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.

Showing 1-3 of 3 results.

A204820 a(n) = -4*a(n-1)*A001505(n-2), with a(1)=8.

Original entry on oeis.org

8, -192, 161280, -638668800, 6974263296000, -162193467211776000, 6893871130369327104000, -483949753351926762700800000, 52208499391605859160162304000000, -8200911084433448356878294712320000000
Offset: 1

Views

Author

John M. Campbell, Jan 19 2012

Keywords

Comments

Sums of coefficients from (4n+1)th moments of binomial(m,k) * binomial(3*m,k); see Maple code below.

Examples

			The evaluation of sum(binomial(n, k)*binomial(3*n, k)*k^9, k=0..n) involves the polynomial 2187*n^11+6561*n^10-45927*n^9-28431*n^8+322947*n^7-257985*n^6-473445*n^5+726003*n^4-110482*n^3-189924*n^2+52624*n-4320, the sum of the coefficients of which is -192 = a(2).

Links

Crossrefs

Cf. A203778, A015219, A202948, A202946.

Programs

  • Maple
    with(PolynomialTools); polyn:=q->expand(simplify((1/(GAMMA(n-((2*floor((q+1)/4)-1))/(2))))*(1/sqrt(3))*GAMMA(n+1/3)*GAMMA(n+2/3)*(1/3)*(1/(27^(-n)))*GAMMA(n)*1/64^n*sum(binomial(n, k)*binomial(3*n, k)*k^q, k=0..n)*(1/(GAMMA(2*n-((2*floor(q/2)-1)/(2)))))*(2^((floor((1/2)*q+1/2)-1)+q)))); coefl:=h->CoefficientList(expand(polyn(h)), n); coe:=(d, b)->coefl(d)[b];seq(sum(coe((4*d+1),b),b=1..(4*d+1)+floor(((4*d+1)+1)/4)+floor((4*d+1)/2)),d=1..6);seq(-(1/8)*GAMMA(2*n-3/2)*GAMMA(n-1/2)*(-1)^n*64^n/Pi,n=1..6);

Formula

a(n)=-(1/8)*GAMMA(2*n-3/2)*GAMMA(n-1/2)*(-1)^n*64^n/Pi

A015219 Odd tetrahedral numbers: a(n) = (4*n+1)*(4*n+2)*(4*n+3)/6.

Original entry on oeis.org

1, 35, 165, 455, 969, 1771, 2925, 4495, 6545, 9139, 12341, 16215, 20825, 26235, 32509, 39711, 47905, 57155, 67525, 79079, 91881, 105995, 121485, 138415, 156849, 176851, 198485, 221815, 246905, 273819, 302621, 333375, 366145, 400995, 437989, 477191, 518665
Offset: 0

Views

Author

Mohammad K. Azarian

Keywords

Links

Crossrefs

Cf. A000292, A000447, A001505.

Programs

  • Magma
    [(4*n+1)*(4*n+2)*(4*n+3)/6: n in [0..40]]; // Vincenzo Librandi, Jan 25 2016
  • Mathematica
    LinearRecurrence[{4, -6, 4, -1}, {1, 35, 165, 455}, 35] (* Ant King, Oct 19 2012 *)
Table[(4 n + 1) (4 n + 2) (4 n + 3)/6, {n, 0, 40}] (* Vincenzo Librandi, Jan 25 2016 *)
  • PARI
    a(n)=binomial(4*n+3,3) \\ Charles R Greathouse IV, Jan 16 2013

Formula

From Jaume Oliver Lafont, Oct 20 2009: (Start)
G.f.: (1+x)*(1+30*x+x^2)/(1-x)^4.
Sum_{n>=0} 1/a(n) = (3/2)*log(2). (End)
From Ant King, Oct 19 2012: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 64 + 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
a(n) = A000292(4*n+1). - L. Edson Jeffery, Jan 16 2013
a(n) = A000447(2*n+1). - Michel Marcus, Jan 25 2016
Sum_{n>=0} (-1)^n/a(n) = 3*(sqrt(2)-1)*Pi/4. - Amiram Eldar, Jan 04 2022
a(n) = A001505(n)/6. - R. J. Mathar, Apr 17 2024
E.g.f.: exp(x)*(3 + 102*x + 144*x^2 + 32*x^3)/3. - Elmo R. Oliveira, Aug 15 2025

Extensions

More terms from Erich Friedman

A379101 Decimal expansion of log(2)/4.

Original entry on oeis.org

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

Views

Author

Stefano Spezia, Dec 15 2024

Keywords

Examples

			0.17328679513998632735430803036454414201887503359006...

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.24.2, p. 414.

Crossrefs

Cf. A001505, A002162, A010767, A016655.

Programs

  • Mathematica
    RealDigits[Log[2]/4, 10, 100][[1]]
  • PARI
    log(2)/4 \\ Amiram Eldar, Aug 19 2025

Formula

Equals log(A010767) = A016655/20. - Hugo Pfoertner, Dec 15 2024
From Amiram Eldar, Aug 19 2025: (Start)
Equals -Sum_{k>=0} zeta(2*k)/(2^(2*k+1)*(2*k+1)).
Equals Sum_{k>=0} 1/((4*k + 1)*(4*k + 2)*(4*k + 3)) = Sum_{k>=0} 1/A001505(k). (End)
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.