cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-16 of 16 results.

A127974 Numerators in expansion of (1-x)^(-2/3).

Original entry on oeis.org

1, 2, 5, 40, 110, 308, 2618, 7480, 21505, 559130, 1621477, 4717024, 41273960, 120646960, 353323240, 3109244512, 9133405754, 26862958100, 711868389650, 2098138411600, 6189508314220, 54821359354520, 161972198092900
Offset: 0

Views

Author

Paul Barry, Feb 09 2007

Keywords

Comments

Numerators of n!/A008544(n) are A127975.

Crossrefs

Programs

  • Mathematica
    Numerator[CoefficientList[Series[(1 - x)^(-2/3), {x, 0, 50}], x]] (* G. C. Greubel, May 07 2018 *)

Formula

a(n) = denominator(n!/A008544(n)).
a(n) = denominator(n!/(Product_{k=0..n-1} (2+3*k))).
Conjecture: a(n-1) = the numerator of (1/(2*sqrt(3)*Pi)) * Integral_{x >= 0} 1/(1 + x^3)^n. - Peter Bala, Jun 11 2024

A197374 Pi(3): fundamental real period of the Dixonian elliptic functions sm(z) and cm(z).

Original entry on oeis.org

5, 2, 9, 9, 9, 1, 6, 2, 5, 0, 8, 5, 6, 3, 4, 9, 8, 7, 1, 9, 4, 1, 0, 6, 8, 4, 9, 8, 9, 4, 5, 3, 1, 6, 1, 0, 7, 7, 1, 5, 6, 0, 5, 6, 1, 4, 6, 0, 7, 6, 7, 2, 5, 9, 0, 3, 8, 0, 7, 1, 5, 7, 2, 5, 5, 0, 6, 3, 5, 9, 0, 0, 5, 1, 8, 4, 3, 2, 3, 7, 4, 0, 8, 1, 6, 4, 6, 0, 9, 8, 0, 0, 0, 0, 1, 5, 0, 7, 6, 1, 6, 5
Offset: 1

Views

Author

Peter Bala, Mar 04 2012

Keywords

Comments

Pi(3) = 5.29991 62508 56349 87194 ... is the real period of the doubly-periodic Dixonian elliptic functions sm(z) (A104133) and cm(z) (A104134): sm(z+Pi(3)) = sm(z); cm(z+Pi(3)) = cm(z). The other period equals Pi(3)*w, where w = exp(2*I*Pi/3).

Examples

			5.2999162508563498719410684989453161077156056146...
		

References

  • A. C. Dixon, On the doubly periodic functions arising out of the curve x^3 + y^3 - 3 alpha xy = 1, Quarterly J. Pure Appl. Math. 24 (1890), 167-233.

Crossrefs

Programs

Formula

Pi(3) = 3*int {0..1} 1/(1-t^3)^(2/3) dt = B(1/3,1/3) = Gamma(1/3)^2/Gamma(2/3) = sqrt(3)/(2*Pi)*Gamma(1/3)^3.
Equals Beta(1/3,1/3) (see Shamos). - Stefano Spezia, Jun 03 2025

A381359 E.g.f. A(x) satisfies 1 - A'(x)^2 + 4*A(x)^3 = 0.

Original entry on oeis.org

1, 12, 720, 129600, 51321600, 37977984000, 47113228800000, 90796614543360000, 256892229695692800000, 1021474451008342425600000, 5513370502054734544896000000, 39267642006336798923489280000000, 360478517037545726209161953280000000, 4181620210850033164370074219315200000000
Offset: 0

Views

Author

Paul D. Hanna, Mar 06 2025

Keywords

Comments

E.g.f. A(x) equals a series trisection of the e.g.f. F(x) of A381360, which satisfies 1 + 3*F'(x)^2 - 4*F(x)^3 = 0. See A381360 for more formulas in which the trisection T1 denotes the e.g.f. A(x) of this sequence.

Examples

			E.g.f.: A(x) = x + 12*x^4/4! + 720*x^7/7! + 129600*x^10/10! + 51321600*x^13/13! + 37977984000*x^16/16! + ... + a(n)*x^(3*n+1)/(3*n+1)! + ... where A'(x) = sqrt(1 + 4*A(x)^3).
The ordinary generating function begins
O.g.f.: G(x) = 1 + 12*x + 720*x^2 + 129600*x^3 + 51321600*x^4 + 37977984000*x^5 + 47113228800000*x^6 + ...
Continued fraction representations for the o.g.f. are as follows.
O.g.f.: G(x) = 1/(1 - 12*x/(1 - 48*x/(1 - 150*x/(1 - 294*x/(1 - 576*x/(1 - 900*x/(1 - ...))))))).
O.g.f.: G(x) = 1/(1-12*x - 24^2*x^2/(1-198*x - 210^2*x^2/(1-870*x - 720^2*x^2/(1-2352*x - 1716^2*x^2/(1-4968*x - ...))))).
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A = serreverse( intformal( 1/sqrt(1 + 4*x^3 + x*O(x^(3*n+1))) ) )); (3*n+1)!*polcoef(A,3*n+1)}
    for(n=0,20, print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^(3*n+1)/(3*n+1)! satisfies the following formulas.
(1) A'(x) = sqrt(1 + 4*A(x)^3).
(2) A'(x) = 1 + 2*Series_Reversion( Integral 1/(1 + x^3)^(2/3) dx )^3.
(3.a) A'(x) = -1 - 2*A(x)/A(-x).
(3.b) A'(x) = -1 + 2*exp( 3*Integral -A(x)*A(-x) dx ).
(4.a) A''(x) = 6*A(x)^2.
(4.b) A'''(x) = 12*A(x)*A'(x).
(4.c) A''''(x) = 12 + 120*A(x)^3.
(5) A(x) = Series_Reversion( Integral 1/sqrt(1 + 4*x^3) dx ).
(6) A(x) = x + 2*Integral Series_Reversion( Integral 1/(1 + x^3)^(2/3) dx )^3 dx.
(7) A(x) = x + Integral Integral 6*A(x)^2 dx dx.
(8.a) A(x) = -x + Integral -2*A(x)/A(-x) dx.
(8.b) A(x) = -A(-x) * exp( 3*Integral -A(x)*A(-x) dx ).
(8.c) A(x) = x * exp( Integral -2/A(-x) - 1/A(x) - 1/x dx ).
(8.d) A(x) = x * exp( Integral sqrt(1 + 4*A(x)^3)/A(x) - 1/x dx ).
(9.a) A(x) = -cm(-x)*sm(-x) where cm(x) and sm(x) are Dixon elliptic functions (A104134 and A104133) satisfying cm(x)^3 + sm(x)^3 = 1.
(9.b) cm(-x) = exp( (1/2)*Integral (A'(x) - 1)/A(x) dx ).
(9.c) sm(-x) = -x*exp( (1/2)*Integral (A'(x) + 1)/A(x) - 2/x dx ).
From Thomas Scheuerle and Paul D. Hanna, Mar 07 2025: (Start)
O.g.f.: 1/T(1, x), where T(k, x) = (1 - 3*k*(3*k-1)^2*x/(1 - 3*k*(3*k+1)^2*x/T(k+1, x))) (continued fraction).
O.g.f.: 1/T(1, x), where T(k, x) = (1 - 3*(2*k - 1)*(9*k^2 - 9*k + 4)*x - ((3*k + 1)*(3*k)*(3*k - 1))^2*x^2/T(k+1, x)) (continued fraction). (End)

A381361 E.g.f. satisfies A(x) = exp( Integral abs(1/A(x)^2) dx ), where abs(F(x)) equals the series expansion formed by the unsigned coefficients in F(x).

Original entry on oeis.org

1, 1, 3, 7, 41, 225, 1435, 11815, 108945, 1062145, 12367475, 154967175, 2052583225, 30729193825, 489419016075, 8162349262375, 149689173742625, 2891193923066625, 58124967786683875, 1262035025301370375, 28655826402568055625, 674057118324247590625, 16913418325411880586875
Offset: 0

Views

Author

Paul D. Hanna, Feb 26 2025

Keywords

Comments

All terms appear to be odd.
Conjecture: for n > 4, a(n) == 0 (mod 5).

Examples

			E.g.f: A(x) = 1 + x/1! + 3*x^2/2! + 7*x^3/3! + 41*x^4/4! + 225*x^5/5! + 1435*x^6/6! + 11815*x^7/7! + 108945*x^8/8! + 1062145*x^9/9! + 12367475*x^10/10! + ...
where A(x) = exp( Integral abs(1/A(x)^2) dx ).
RELATED SERIES.
Compare the signed coefficients of 1/A(x)^2 given by
1/A(x)^2 = 1 - 2*x + 16*x^3/3! - 64*x^4/4! + 2560*x^6/6! - 20480*x^7/7! + 1884160*x^9/9! - 21299200*x^10/10! + 3604480000*x^12/12! + ...
to the unsigned coefficients in log(A(x)) given by
log(A(x)) = x + 2*x^2/2! + 16*x^4/4! + 64*x^5/5! + 2560*x^7/7! + 20480*x^8/8! + 1884160*x^10/10! + 21299200*x^11/11! + 3604480000*x^13/13! + ...
to see that log(A(x)) = Integral abs(1/A(x)^2) dx.
Let B(x) be the e.g.f. of A381360, which starts as
B(x) = A(x) = 1 + x + 2*x^2/2! + 4*x^3/3! + 12*x^4/4! + 40*x^5/5! + 160*x^6/6! + 720*x^7/7! + 3680*x^8/8! + ... + A381360(n)*x^n/n! + ...
then A(x) = B(2*x)^(1/2).
SERIES TRISECTIONS.
The series trisections of A(x) = T0 + T1 + T2 begin
T0 = 1 + 7*x^3/3! + 1435*x^6/6! + 1062145*x^9/9! + 2052583225*x^12/12! + 8162349262375*x^15/15! + 58124967786683875*x^18/18! + ...
T1 = x + 41*x^4/4! + 11815*x^7/7! + 12367475*x^10/10! + 30729193825*x^13/13! + 149689173742625*x^16/16! + 1262035025301370375*x^19/19! + ...
T2 = 3*x^2/2! + 225*x^5/5! + 108945*x^8/8! + 154967175*x^11/11! + 489419016075*x^14/14! + 2891193923066625*x^17/17! + 28655826402568055625*x^20/20! + ...
where (T2^2 + 2*T0*T1)^2 = (T0^2 + 2*T1*T2) * (T1^2 + 2*T0*T2),
also, (T2^2 - T0*T1)^2  = -2 * (T0^2 - T1*T2) * (T1^2 - T0*T2).
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A = sqrt(1 + 2*serreverse( intformal( 1/sqrt(1 + 8*x + 16*x^2 + 32*x^3/3 + x*O(x^n))) ) )); n!*polcoef(A,n)}
    for(n=0,25, print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = sqrt( 1 + 2*Series_Reversion( Integral 1/sqrt(1 + 8*x + 16*x^2 + 32*x^3/3) dx ) ).
(2) A(x) = sqrt( 1 + Integral 2*sqrt( (4*A(x)^6 - 1)/3 ) dx ).
(3) A(x) = sqrt( 1 + 2*x + Integral Integral 8*A(x)^4 dx dx ).
(4) A(x) = exp( x + Integral Integral (4*A(x)^6 + 2)/(3*A(x)^4) dx dx ).
(5) A(x)^4 = (A(x)^2)''/8.
(6.a) 0 = A'(x)^2 + A(x)*A''(x) - 4*A(x)^4.
(6.b) 0 = 1 + 3*A(x)^2*A'(x)^2 - 4*A(x)^6.
(6.c) 0 = 1 - 3*A(x)^3*A''(x) + 8*A(x)^6.
(6.d) 0 = 1 + 2*A(x)^2*A'(x)^2 - A(x)^3*A''(x).
Define the trisections of A(x) = T0 + T1 + T2 by
T0 = Sum{n>=0} a(3*n)*x^(3*n)/(3*n)!,
T1 = Sum{n>=0} a(3*n+1)*x^(3*n+1)/(3*n+1)!,
T2 = Sum{n>=0} a(3*n+2)*x^(3*n+2)/(3*n+2)!,
then these series obey the following formulas.
(7.a) (T2^2 + 2*T0*T1)^2 = (T0^2 + 2*T1*T2) * (T1^2 + 2*T0*T2).
(7.b) (T2^2 - T0*T1)^2 = -2 * (T0^2 - T1*T2) * (T1^2 - T0*T2).
(8) A(x) = 1/sqrt(1 - 2*Integral (T0 - T1) * (A(x) - 3*T2) dx).
(9) A(x) = exp( x + 2*Integral Integral A(x)^2 - (T2^2 + 2*T0*T1) dx dx ).
(10.a) T0^2 + 2*T1*T2 = ((A(x)*A'(x) - 1)/A(x)^2)'/4.
(10.b) T2^2 + 2*T0*T1 = A(x)^2 - (A'(x)/A(x))'/2.
(10.c) T1^2 + 2*T0*T2 = ((A(x)*A'(x) + 1)/A(x)^2)'/4.
(11.a) T0^2 + 2*T1*T2 = A(x)^2/3 + (1 + 3*A(x)*A'(x))/(6*A(x)^4).
(11.b) T2^2 + 2*T0*T1 = A(x)^2/3 - 1/(3*A(x)^4).
(11.c) T1^2 + 2*T0*T2 = A(x)^2/3 + (1 - 3*A(x)*A'(x))/(6*A(x)^4).
(12) A(x) = B(2*x)^(1/2) where B(x) is the e.g.f. of A381360.

A158111 E.g.f.: sm^-1(x) = Sum_{n>=0} a(n)*x^(3n+1)/(3n+1)!; a(n) = coefficient of x^(3n+1)/(3n+1)! in the Maclaurin expansion of the inverse of the Dixon elliptic function sm(x,0).

Original entry on oeis.org

1, 4, 400, 179200, 216832000, 552487936000, 2554704216064000, 19415752042086400000, 225960522265801523200000, 3818732826292045742080000000, 89923520593525093134499840000000, 2854532237720860556461562920960000000, 118891267701073842176624095657984000000000
Offset: 0

Views

Author

Paul D. Hanna, Mar 18 2009

Keywords

Comments

sm(x) = sm(x,0) satisfies: Integral_{y=0..sm(x,0)} dy/(1-y^3)^(2/3) = x.

Examples

			E.g.f.: 1/(1-x^3)^(2/3) = 1 + 4*x^3/3! + 400*x^6/6! + 179200*x^9/9! + ...
E.g.f.: sm^-1(x) = x + 4*x^4/4! + 400*x^7/7! + 179200*x^10/10! + ...
sm(x) = x - 4*x^4/4! + 160*x^7/7! - 20800*x^10/10! + 6476800*x^13/13! + ...
		

Crossrefs

Programs

  • Maple
    a(n):= mul(k-0^(mod(k,3)),k=1..3*n):seq(a(n), n = 0 .. 12);
    # Peter Bala, Feb 22 2015
  • Mathematica
    Join[{1},Table[Product[(3k-2)(3k-1)^2,{k,n}],{n,14}]] (* Harvey P. Dale, May 19 2012 *)
    a[k_] := Pochhammer[2/3, k] (3 k)!/k!; Array[a, 15, 0] (* Jan Mangaldan, Jan 06 2017 *)
  • PARI
    a(n)=prod(k=1,n,(3*k-2)*(3*k-1)^2)

Formula

a(n) = Product_{k=1..n} (3k-2)*(3k-1)^2 for n > 0 with a(0)=1.
E.g.f.: Sum_{n>=0} a(n)*x^(3m)/(3m)! = 1/(1-x^3)^(2/3).
From Peter Bala, Feb 22 2015: (Start)
a(n) = (n - 1/3)! * (3*n)!/( (-1/3)! * n! ).
a(n) = Product {k = 1..3*n} (k - 0^(k mod 3)), where we apply the usual convention that 0^0 = 1. Cf. A255406. (End)
a(n) ~ Gamma(1/3) * 3^(3*n + 1) * n^(3*n + 1/6) / (sqrt(2*Pi) * exp(3*n)). - Vaclav Kotesovec, Apr 10 2018

Extensions

More terms from Harvey P. Dale, May 19 2012

A261745 Decimal expansion of -sm(-1), where sm(t) is the Dixonian elliptic function sm(t).

Original entry on oeis.org

1, 2, 0, 5, 4, 1, 5, 1, 5, 1, 4, 0, 2, 9, 8, 3, 1, 5, 4, 8, 3, 1, 4, 1, 1, 3, 7, 5, 7, 8, 4, 4, 8, 8, 0, 1, 2, 0, 7, 2, 7, 0, 4, 1, 9, 1, 8, 8, 2, 2, 4, 9, 5, 8, 1, 0, 9, 3, 2, 7, 1, 8, 2, 3, 5, 4, 4, 7, 6, 4, 8, 8, 1, 0, 6, 5, 5, 1, 1, 2, 5, 5, 6, 3, 2, 1, 7, 0, 3, 6, 5, 2, 2, 9, 3, 7, 9, 8, 9, 0, 8, 0, 6, 7, 9
Offset: 1

Views

Author

Jean-François Alcover, Aug 30 2015

Keywords

Comments

In the context of particle physics and in the case of a Yule branching process with two types of particles, this constant appears in the asymptotic expression of the probability that all particles be of the second type at time t as exp(-t)*smh(1).

Examples

			1.205415151402983154831411375784488012072704191882249581...
		

Crossrefs

Programs

  • Mathematica
    sm[z_] := 6*WeierstrassP[z, {0, 1/27}]/(1 - 3*WeierstrassPPrime[z, {0, 1/27}]); N[-sm[-1], 105] // RealDigits // First
    (* or, without using the Weierstrass P function: *) nint[y_?NumericQ] := NIntegrate[1/(1 + w^3)^(2/3), {w, 0, y}, WorkingPrecision -> 105]; smh[t_] := y /. FindRoot[nint[y] == t, {y, t}, WorkingPrecision -> 105]; N[smh[1], 105] // RealDigits // First
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