A258994 -id:A258994 - cp's OEIS frontend

A000831 Expansion of e.g.f. (1 + tan(x))/(1 - tan(x)).

1, 2, 4, 16, 80, 512, 3904, 34816, 354560, 4063232, 51733504, 724566016, 11070525440, 183240753152, 3266330312704, 62382319599616, 1270842139934720, 27507470234550272, 630424777638805504, 15250953398036463616, 388362339077351014400, 10384044045105304174592

Offset: 0

N. J. A. Sloane

nonn

			(1+tan x)/(1-tan x) = 1 + 2*x/1! + 4*x^2/2! + 16*x^3/3! + 80*x^4/4! + 512*x^5/5! + ...

Alois P. Heinz, Table of n, a(n) for n = 0..432 (first 84 terms from R. J. Mathar)
William Y. C. Chen and Amy M. Fu, The Dumont Ansatz for the Eulerian Polynomials, Peak Polynomials and Derivative Polynomials, arXiv:2204.01497 [math.CO], 2022.
D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers, Adv. Appl. Math., 16 (1995), 275-296.
M. S. Tokmachev, Correlations Between Elements and Sequences in a Numerical Prism, Bulletin of the South Ural State University, Ser. Mathematics. Mechanics. Physics, 2019, Vol. 11, No. 1, 24-33.

Bisections: A002436 and A012393.

Cf. A000111, A000182, A155100, A258880, A258901, A258994.

m:=30; R:=PowerSeriesRing(Rationals(), m);
Coefficients(R!(Laplace( (1+Tan(x))/(1-Tan(x)) ))); // G. C. Greubel, Mar 21 2019; Apr 28 2023

A000831 := (1+tan(x))/(1-tan(x)) : for n from 0 to 200 do printf("%d %d ",n,n!*coeftayl(A000831,x=0,n)) ; end: # R. J. Mathar, Nov 19 2006
A000831 := n -> `if`(n=0,1,(-1)^((n^2-n)/2)*4^n*(euler(n,1/2)+euler( n,1))): # Peter Luschny, Nov 24 2010
# third Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, 2*add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Sep 02 2020

Range[0, 18]! CoefficientList[Series[(1+Tan[x])/(1-Tan[x]), {x,0,18}], x] (* Robert G. Wilson v, Apr 16 2011 *)
b[u_, o_] := b[u, o] = If[u+o == 0, 1, 2*Sum[b[o-1+j, u-j], {j, 1, u}]];
a[n_] := b[n, 0];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Dec 02 2023, after Alois P. Heinz *)

a(n):=sum(if evenp(n+k) then ((-1)^((n+k)/2)*sum(j!*stirling2(n,j)*2^(n-j+1)*(-1)^(j)*binomial(j-1,k-1),j,k,n)) else 0,k,1,n); /* Vladimir Kruchinin, Aug 19 2010 */

a(n) = if( n<1, n==0, n! * polcoeff( 1 + 2 / ( 1 / tan( x + x * O(x^n)) - 1), n)) /* Michael Somos, Apr 16 2011 */

a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( (cos(x + A) + sin(x + A)) / (cos(x + A) - sin(x + A)), n)) /* Michael Somos, Apr 16 2011 */

@CachedFunction
def sp(n,x) :
    if n == 0 : return 1
    return -add(2^(n-k)*sp(k,1/2)*binomial(n,k) for k in range(n)[::2])
A000831 = lambda n : abs(sp(n,x))
[A000831(n) for n in (0..21)]     # Peter Luschny, Jul 30 2012

def A000831_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1+tan(x))/(1-tan(x)) ).egf_to_ogf().list()
A000831_list(40) # G. C. Greubel, Mar 21 2019; Apr 28 2023

E.g.f.: tan(x+Pi/4).
a(n) = Sum_{k=1..n} (if even(n+k) ( (-1)^((n+k)/2)*Sum_{j=k..n} (j!*stirling2(n,j)*2^(n-j+1)*(-1)^(j)*binomial(j-1,k-1) ), n>0. - Vladimir Kruchinin, Aug 19 2010
a(n) = 4^n*(E_{n}(1/2) + E_{n}(1))*(-1)^((n^2-n)/2) for n > 0, where E_{n}(x) is an Euler polynomial. - Peter Luschny, Nov 24 2010
a(n) = 2^n * A000111(n). - Gerard P. Michon, Feb 24 2011
From Sergei N. Gladkovskii, Dec 01 2011 - Jan 24 2014: (Start)
Continued fractions:
E.g.f.: -1 + 2/(1-x*G(0)); G(k) = 1 - (x^2)/((x^2) - (2*k + 1)*(2*k + 3)/G(k+1)).
E.g.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 4*k+1 + x/(1+x/ (4*k+3 - x/(1- x/U(k+1)))).
E.g.f.: 1 + 2*x/(G(0)-x) where G(k) =  2*k+1 - x^2/G(k+1).
G.f.: 1 + 2*x/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/( 1 - 2*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1)).
E.g.f.: tan(2*x) + sec(2*x) = (x-1)/(x+1) - 2*(2*x^2+3)/( T(0)*3*x*(1+x)- 2*x^2-3)/(x+1), where T(k) = 1 - x^4*(4*k-1)*(4*k+7)/( x^4*(4*k-1)*(4*k+7) - (4*k+1)*(4*k+5)*(16*k^2 + 8*k - 2*x^2 - 3)*(16*k^2 + 40*k - 2*x^2 + 21)/T(k+1)).
E.g.f.: 1 + 2*x/Q(0), where Q(k) = 4*k+1 -x/(1 - x/( 4*k+3 + x/(1 + x/Q(k+1)))).
E.g.f.: tan(2*x) + sec(2*x) = 2*R(0)-1, where R(k) = 1 + x/( 4*k+1 - x/(1 - x/( 4*k+3 + x/R(k+1)))).
G.f.: 1+ G(0)*2*x/(1-2*x), where G(k) = 1 - 2*x^2*(k+1)*(k+2)/(2*x^2*(k+1)*(k+2) - (1-2*x*(k+1))*(1-2*x*(k+2))/G(k+1)). (End)
a(n) ~ n! * (4/Pi)^(n+1). - Vaclav Kotesovec, Jun 15 2015
a(0) = 1; a(n) = 2 * Sum_{k=0..n-1} binomial(n-2,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jun 11 2020

A258927 E.g.f. satisfies: A(x) = Integral 1 + A(x)^6 dx.

Original entry on oeis.org

1, 720, 410572800, 4492717498368000, 348990783113936240640000, 118162808964225967251573964800000, 130226468530398571130647349959852032000000, 384446125794905598149974467971605129718661120000000, 2644398446216951886577241780697447635225293650237849600000000

Offset: 0

Paul D. Hanna, Jun 15 2015

nonn

From Vaclav Kotesovec, Jun 17 2015: (Start)
In general, for k>2, if e.g.f. satisfies A(x) = Integral 1 + A(x)^k dx, then a(n) ~ k^(k/(k-1)) * n^(1/(k-1)) * (k*n)! * (k*sin(Pi/k)/Pi)^(k*n + k/(k-1)) / ((k-1)^(1/(k-1)) * Gamma(1/(k-1))).
(End)

			E.g.f.: A(x) = x + 720*x^7/7! + 410572800*x^13/13! + 4492717498368000*x^19/19! +...
where Series_Reversion(A(x)) = x - x^7/7 + x^13/13 - x^19/19 + x^25/25 +...
Also, A(x) = S(x)/C(x) where
S(x) = x - 120*x^7/7! - 21859200*x^13/13! - 131273353728000*x^19/19! +...+ A258926(n)*x^(6*n+1)/(6*n+1)! +...
C(x) = 1 - 120*x^6/6! - 21859200*x^12/12! - 131273353728000*x^18/18! +...+ A258926(n)*x^(6*n)/(6*n)! +...
such that C(x)^6 + S(x)^6 = 1.

Vaclav Kotesovec, Table of n, a(n) for n = 0..75

Cf. A000182(n-1) (k=2), A258880 (k=3), A258901 (k=4), A258925 (k=5), A259112 (k=7), A259113 (k=8), A258926, A258994.

/* E.g.f. Series_Reversion( Integral 1/(1+x^6) dx ): */
{a(n) = local(A=x); A = serreverse( intformal( 1/(1 + x^6 + O(x^(6*n+2))) ) ); (6*n+1)!*polcoeff(A, 6*n+1)}
for(n=0, 20, print1(a(n), ", "))

/* E.g.f. A(x) = Integral 1 + A(x)^6 dx.: */
{a(n) = local(A=x); for(i=1, n+1, A = intformal( 1 + A^6 + O(x^(6*n+2)) )); (6*n+1)!*polcoeff(A, 6*n+1)}
for(n=0, 20, print1(a(n), ", "))

E.g.f. A(x) satisfies:
(1) A(x) = Series_Reversion( Integral 1/(1+x^6) dx ).
(2) A(x)^3 = tan( 3 * Integral A(x)^2 dx ).
Let C(x) = S'(x) such that S(x) = Series_Reversion( Integral 1/(1-x^6)^(1/6) dx ) is the e.g.f. of A258926, then e.g.f. A(x) of this sequence satisfies:
(3) A(x) = S(x)/C(x),
(4) A(x) = Integral 1/C(x)^6 dx,
(5) A(x)^3 = S(x)^3/C(x)^3 = tan( 3 * Integral S(x)^2/C(x)^2 dx ).
a(n) ~ 2^(6/5) * 3^(6*n+12/5) * (6*n)! * n^(1/5) / (5^(1/5) * Gamma(1/5) * Pi^(6*n+6/5)). - Vaclav Kotesovec, Jun 18 2015

A258969 E.g.f.: A'(x) = 1 + A(x)^3, with A(0)=1.

Original entry on oeis.org

1, 2, 6, 42, 390, 4698, 69174, 1203498, 24163110, 549811962, 13982486166, 393026414922, 12099527531910, 404881353252378, 14632253175107574, 567974815524008298, 23567351945550373350, 1040985881615266375482, 48767788927611416600406, 2415210691383917131432842

Offset: 0

Vaclav Kotesovec, Jun 15 2015

nonn

Conjecture: A227250(n+1) = a(n).

			A(x) = 1 + 2*x + 6*x^2/2! + 42*x^3/3! + 390*x^4/4! + 4698*x^5/5! + ...
A'(x) = 2 + 6*x + 21*x^2 + 65*x^3 + 783*x^4/4 + 11529*x^5/20 + ...
1 + A(x)^3 = 2 + 6*x + 21*x^2 + 65*x^3 + 783*x^4/4 + 11529*x^5/20 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..200

Cf. A000831, A024396, A193534, A227250, A258970, A258971, A258880, A258994.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^3-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/((2+x)*(1+x+x^2) +x*O(x^n)) )); n!*polcoeff(A,n)}
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ (3/(Pi/sqrt(3)-log(2)))^(n+1/2) * n^n / exp(n).

E.g.f.: 1 + Series_Reversion( Integral 1/((2+x)*(1+x+x^2)) dx ). - Paul D. Hanna, Jun 16 2015

A258970 E.g.f.: A'(x) = 1 + A(x)^4, with A(0)=1.

Original entry on oeis.org

1, 2, 8, 80, 1088, 19328, 422912, 10987520, 330555392, 11300913152, 432717037568, 18344259092480, 852932666851328, 43157160112160768, 2360748463307620352, 138821061188696145920, 8732741520836633034752, 585172975239737913638912, 41612642758392039581155328

Offset: 0

Vaclav Kotesovec, Jun 15 2015

nonn

			A(x) = 1 + 2*x + 8*x^2/2! + 80*x^3/3! + 1088*x^4/4! + 19328*x^5/5! + ...
A'(x) = 2 + 8*x + 40*x^2 + 544*x^3/3 + 2416*x^4/3 + 52864*x^5/15 + ...
1 + A(x)^4 = 2 + 8*x + 40*x^2 + 544*x^3/3 + 2416*x^4/3 + 52864*x^5/15 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..100

Cf. A000831, A258969, A258971, A258994, A258901.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^4-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^4 +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ n! * d^(n+1/3) / (3^(1/3) * GAMMA(1/3) * n^(2/3)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(4*j-1) = 4*sqrt(2)/(Pi + log(3-2*sqrt(2))) = 4.10260201986929...

E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^4) dx ). - Paul D. Hanna, Jun 16 2015

A258971 E.g.f.: A'(x) = 1 + A(x)^5, with A(0)=1.

Original entry on oeis.org

1, 2, 10, 130, 2330, 54770, 1591690, 55065250, 2209888250, 100922263250, 5167670934250, 293215490277250, 18260340583516250, 1238269550334211250, 90824251513716786250, 7164531681653318001250, 604824006980892825496250, 54406894886223009690031250

Offset: 0

Vaclav Kotesovec, Jun 15 2015

nonn

In general, for k>1, if e.g.f. satisfies A'(x) = 1 + A(x)^k, with A(0)=1, then a(n) ~ n! * d^(n + 1/(k-1)) / ((k-1)^(1/(k-1)) * Gamma(1/(k-1)) * n^(1-1/(k-1))), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(k*j-1).

			A(x) = 1 + 2*x + 10*x^2/2! + 130*x^3/3! + 2330*x^4/4! + 54770*x^5/5! + ...
A'(x) = 2 + 10*x + 65*x^2 + 1165*x^3/3 + 27385*x^4/12 + 159169*x^5/12 + ...
1 + A(x)^5 = 2 + 10*x + 65*x^2 + 1165*x^3/3 + 27385*x^4/12 + 159169*x^5/12 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..80

Cf. A000831 (k=2), A258969 (k=3), A258970 (k=4), A258994 (k=6), A258925.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^5-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^5 +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ n! * d^(n+1/4) / (4^(1/4) * Gamma(1/4) * n^(3/4)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(5*j-1) = 40*sqrt(5-sqrt(5)) / (8*sqrt(2)*Pi + sqrt(5+sqrt(5)) * ((9-5*sqrt(5))*log(2) + (sqrt(5)-5)*log(7+3*sqrt(5)))) = 5.53569595526739362969262739469167643400611216649309306882558956...

E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^5) dx ). - Paul D. Hanna, Jun 16 2015

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A000831 Expansion of e.g.f. (1 + tan(x))/(1 - tan(x)).

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A258927 E.g.f. satisfies: A(x) = Integral 1 + A(x)^6 dx.

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A258969 E.g.f.: A'(x) = 1 + A(x)^3, with A(0)=1.

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A258970 E.g.f.: A'(x) = 1 + A(x)^4, with A(0)=1.

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A258971 E.g.f.: A'(x) = 1 + A(x)^5, with A(0)=1.

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