} a(n)*x^n/(2*n)!, then F(x)=2/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)).
E.g.f.: 1/cos(x) = 1 + x^2/(2-x^2)*Q(0), where Q(k) = 1 - 2*x^2*(k+1)*(2*k+1)/( 2*x^2*(k+1)*(2*k+1)+ (12-x^2 + 14*k + 4*k^2)*(2-x^2 + 6*k + 4*k^2)/Q(k+1)). (End)
a(n) = Sum_{k=1..2*n} (Sum_{i=0..k-1} (i-k)^(2*n)*binomial(2*k,i)*(-1)^(i+k+n)) / 2^(k-1) for n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
It appears that a(n) = 3*A076552(n -1) + 2*(-1)^n for n >= 1. Conjectural congruences: a(2*n) == 5 (mod 60) for n >= 1 and a(2*n+1) == 1 (mod 60) for n >= 0. - Peter Bala, Jul 26 2013
From Peter Bala, Mar 09 2015: (Start)
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - sqrt(-x)*(2*k + 1)) = Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(2*k + 1)^2).
O.g.f. is 1 + x*d/dx(log(F(x))), where F(x) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ... is the o.g.f. for A255881. (End)
Sum_(n >= 1, A034947(n)/n^(2d+1)) = a(d)*Pi^(2d+1)/(2^(2d+2)-2)(2d)! for d >= 0; see Allouche and Sondow, 2015. - Jonathan Sondow, Mar 21 2015
Asymptotic expansion: 4*(4*n/(Pi*e))^(2*n+1/2)*exp(1/2+1/(24*n)-1/(2880*n^3) +1/(40320*n^5)-...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
a(n) = 2*(-1)^n*Im(Li_{-2n}(i)), where Li_n(x) is polylogarithm, i=sqrt(-1). - Vladimir Reshetnikov, Oct 22 2015
Limit_{n->infinity} ((2*n)!/a(n))^(1/(2*n)) = Pi/2. - Stanislav Sykora, Oct 07 2016
O.g.f.: 1/(1 + x - 2*x/(1 - 2*x/(1 + x - 12*x/(1 - 12*x/(1 + x - 30*x/(1 - 30*x/(1 + x - ... - (2*n - 1)*(2*n)*x/(1 - (2*n - 1)*(2*n)*x/(1 + x - ... ))))))))). - Peter Bala, Nov 09 2017
For n>0, a(n) = (-PolyGamma(2*n, 1/4) / 2^(2*n - 1) - (2^(2*n + 2) - 2) * Gamma(2*n + 1) * zeta(2*n + 1)) / Pi^(2*n + 1). - Vaclav Kotesovec, May 04 2020
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)) * exp(Sum_{k>=1} bernoulli(k+1) / (k*(k+1)*2^k*n^k)). - Vaclav Kotesovec, Mar 05 2021
Peter Bala's conjectured congruences, a(2n) == 5 (mod 60) for n >= 1 and a(2n + 1) == 1 (mod 60), hold due to the results of Stern (mod 4) and Knuth & Buckholtz (mod 3 and 5). - Charles R Greathouse IV, Mar 23 2022

Typo in name corrected by Anders Claesson, Dec 01 2015

A014641 Odd octagonal numbers: (2n+1)*(6n+1).

Original entry on oeis.org

1, 21, 65, 133, 225, 341, 481, 645, 833, 1045, 1281, 1541, 1825, 2133, 2465, 2821, 3201, 3605, 4033, 4485, 4961, 5461, 5985, 6533, 7105, 7701, 8321, 8965, 9633, 10325, 11041, 11781, 12545, 13333, 14145, 14981, 15841, 16725, 17633, 18565, 19521, 20501, 21505

Offset: 0

Mohammad K. Azarian, Dec 11 1999

Sequence found by reading the line from 1, in the direction 1, 21, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

Muniru A Asiru, Table of n, a(n) for n = 0..5000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.2.
Leo Tavares, Illustration: Square Block Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001082, A014642, A014793, A014794, A243201, A289873.
Cf. A160485, A245244.
Cf. A016754, A014105.

List([0..50],n->(2*n+1)*(6*n+1)); # Muniru A Asiru, Feb 05 2019

[ (2*n+1)*(6*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 08 2014

A014641:=n->(2*n+1)*(6*n+1); seq(A014641(n), n=0..50); # Wesley Ivan Hurt, Jun 08 2014

Table[(2n + 1)(6n + 1), {n, 0, 49}]  (* Harvey P. Dale, Mar 24 2011 *)

a(n)=(2*n+1)*(6*n+1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = a(n-1) + 24*n - 4, with n > 0, a(0)=1. - Vincenzo Librandi, Dec 28 2010
G.f.: (1 + 18*x + 5*x^2)/(1 - 3*x + 3*x^2 - x^3). - Colin Barker, Jan 06 2012
a(n) = A289873(6*n+2). - Hugo Pfoertner, Jul 15 2017
From Peter Bala, Jan 22 2018: (Start)
This is the polynomial Qbar(2,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials.
a(n) = (1/4^n) * Sum_{k = 0..n} (2*k + 1)^4*binomial(2*n + 1, n - k).
a(n-1) = (2/4^n) * binomial(2*n,n) * ( 1 + 3^4*(n - 1)/(n + 1) + 5^4*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^4*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*log(3))/8.
Sum_{n>=0} (-1)^n/a(n) = Pi/8 + sqrt(3)*log(2+sqrt(3))/4. (End)
E.g.f.: exp(x)*(1 + 20*x + 12*x^2). - Stefano Spezia, Apr 16 2022
a(n) = A016754(n) + 4*A014105(n). - Leo Tavares, May 20 2022

More terms from Patrick De Geest

Better description from N. J. A. Sloane

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

A160480 The Beta triangle read by rows.

Original entry on oeis.org

-1, -11, 1, -299, 36, -1, -15371, 2063, -85, 1, -1285371, 182474, -8948, 166, -1, -159158691, 23364725, -1265182, 29034, -287, 1, -27376820379, 4107797216, -237180483, 6171928, -77537, 456, -1

Offset: 2

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

The coefficients of the BS1 matrix are defined by BS1[2*m-1,n] = int(y^(2*m-1)/(cosh(y))^(2*n-1),y=0..infinity)/factorial(2*m-1) for m = 1, 2, ... and n = 1, 2, ... .
This definition leads to BS1[2*m-1,n=1] = 2*beta(2*m), for m = 1, 2, ..., 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) which we used to extend our definition of the BS1 matrix coefficients to m = 0, -1, -2, ... . We discovered that BS1[ -1,n] = 1 for n = 1, 2, ... . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity).
The coefficients in the columns of the BS1 matrix, for m = 1, 2, 3, ..., and n = 2, 3, 4, ..., can be generated with the GK(z;n) polynomials for which we found the following general expression GK(z;n) = ((-1)^(n+1)*CFN2(z;n)*GK(z;n=1) + BETA(z;n))/p(n).
The CFN2(z;n) polynomials depend on the central factorial numbers A008956.
The BETA(z;n) are the Beta polynomials which lead to the Beta triangle.
The zero patterns of the Beta polynomials resemble a UFO. These patterns resemble those of the Eta, Zeta and Lambda polynomials, see A160464, A160474 and A160487.
The first Maple algorithm generates the coefficients of the Beta triangle. The second Maple algorithm generates the BS1[2*m-1,n] coefficients for m = 0, -1, -2, -3, ... .
Some of our results are conjectures based on numerical evidence, see especially A160481.

			The first few rows of the triangle BETA(n,m) with n=2,3,... and m=1,2,... are
  [ -1],
  [ -11, 1],
  [ -299, 36, -1],
  [ -15371, 2063 -85, 1].
The first few BETA(z;n) polynomials are
  BETA(z;n=2) = -1,
  BETA(z;n=3) = -11 + z^2,
  BETA(z;n=4) = -299 + 36*z^2 - z^4.
The first few CFN1(z;n) polynomials are
  CFN2(z;n=2) = (z^2 - 1),
  CFN2(z;n=3) = (z^4 - 10*z^2 + 9),
  CFN2(z;n=4) = (z^6 - 35*z^4 + 259*z^2 - 225).
The first few generating functions GK(z;n) are
  GK(z;n=2) = ((-1)*(z^2-1)*GK(z,n=1) + (-1))/2,
  GK(z;n=3) = ((z^4 - 10*z^2 + 9)*GK(z,n=1)+ (-11 + z^2))/24,
  GK(z;n=4) = ((-1)*(z^6 - 35*z^4 + 259*z^2 - 225)*GK(z,n=1) + (-299 + 36*z^2 - z^4))/720.

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.
J. M. Amigo, Relations among Sums of Reciprocal Powers Part II, International Journal of Mathematics and Mathematical Sciences , Volume 2008 (2008), pp. 1-20.
Johannes W. Meijer, The zeros of the Eta, Zeta, Beta and Lambda polynomials, jpg and pdf, Mar 03 2013.

A160481 equals the rows sums.
A101269 and A160482 equal the first and second left hand columns.
A160483 and A160484 equal the second and third right hand columns.
A160485 and A160486 are two related triangles.
The CFN2(z, n) and the cfn2(n, k) lead to A008956.
Cf. the Eta, Zeta and Lambda triangles: A160464, A160474 and A160487.
Cf. A162443 (BG1 matrix).

nmax := 8; mmax := nmax: for n from 1 to nmax do BETA(n, n) := 0 end do: m := 1: for n from m+1 to nmax do BETA(n, m) := (2*n-3)^2*BETA(n-1, m) - (2*n-4)! od: for m from 2 to mmax do for n from m+1 to nmax do BETA(n, m) := (2*n-3)^2*BETA(n-1, m) - BETA(n-1, m-1) od: od: seq(seq(BETA(n, m), m=1..n-1), n= 2..nmax);
# End first program
nmax1 := 25; m := 1; BS1row := 1-2*m; for n from 0 to nmax1 do cfn2(n, 0) := 1: cfn2(n, n) := (doublefactorial(2*n-1))^2 od: for n from 1 to nmax1 do for k from 1 to n-1 do cfn2(n, k) := (2*n-1)^2*cfn2(n-1, k-1) + cfn2(n-1, k) od: od: mmax1 := nmax1: for m1 from 1 to mmax1 do BS1[1-2*m1, 1] := euler(2*m1-2) od: for n from 2 to nmax1 do for m1 from 1 to mmax1-n+1 do BS1[1-2*m1, n] := (-1)^(n+1)*sum((-1)^(k1+1)*cfn2(n-1, k1-1) * BS1[2*k1-2*n-2*m1+1, 1], k1 =1..n)/(2*n-2)! od: od: seq(BS1[1-2*m, n], n=1..nmax1-m+1);
# End second program

BETA[2, 1] = -1;
BETA[n_, 1] := BETA[n, 1] = (2*n - 3)^2*BETA[n - 1, 1] - (2*n - 4)!;
BETA[n_ /; n > 2, m_ /; m > 0] /; 1 <= m <= n := BETA[n, m] = (2*n - 3)^2*BETA[n - 1, m] - BETA[n - 1, m - 1];
BETA[, ] = 0;
Table[BETA[n, m], {n, 2, 9}, {m, 1, n - 1}] // Flatten (* Jean-François Alcover, Dec 13 2017 *)

We discovered a relation between the Beta triangle coefficients BETA(n,m) = (2*n-3)^2* BETA(n-1,m)- BETA(n-1,m-1) for n = 3, 4, ... and m = 2, 3, ... with BETA(n,m=1) = (2*n-3)^2*BETA(n-1,m=1) - (2*n-4)! for n = 2, 3, ... and BETA(n,n) = 0 for n = 1, 2, ... .
The generating functions GK(z;n) of the coefficients in the matrix columns are defined by
GK(z;n) = sum(BS1[2*m-1,n]*z^(2*m-2), m=1..infinity) with n = 1, 2, ... .
This definition leads to GK(z;n=1) = 1/(z*cos(Pi*z/2))*int(sin(z*t)/sin(t),t=0..Pi/2).
Furthermore we discovered that GK(z;n) = GK(z;n-1)*((2*n-3)/(2*n-2)-z^2/((2*n-2)*(2*n-3)))-1/((2*n-2)*(2*n-3)) for n = 2, 3, ... .
We found the following general expression for the GK(z;n) polynomials, for n = 2, 3, ...,
GK(z;n) = ((-1)^(n+1)*CFN2(z;n)*GK(z;n=1) + BETA(z;n))/p(n) with p(n) = (2*n-2)!.

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

A162443 Numerators of the BG1[ -5,n] coefficients of the BG1 matrix.

Original entry on oeis.org

5, 66, 680, 2576, 33408, 14080, 545792, 481280, 29523968, 73465856, 27525120, 856162304, 1153433600, 18798870528, 86603988992, 2080374784, 2385854332928, 3216930504704, 71829033058304, 7593502179328, 281749854617600

Offset: 1

Johannes W. Meijer, Jul 06 2009

The BG1 matrix coefficients are defined by BG1[2m-1,1] = 2*beta(2m) and the recurrence relation BG1[2m-1,n] = BG1[2m-1,n-1] - BG1[2m-3,n-1]/(2*n-3)^2 with m = .. , -2, -1, 0, 1, 2, .. and n = 1, 2, 3, .. . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). For the BG2 matrix, the even counterpart of the BG1 matrix, see A008956.
We discovered that the n-th term of the row coefficients can be generated with BG1[1-2*m,n] = RBS1(1-2*m,n)* 4^(n-1)*((n-1)!)^2/ (2*n-2)! for m >= 1. For the BS1(1-2*m,n) polynomials see A160485.
The coefficients in the columns of the BG1 matrix, for m >= 1 and n >= 2, can be generated with GFB(z;n) = ((-1)^(n+1)*CFN2(z;n)*GFB(z;n=1) + BETA(z;n))/((2*n-3)!!)^2 for n >= 2. For the CFN2(z;n) and the Beta polynomials see A160480.
The BG1[ -5,n] sequence can be generated with the first Maple program and the BG1[2*m-1,n] matrix coefficients can be generated with the second Maple program.
The BG1 matrix is related to the BS1 matrix, see A160480 and the formulas below.

			The first few formulas for the BG1[1-2*m,n] matrix coefficients are:
BG1[ -1,n] = (1)*4^(n-1)*(n-1)!^2/(2*n-2)!
BG1[ -3,n] = (1-2*n)*4^(n-1)*(n-1)!^2/(2*n-2)!
BG1[ -5,n] = (1-8*n+12*n^2)*4^(n-1)*(n-1)!^2/(2*n-2)!
The first few generating functions GFB(z;n) are:
GFB(z;2) = ((-1)*(z^2-1)*GFB(z;1) + (-1))/1
GFB(z;3) = ((+1)*(z^4-10*z^2+9)*GFB(z;1) + (-11 + z^2))/9
GFB(z;4) = ((-1)*( z^6- 35*z^4+259*z^2-225)*GFB(z;1) + (-299 + 36*z^2 - z^4))/225

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.
J. M. Amigo, Relations among Sums of Reciprocal Powers Part II, International Journal of Mathematics and Mathematical Sciences , Volume 2008 (2008), pp. 1-20.

A162444 are the denominators of the BG1[ -5, n] matrix coefficients.
The BG1[ -3, n] equal (-1)*A002595(n-1)/A055786(n-1) for n >= 1.
The BG1[ -1, n] equal A046161(n-1)/A001790(n-1) for n >= 1.
The cs(n) equal A046161(n-2)/A001803(n-2) for n >= 2.
The BETA(z, n) polynomials and the BS1 matrix lead to the Beta triangle A160480.
The CFN2(z, n), the t2(n, m) and the BG2 matrix lead to A008956.
Cf. A000364, A001818 and A160485.
Cf. A162443 (BG1 matrix), A162446 (ZG1 matrix) and A162448 (LG1 matrix).

a := proc(n): numer((1-8*n+12*n^2)*4^(n-1)*(n-1)!^2/(2*n-2)!) end proc: seq(a(n), n=1..21);
# End program 1
nmax1 := 5; coln := 3; Digits := 20: mmax1 := nmax1: for n from 0 to nmax1 do t2(n, 0) := 1 od: for n from 0 to nmax1 do t2(n, n) := doublefactorial(2*n-1)^2 od: for n from 1 to nmax1 do for m from 1 to n-1 do t2(n, m) := (2*n-1)^2* t2(n-1, m-1) + t2(n-1, m) od: od: for m from 1 to mmax1 do BG1[1-2*m, 1] := euler(2*m-2) od: for m from 1 to mmax1 do BG1[2*m-1, 1] := Re(evalf(2*sum((-1)^k1/(1+2*k1)^(2*m), k1=0..infinity))) od: for m from -mmax1 +coln to mmax1 do BG1[2*m-1, coln] := (-1)^(coln+1)*sum((-1)^k1*t2(coln-1, k1)*BG1[2*m-(2*coln-1)+2*k1, 1], k1=0..coln-1)/doublefactorial(2*coln-3)^2 od;
# End program 2
# Maple programs edited by Johannes W. Meijer, Sep 25 2012

a(n) = numer(BG1[ -5,n]) and A162444(n) = denom(BG1[ -5,n]) with BG1[ -5,n] = (1-8*n+12*n^2)*4^(n-1)*(n-1)!^2/(2*n-2)!.
The generating functions GFB(z;n) of the coefficients in the matrix columns are defined by
GFB(z;n) = sum(BG1[2*m-1,n]*z^(2*m-2), m=1..infinity).
GFB(z;n) = (1-z^2/(2*n-3)^2)*GFB(n-1) - 4^(n-2)*(n-2)!^2/((2*n-4)!*(2*n-3)^2) for n => 2 with GFB(z;n=1) = 1/(z*cos(Pi*z/2))*int(sin(z*t)/sin(t),t=0..Pi/2).
The column sums cs(n) = sum(BG1[2*m-1,n]*z^(2*m-2), m=1..infinity) = 4^(n-1)/((2*n-2)*binomial(2*n-2,n-1)) for n >= 2.
BG1[2*m-1,n] = (n-1)!^2*4^(n-1)*BS1[2*m-1,n]/(2*n-2)!

A160486 Triangle of polynomial coefficients related to the o.g.f.s. of the RBS1 polynomials.

Original entry on oeis.org

1, 1, 1, 1, 18, 5, 1, 179, 479, 61, 1, 1636, 18270, 19028, 1385, 1, 14757, 540242, 1949762, 1073517, 50521, 1, 132854, 14494859, 137963364, 241595239, 82112518, 2702765

Offset: 1

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

As we showed in A160485 the n-th term of the coefficients of matrix row BS1[1-2*m,n] for m = 1 , 2, 3, .. , can be generated with the RBS1(1-2*m,n) polynomials.
We define the o.g.f.s. of these polynomials by GFRBS1(z,1-2*m) = sum(RBS1(1-2*m,n)*z^(n-1), n=1..infinity) for m = 1, 2, 3, .. . The general expression of these o.g.f.s. is GFRBS1(z,1-2*m) = (-1)*RB(z,1-2*m)/(z-1)^m.
The RB(z,1-2*m) polynomials lead to a triangle that is a subtriangle of the 'double triangle' A008971. The even rows of the latter triangle are identical to the rows of our triangle.
The Maple program given below is derived from the one given in A008971.

			The first few rows of the triangle are:
[1]
[1, 1]
[1, 18, 5]
[1, 179, 479, 61]
[1, 1636, 18270, 19028, 1385]
The first few RB(z,1-2*m) polynomials are:
RB(z,-1) = 1
RB(z,-3) = z+1
RB(z,-5) = z^2+18*z+5
RB(z,-7) = z^3+179*z^2+479*z+61
The first few GFRBS1(z,1-2*m) are:
GFRBS1(z,-1) = (-1)*(1)/(z-1)
GFRBS1(z,-3) = (-1)*(z+1)/(z-1)^2
GFRBS1(z,-5) = (-1)*(z^2+18*z+5)/(z-1)^3
GFRBS1(z,-7) = (-1)*(z^3+179*z^2+479*z+61)/(z-1)^4

Cf. A160480 and A160485.
The row sums equal A010050.
This triangle is a sub-triangle of A008971.
A000340(2*n-2), A000363(2*n+2) and A000507(2*n+4) equal the second, third and fourth left hand columns.
The first right hand column equals the Euler numbers A000364.

nmax:=15; G := sqrt(1-t)/(sqrt(1-t)*cosh(x*sqrt(1-t))-sinh(x*sqrt(1-t))): Gser := simplify(series(G, x=0, nmax+1)): for m from 0 to nmax do P[m] := sort(expand(m!* coeff(Gser, x, m))) od: nmx := floor(nmax/2); for n from 0 to nmx do for k from 0 to nmx-1 do A(n+1, n+1-k) := coeff(P[2*n], t, n-k) od: od: seq(seq(A(n,m), m=1..n), n=1..nmx);

A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

Original entry on oeis.org

1, 183, 1205, 3787, 8649, 16511, 28093, 44115, 65297, 92359, 126021, 167003, 216025, 273807, 341069, 418531, 506913, 606935, 719317, 844779, 984041, 1137823, 1306845, 1491827, 1693489, 1912551, 2149733, 2405755, 2681337, 2977199, 3294061, 3632643, 3993665

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(3,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Jan 22 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A014641, A158673, A160485, A245244, A272127, A272129.

[120*n^3 + 60*n^2 + 2*n + 1: n in [0..50]];

Table[120 n^3 + 60 n^2 + 2 n + 1, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{1,183,1205,3787},40] (* Harvey P. Dale, Nov 08 2020 *)

a(n) = 120*n^3 + 60*n^2 + 2*n + 1; \\ Altug Alkan, Apr 30 2016

O.g.f.: (1 + 179*x + 479*x^2 + 61*x^3)/(1-x)^4.
E.g.f.: (1 + 182*x + 420*x^2 + 120*x^3)*exp(x).
a(n) = (2*n+1)*(60*n^2+1).
a(n) = (2*n+1) * A158673(n).
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.
See page 7 in Brent's paper:
a(n) = (2*n+1)^2*A014641(n) - 2*n*(2*n+1)*A014641(n-1).
A272127(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^6 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^6*(n - 1)/(n + 1) + 5^6*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^6*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

Original entry on oeis.org

1, 1641, 26465, 134953, 427905, 1046441, 2172001, 4026345, 6871553, 11010025, 16784481, 24577961, 34813825, 47955753, 64507745, 85014121, 110059521, 140268905, 176307553, 218881065, 268735361, 326656681, 393471585, 470046953, 557289985, 656148201, 767609441

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(4,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Jan 21 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A014641, A272126, A160485, A245244, A272129.

[1680*n^4-168*n^2+128*n+1: n in [0..50]];

Table[1680 n^4 - 168 n^2 + 128 n + 1, {n, 0, 30}]
LinearRecurrence[{5,-10,10,-5,1},{1,1641,26465,134953,427905},30] (* Harvey P. Dale, Nov 27 2017 *)

a(n) = 1680*n^4 - 168*n^2 + 128*n + 1; \\ Altug Alkan, Apr 30 2016

O.g.f.: (1+1636*x+18270*x^2+19028*x^3+1385*x^4)/(1-x)^5.
E.g.f.: (1+1640*x+11592*x^2+10080*x^3+1680*x^4)*exp(x).
a(n) = (2*n+1)*(840*n^3-420*n^2+126*n+1).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), for n>4
See page 7 in Brent's paper:
a(n) = (2*n+1)^2*A272126(n) - 2*n*(2*n+1)*A272126(n-1).
A272128(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^8 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^8*(n - 1)/(n + 1) + 5^8*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^8*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

A272128 a(n) = 30240n^5-25200n^4+5040n^3+7320n^2-2638*n+1.

Original entry on oeis.org

1, 14763, 628805, 5501167, 24943689, 79549811, 203823373, 449807415, 890712977, 1624547899, 2777745621, 4508793983, 7011864025, 10520438787, 15310942109, 21706367431, 30079906593, 40858578635, 54526858597, 71630306319, 92779195241, 118652141203

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(5,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Feb 01 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A014641, A272126, A272127.

[30240*n^5-25200*n^4+5040*n^3+7320*n^2-2638*n+1: n in [0..40]];

Table[30240 n^5 - 25200 n^4 + 5040 n^3 + 7320 n^2 - 2638 n + 1, {n, 0, 30}]

a(n)=30240*n^5-25200*n^4+5040*n^3+7320*n^2-2638*n+1 \\ Charles R Greathouse IV, Apr 30 2016

O.g.f.: (1 + 14757*x + 540242*x^2 + 1949762*x^3 + 1073517*x^4 + 50521*x^5)/(1-x)^6.
E.g.f.: (1 + 14762*x + 299640*x^2 + 609840*x^3 + 277200*x^4 + 30240*x^5)*exp(x).
a(n) = (2*n+1)*(15120*n^4-20160*n^3+12600*n^2-2640*n+1).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
See page 7 in Brent's paper: a(n) = (2*n+1)^2*A272127(n) - 2*n*(2*n+1)*A272127(n-1).
From Peter Bala, Feb 01 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^10 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^10*(n - 1)/(n + 1) + 5^10*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^10*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

cp's OEIS Frontend

A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

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A014641 Odd octagonal numbers: (2n+1)*(6n+1).

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

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A160480 The Beta triangle read by rows.

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

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A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

A272128 a(n) = 30240n^5-25200n^4+5040n^3+7320n^2-2638*n+1.