A002439 -id:A002439 - cp's OEIS frontend

A272467 E.g.f.: (sin(2x) + sin(3x)) / sin(5*x).

1, 6, 186, 14166, 2009946, 458225526, 153212718906, 70632832168086, 42939614599671066, 33282798350926545846, 32036398991671262130426, 37490905748197466281582806, 52420996658289450763331680986, 86309558223400912513674314622966, 165280246718130394753827229389826746, 364233987506387128128991081880073730326, 915234544675507984101674168382043517591706

Offset: 0

Paul D. Hanna, Apr 30 2016

nonn

			E.g.f.: A(x) = 1 + 6*x^2/2! + 186*x^4/! + 14166*x^6/6! + 2009946*x^8/8! + 458225526*x^10/10! + 153212718906*x^12/12! +...
such that A(x) = (sin(2*x) + sin(3*x)) / sin(5*x).
O.g.f.: F(x) = 1 + 6*x + 186*x^2 + 14166*x^3 + 2009946*x^4 + 458225526*x^5 + 153212718906*x^6 + 70632832168086*x^7 + 42939614599671066*x^8 +...
such that the o.g.f. can be expressed as the continued fraction:
F(x) = 1/(1 - 2*3*x/(1 - 5^2*x/(1 - 7*8*x/(1 - 10^2*x/(1 - 12*13*x/(1 - 15^2*x/(1 - 17*18*x/(1 - 20^2*x/(1 - 22*23*x/(1 - 25^2*x/(1 - 27*28*x/(1 - ...)))))))))))).

Paul D. Hanna, Table of n, a(n) for n = 0..200
P. Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)

Cf. A272158, A272468, A156185.

With[{nn=40},Take[CoefficientList[Series[(Sin[2x]+Sin[3x])/Sin[5x],{x,0,nn}],x] Range[ 0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Jun 12 2022 *)

{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (sin(2*X) + sin(3*X))/sin(5*X), 2*n)}
for(n=0, 20, print1(a(n), ", "))

{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (cos(2*X) + cos(3*X))/(1 + cos(5*X)), 2*n)}
for(n=0, 20, print1(a(n), ", "))

{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (exp(2*I*X) + exp(3*I*X))/(1 + exp(5*I*X)), 2*n)}
for(n=0, 20, print1(a(n), ", "))

E.g.f.: cos(x/2) / cos(5*x/2).
E.g.f.: (cos(2*x) + cos(3*x)) / (1 + cos(5*x)).
E.g.f.: (exp(2*i*x) + exp(3*i*x)) / (1 + exp(5*i*x)), where i^2 = -1.
E.g.f.: exp(2*i*x)/(1 + exp(5*i*x)) + exp(-2*i*x)/(1 + exp(-5*i*x)), where i^2 = -1.
O.g.f.: 1/(1 - 2*3*x/(1 - 5^2*x/(1 - 7*8*x/(1 - 10^2*x/(1 - ... - (5*n+2)*(5*n+3)*x/(1 - (5*n+5)^2*x/(1 - ...))))))), a continued fraction.
a(n) = 6 (mod 10) for n>0.
a(n) ~ (2*n)! * sqrt(2*(5 + sqrt(5))) * 5^(2*n) / Pi^(2*n+1). - Vaclav Kotesovec, Apr 30 2016
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + 4*x - 10*x/(1 - 15*x/(1 + 4*x - 70*x/(1 - 80*x/(1 + 4*x - ... - 5*n*(5*n-3)*x/(1 - 5*n*(5*n-2)*x/(1 + 4*x - ....
G.f.: 1/(1 + 9*x - 15*x/(1 - 10*x/(1 + 9*x - 80*x/(1 - 70*x/(1 + 9*x - ... - 5*n*(5*n-2)*x/(1 - 5*n*(5*n-3)*x/(1 + 9*x - .... (End)

A208680 Kashaev invariant for the (7,2)-torus knot.

Original entry on oeis.org

1, 143, 58081, 48571823, 69471000001, 151763444497103, 470164385248041121, 1960764928973430783983, 10591336845363318048877441, 71933835058256664782546056463, 599982842750416411984319126244961

Offset: 1

Peter Bala, Mar 01 2012

Compare with A156370. For other Kashaev invariants see A002439 ((3,2)-torus knot), A208679 and A208681.

P. Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
K. Hikami, Volume Conjecture and Asymptotic Expansion of q-Series, Experimental Mathematics Vol. 12, Issue 3 (2003).

Cf. A002439 ((3,2)-torus knot), A156370, A208679, A208681.

E.g.f.: 1/2*sin(2*x)/cos(7*x) = x + 143*x^3/3! + 58081*x^5/5! + ....
a(n) = (-1)^n/(4*n+4)*28^(2*n+1)*sum {k = 1..28} X(k)*B(2*n+2,k/28), where B(n,x) is a Bernoulli polynomial and X(n) is a periodic function modulo 28 given by X(n) = 0 except for X(28*n+5) = X(28*n+23) = 1 and X(28*n+9) = X(28*n+19) = -1.
a(n) = 1/2*(-1)^(n+1)*L(-2*n-1,X) in terms of the associated L-series attached to the periodic arithmetical function X.
From Peter Bala, May 16 2017: (Start)
O.g.f. as continued fraction: A(x) = 1/(1 + 25*x - 6*28*x/(1 - 8*28*x/(1 + 25*x -...- n*(7*n-1)*28*x/(1 - n*(7*n+1)*28*x/(1 + 25*x - ... ))))).
Also,  A(x) = 1/(1 + 81*x - 8*28*x/(1 - 6*28*x/(1 + 81*x -...- n*(7*n+1)*28*x/(1 - n*(7*n-1)*28*x/(1 + 81*x - ... ))))). (End)
a(n) ~ sin(Pi/7) * 2^(4*n) * 7^(2*n-1) * n^(2*n-1/2) / (Pi^(2*n-1/2) * exp(2*n)). - Vaclav Kotesovec, May 18 2017

A125054 Central terms of triangle A125053.

Original entry on oeis.org

1, 3, 21, 327, 9129, 396363, 24615741, 2068052367, 225742096209, 31048132997523, 5252064083753061, 1071525520294178007, 259439870666594250489, 73542221109962636293083, 24125551094579137082039181, 9068240688454120376775401247, 3871645204706420218816959159969

Offset: 0

Paul D. Hanna, Nov 21 2006

nonn

Triangle A125053 is a variant of triangle A008301 (enumeration of binary trees) such that the leftmost column is the secant numbers (A000364).
Right edge of triangle A210108.
Apparently all terms (except the initial 1) have 3-valuation 1. - F. Chapoton, Aug 02 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)

Cf. A125053, A125055, A000182.

b[n_]:=n!*SeriesCoefficient[Tan[x],{x,0,n}]; Table[Sum[Binomial[n,k]*b[2*k+1],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 30 2015 *)

Binomial transform of A000182 (e.g.f. tan(x)).
a(n) = Sum_{k=0..n} A130847(n,k)*2^k. - Philippe Deléham, Jul 22 2007
G.f.: 1/(1-sqrt(x))/Q(0), where Q(k)= 1 + sqrt(x) - x*(2*k+1)*(2*k+2)/(1 - sqrt(x) - x*(2*k+2)*(2*k+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
G.f.: Q(0)/(1-3*x), where Q(k) = 1 - 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2/( 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2 - (1 - 8*x*k^2 - 8*x*k -3*x)*(1 - 8*x*k^2 - 24*x*k -19*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 23 2013
G.f.: Q(0)/(1-1*x), where Q(k) = 1 - (2*k+1)*(2*k+2)*x/(x*(2*k+1)*(2*k+2) - (1-x)/(1 - (2*k+2)*(2*k+3)*x/(x*(2*k+2)*(2*k+3) - (1-x)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013
a(n) ~ 2^(4*n+5) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+3/2)). - Vaclav Kotesovec, May 30 2015
From Peter Bala, May 11 2017: (Start)
O.g.f. as an S-fraction: A(x) = 1/(1 - 3*x/(1 - 4*x/(1 - 15*x/(1 - 16*x/(1 - 35*x/(1 - 36*x/(1 - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36, ...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,....
A(x) = 1/(1 + 3*x - 6*x/(1 - 2*x/(1 + 3*x - 20*x/(1 - 12*x/(1 + 3*x - 42*x/(1 - 30*x/(1 + 3*x - ...))))))), , where the unsigned coefficients in the partial numerators [6, 2, 20, 12, 42, 30, ...] are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. (End)

A182825 E.g.f. 1/(cos(sqrt(3)x) - sin(sqrt(3)x)/sqrt(3)).

Original entry on oeis.org

1, 1, 5, 21, 153, 1209, 12285, 140589, 1871217, 27773361, 460041525, 8363802501, 166064229513, 3570030632169, 82674532955565, 2051044762727709, 54279654050034657, 1526205561241263201, 45438086217150617445, 1427921718081647393781, 47235337785416646609273

Offset: 0

Paul Barry, Dec 05 2010

nonn

First column of A182824. Hankel transform is 4^C(n+1,2)*(A000178(n))^2.

Moments of orthogonal polynomials whose coefficient array is A182826.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Shi-Mei Ma, Jun Ma, Yeong-Nan Yeh, On certain combinatorial expansions of descent polynomials and the change of grammars, arXiv:1802.02861 [math.CO], 2018.

nn = 20; Table[n!, {n, 0, nn}] CoefficientList[Series[1/(Cos[Sqrt[3]*x] - Sin[Sqrt[3]*x]/Sqrt[3]), {x, 0, nn}], x] (* T. D. Noe, Jun 28 2011 *)

From Peter Bala, Jan 21 2011: (Start)
By comparing the e.g.f. for this sequence with the e.g.f for the type B Eulerian numbers A060187 we can show that
(1)... a(n) = B(n,w)/(1+w)^(n+1), where w = exp(2*Pi*I/3) and {B(n,x)}n>=1 = [x,x+x^2,x+6*x^2+x^3,x+23*x^2+23*x^3+x^4,...] are the type B Eulerian polynomials.
Equivalently,
(2)... a(n) = (-I*sqrt(3))^n*Sum_{k = 0..n} 2^k*k!*A039755(n,k)*(-1/2+sqrt(3)*I/6)^k,
where A039755(n,k) are the type B analogs of the Stirling numbers of the second kind. We can rewrite this as
(3)... a(n) = (-I*sqrt(3))^n*sum {k = 0..n} (-1/2+sqrt(3)*I/6)^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(4a)... a(p) = 1 (mod p) for prime p = 6*n+1,
(4b)... a(p) = -1 (mod p) for prime p = 6*n+5.
For similar results see A000111. Let u = exp(2*Pi*I/6) = 1/2+sqrt(3)/2*I be a primitive sixth root of unity.
(5)... a(n) = Sum_{k = 0..n+1} u^(n+2-2*k)*Sum_{j = 1..n+1} (-1)^(k-j)*binomial(n+1,k-j)*(2*j-1)^n. Cf. A002439. (End)
G.f.: 1/Q(0), where Q(k) = 1 - x*(2*k+1) - x^2*(2*k+2)^2/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
E.g.f.: 1/E(0), where E(k) = 1 - x/( 2*k+1 - 3*x*(2*k+1)/(3*x + 2*(k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 28 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 4*x^2*(k+1)^2/(4*x^2*(k+1)^2 - (1 -x -2*x*k)*(1 -3*x -2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 10 2013

A225147 a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.

Original entry on oeis.org

-1, 2, 5, -46, -205, 3362, 22265, -515086, -4544185, 135274562, 1491632525, -54276473326, -718181418565, 30884386347362, 476768795646785, -23657073914466766, -417370516232719345, 23471059057478981762, 465849831125196593045, -29279357851856595135406

Offset: 0

Peter Luschny, Apr 30 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Peter Luschny, Generalized Eulerian polynomials.

Cf. A000810 (real part (up to sign)), A212435 (k=2), A122045 (k=1), A002439.

Bisections are A002438 and A000191.

B := proc(n, u, k) option remember;
if n = 1 then if (u < 0) or (u >= 1) then 0 else 1 fi
else k*u*B(n-1, u, k) + k*(n-u)*B(n-1, u-1, k) fi end:
EulerianPolynomial := proc(n, k, x) local m; if x = 0 then RETURN(1) fi;
add(B(n+1, m+1/k, k)*u^m, m = 0..n); subs(u=x, %) end:
seq(Im((1-I)^(1-n)*EulerianPolynomial(n, 3, I)), n=0..19);

CoefficientList[Series[-E^(-2*x)*Sech[3*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Sep 29 2014 after Sergei N. Gladkovskii *)
Table[-6^n EulerE[n,1/6], {n,0,19}] (* Peter Luschny, Nov 16 2016 after Peter Bala *)

from mpmath import mp, polylog, im
mp.dps = 32; mp.pretty = True
def A225147(n): return im(-2*I*(1+add(binomial(n,j)*polylog(-j,I)*3^j for j in (0..n))))
[int(A225147(n)) for n in (0..19)]

a(n) = Im(-2*i*(1+Sum_{j=0..n}(binomial(n,j)*Li{-j}(i)*3^j))).
For a recurrence see the Maple program.
G.f.: conjecture -T(0)/(1+2*x), where T(k) = 1 - 9*x^2*(k+1)^2/(9*x^2*(k+1)^2 + (1+2*x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
a(n) = -(-3)^n*skp(n, 2/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014
G.f.: A225147 = -1/T(0), where T(k) = 1 + 2*x + (k+1)^2*(3*x)^2/ T(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 29 2014
E.g.f.: -exp(-2*x)*sech(3*x). - Sergei N. Gladkovskii, Sep 29 2014
a(n) ~ n! * (sqrt(3)*sin(Pi*n/2) - cos(Pi*n/2)) * 2^(n+1) * 3^n / Pi^(n+1). - Vaclav Kotesovec, Sep 29 2014
From Peter Bala, Nov 13 2016: (Start)
a(n) = - 6^n*E(n,1/6), where E(n,x) denotes the Euler polynomial of order n.
a(2*n) = (-1)^(n+1)*A002438(n); a(2*n+1) = (1/2)*(-1)^n*A002439(n). (End)

A101923 Expansion of 2 * arccot(cos(x)).

Original entry on oeis.org

1, 2, 1, -148, -3719, -20098, 5055961, 403644152, 7831409041, -2707151879398, -472143935754479, -22085804322342748, 9362259685093715401, 2995219209329323622102, 274269338931958691728681, -132963342779629343323496848, -70698673853383423350187244639

Offset: 1

Ralf Stephan, Dec 27 2004

Odd coefficients are zero.

Vincenzo Librandi, Table of n, a(n) for n = 1..250

Cf. A012494, A000364, A000464, A156138, A002439.

Cf. other sequences with a g.f. of the form sin(x)/(1 - k*sin^2(x)): A012494 (k=-1), A000364 (k=1), A000464 (k=2), A156138 (k=3), A002439 (k=4).

with(gfun):
series(sin(x)/(1-(1/2)*sin(x)^2), x, 35):
L := seriestolist(%):
seq(op(2*i, L)*(2*i-1)!, i = 1..floor((1/2)*nops(L)));
# Peter Bala, Feb 06 2017

With[{nn=40},Take[CoefficientList[Series[2ArcCot[Cos[x]],{x,0,nn}],x] Range[0,nn]!,{3,-1,2}]] (* Harvey P. Dale, Nov 17 2014 *) (* adapted by Vincenzo Librandi, Feb 07 2017 *)

2*acot(cos(x)) = Pi/2 + x^2/2! + 2*x^4/4! + x^6/6! - 148*x^8/8! - 3719*x^10/10! -...
2*atan(cos(x)) = Pi/2 - x^2/2! - 2*x^4/4! - x^6/6! + 148*x^8/8! + 3719*x^10/10! +...
G.f. sin(x)/(1 - 1/2*sin(x)^2) = x + 2*x^3/3! + x^5/5! - 148*x^7/7! - ... - Peter Bala, Feb 06 2017

More terms from Harvey P. Dale, Nov 17 2014

Signs of the data entries corrected by Peter Bala, Feb 06 2017

A156138 Q_{2n+1}(sqrt(2))/sqrt(2) (see A104035).

Original entry on oeis.org

1, 17, 901, 99917, 18991081, 5514615017, 2270974911661, 1258937450889317, 903952433274722641, 816101554527859690817, 904827968753139590344021, 1208617989532834039606507517, 1914312457105234828011498655801, 3547500444096776665586928259547417, 7604155838367549221056955383942297981

Offset: 0

N. J. A. Sloane, Nov 06 2009

			G.f. = 1 + 17*x + 901*x^2 + 99917*x^3 + 18991081*x^4 + 5514615017*x^5 + ... - _Michael Somos_, Aug 19 2018

G. C. Greubel, Table of n, a(n) for n = 0..207

Cf. A101923, A000364, A000464, A002439.

Cf. other sequences with a g.f. of the form sin(x)/(1 - k*sin^2(x)): A101923 (k=1/2), A000364 (k=1), A000464 (k=2), A002439 (k=4).

with(gfun):
series(sin(x)/(1-3*sin(x)^2), x, 30):
L := seriestolist(%):
seq(op(2*i, L)*(2*i-1)!, i = 1..floor((1/2)*nops(L)));
# Peter Bala, Feb 06 2017

With[{nmax = 50}, CoefficientList[Series[Sin[x]/(1 - 3*Sin[x]^2), {x, 0, nmax}], x]*Range[0, nmax]!][[2 ;; ;; 2]] (* G. C. Greubel, Aug 17 2018 *)

x='x+O('x^50); v=Vec(serlaplace(sin(x)/(1 - 3*sin(x)^2))); vector((#v-1)\2 ,n,v[2*n-1]) \\ G. C. Greubel, Aug 17 2018

E.g.f.: sin(x)/(1 - 3*sin(x)^2) = x + 17*x^3/3! + 901*x^5/5! + 99917*x^7/7! + ... - Peter Bala, Feb 06 2017

A156187 Numerator of Euler(n, 1/6).

Original entry on oeis.org

1, -1, -5, 23, 205, -1681, -22265, 257543, 4544185, -67637281, -1491632525, 27138236663, 718181418565, -15442193173681, -476768795646785, 11828536957233383, 417370516232719345, -11735529528739490881, -465849831125196593045, 14639678925928297567703

Offset: 0

N. J. A. Sloane, Nov 07 2009

The (unsigned) odd-indexed terms of the sequence appear to give A002439. - Peter Bala, May 16 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p. See Eq. (13).

For denominators see A156189. Cf. A002439.

Numerator[EulerE[Range[0,20],1/6]] (* Vincenzo Librandi, May 04 2012 *)

A273033 E.g.f.: (sin(3x) + sin(4x)) / sin(7*x).

Original entry on oeis.org

1, 12, 732, 109332, 30406812, 13587056052, 8904250650492, 8045727017033172, 9586782871360007772, 14564334832981893064692, 27477080512619965247054652, 63024425641459625896776174612, 172720667970739808701108304367132, 557383361208023769780400587942586932, 2092050338949043346342979863638489321212, 9036239176876728629700436615577988154925652

Offset: 0

Paul D. Hanna, May 13 2016

nonn

			E.g.f.: A(x) = 1 + 12*x^2/2! + 732*x^4/4! + 109332*x^6/6! + 30406812*x^8/8! + 13587056052*x^10/10! + 8904250650492*x^12/12! +...
such that A(x) = (sin(3*x) + sin(4*x)) / sin(7*x).
O.g.f.: F(x) = 1 + 12*x + 732*x^2 + 109332*x^3 + 30406812*x^4 + 13587056052*x^5 + 8904250650492*x^6 + 8045727017033172*x^7 +...
such that the o.g.f. can be expressed as the continued fraction:
F(x) = 1/(1 - 3*4*x/(1 - 7^2*x/(1 - 10*11*x/(1 - 14^2*x/(1 - 17*18*x/(1 - 21^2*x/(1 - 24*25*x/(1 - 28^2*x/(1 - 31*32*x/(1 - 35^2*x/(1 - 38*39*x/(1 - ...)))))))))))).

P. Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)

Cf. A272158, A272467, A273031, A273032, A156196.

With[{nn=40},Take[CoefficientList[Series[(Sin[3x]+Sin[4x])/Sin[7x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Sep 23 2019 *)

{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (sin(3*X) + sin(4*X))/sin(7*X), 2*n)}
for(n=0, 20, print1(a(n), ", "))

{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (cos(3*X) + cos(4*X))/(1 + cos(7*X)), 2*n)}
for(n=0, 20, print1(a(n), ", "))

{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (exp(3*I*X) + exp(4*I*X))/(1 + exp(7*I*X)), 2*n)}
for(n=0, 20, print1(a(n), ", "))

E.g.f.: cos(x/2) / cos(7*x/2).
E.g.f.: (cos(3*x) + cos(4*x)) / (1 + cos(7*x)).
E.g.f.: (exp(3*i*x) + exp(4*i*x)) / (1 + exp(7*i*x)), where i^2 = -1.
E.g.f.: exp(3*i*x)/(1 + exp(7*i*x)) + exp(-3*i*x)/(1 + exp(-7*i*x)), where i^2 = -1.
O.g.f.: 1/(1 - 3*4*x/(1 - 7^2*x/(1 - 10*11*x/(1 - 14^2*x/(1 - ... - (7*n+3)*(7*n+4)*x/(1 - (7*n+7)^2*x/(1 - ...))))))), a continued fraction.
a(n) ~ (2*n)! * 4*cos(Pi/14) * 7^(2*n) / Pi^(2*n+1). - Vaclav Kotesovec, May 14 2016
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + 9*x - 21*x/(1 - 28*x/(1 + 9*x - 140*x/(1 - 154*x/(1 + 9*x - ... - 7*n*(7*n-4)*x/(1 - 7*n*(7*n-3)*x/(1 + 9*x - ...
G.f.: 1/(1 + 16*x - 28*x/(1 - 21*x/(1 + 16*x - 154*x/(1 - 140*x/(1 + 16*x - ... - 7*n*(7*n-3)*x/(1 - 7*n*(7*n-4)*x/(1 + 16*x - .... (End)

A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).

Original entry on oeis.org

0, -15, -3, 2, 3, 6, -7, 5, 12, 15, 20, 9, 21, 30, 35, 42, 33, 45, 56, 63, 72, 65, 77, 90, 99, 110, 105, 117, 132, 143, 156, 153, 165, 182, 195, 210, 209, 221, 240, 255, 272, 273, 285, 306, 323, 342, 345, 357, 380, 399, 420, 425, 437

Offset: 0

Paul Curtz, Jun 05 2013

The sequence is the fifth row of the following array:
0,   6,  20, 42, 72, 110, 156, 210, 272, ...   A002943
0,   3,   6, 15, 20,  35,  42,  63,  72, ...   bisections A002943, A000466
0,   2,   3,  6, 12,  15,  20,  30,  35, ...   A226023 (trisections A002943, A000466, A002439)
0,  -3,   2,  3,  6,   5,  12,  15,  20, ...   A214297 (quadrisections A078371)
0, -15,  -3,  2,  3,   6,  -7,   5,  12, ...   a(n)
0, -63, -15, -3,  2,   3,   6, -55,  -7, ...
The principle of construction is that (i) the lower left triangular portion has constant values down the diagonals (6, 3, 2, -3, -15, ...), defined from row 4 on by the negated values of A024036. (ii) The extension along the rows is defined by maintaining bisections, trisections, quadrisections etc of the form (2*n+x)*(2*n+y) with some constants x and y. In the fifth line this needs the quintisections shown in the NAME.
Each row in the array has the subsequences of the previous row plus another subsequence of the format (2*n+1)*(2*n+y) shuffled in; the first A002943, the second also A000466, the third also A002439, the fourth also A078371, and the fifth (2*n+3)*(2*n-5).
Only the first three rows are monotonically increasing everywhere.
a(n) is divisible by A226203(n).
Numerators of: 0, -15/4, -3/4, 2/9, 3/16, 6/25, -7/36, 5/36, 12/49, 15/64, 20/81, ... = a(n)/A226096(n). A permutation of A225948(n+1)/A226008(n+1).
Is the sequence increasing monotonically from 221 on?

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2,-2,0,0,0,-1,1).

Cf. A000466, A002439, A002939, A002943, A024036, A078371, A145923.

Cf. A214297, A225948, A226008, A226023, A226096, A226203.

R:=PowerSeriesRing(Integers(), 60); [0] cat Coefficients(R!( -x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9)/((1-x^5)^2*(1-x)) )); // G. C. Greubel, Mar 23 2024

CoefficientList[Series[x*(15 - 12*x - 5*x^2 - x^3 - 3*x^4 - 17*x^5 + 12*x^6 + 3*x^7 - x^8 + x^9)/((x^4 + x^3 + x^2 + x + 1)^2*(x - 1)^3), {x, 0, 80}], x] (* Wesley Ivan Hurt, Oct 03 2017 *)

def A226379_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( -x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9)/((1-x^5)^2*(1-x)) ).list()
A226379_list(50) # G. C. Greubel, Mar 23 2024

4*a(n) = A226096(n) - period 5: repeat [1, 64, 16, 1, 4].
G.f.: x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9) / ( (x^4+x^3+x^2+x+1)^2 *(x-1)^3 ). - R. J. Mathar, Jun 13 2013
a(n) = a(n-1)+2*a(n-5)-2*a(n-6)-a(n-10)+a(n-11) for n > 10. - Wesley Ivan Hurt, Oct 03 2017

cp's OEIS Frontend

A272467 E.g.f.: (sin(2*x) + sin(3*x)) / sin(5*x).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A208680 Kashaev invariant for the (7,2)-torus knot.

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A125054 Central terms of triangle A125053.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A182825 E.g.f. 1/(cos(sqrt(3)*x) - sin(sqrt(3)*x)/sqrt(3)).

Views

Author

Keywords

Comments

Links

Programs

Mathematica

Formula

A225147 a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

A101923 Expansion of 2 * arccot(cos(x)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A156138 Q_{2n+1}(sqrt(2))/sqrt(2) (see A104035).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A156187 Numerator of Euler(n, 1/6).

Views

Author

Keywords

A272467 E.g.f.: (sin(2x) + sin(3x)) / sin(5*x).

A182825 E.g.f. 1/(cos(sqrt(3)x) - sin(sqrt(3)x)/sqrt(3)).

A273033 E.g.f.: (sin(3x) + sin(4x)) / sin(7*x).

A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).