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.

A024396 a(n) = ( Product {k = 1..n} 3*k - 1 ) * ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 1) ).

Original entry on oeis.org

1, 3, 34, 294, 4996, 72612, 1661680, 34029840, 981118240, 25947526560, 902963019520, 29279156256000, 1193967167680000, 45861003136704000, 2144641818280192000, 95220827527499520000, 5023176259163442688000, 253121597596239128064000, 14869466904778827894784000
Offset: 1

Views

Author

Clark Kimberling

Keywords

Comments

Original name: s(1)*s(2)*...*s(n)*(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 3k-1 for k = 1,2,3,...

Links

Crossrefs

Cf. A008544, A024217, A193534.

Programs

  • Magma
    I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-4)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015
  • Maple
    a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-4)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # Peter Bala, Feb 20 2015
  • Mathematica
    Table[Product[3*k-1,{k,1,n}] * Sum[(-1)^(k+1)/(3*k-1),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Feb 21 2015 *)
nxt[{n_,a_,b_}]:={n+1,b,3b+a*(3n-1)^2}; NestList[nxt,{2,1,3},20][[;;,2]] (* Harvey P. Dale, Jun 07 2023 *)

Formula

From Peter Bala, Feb 20 2015: (Start)
Recurrence: a(n+1) = 3*a(n) + (3*n - 1)^2*a(n-1) with a(1) = 1 and a(2) = 3.
The triple factorial numbers A008544 satisfy the same second-order recurrence equation. This leads to the continued fraction representation a(n)/A008544(n) = 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + ... + (3*n - 1)^2/(3 )))).
Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 1) = 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + 11^2/(3 + ... ))))) due to Euler. The alternating sum has the value 1/3*( Pi/sqrt(3) - log(2) ) = A193534. Cf. A024217. (End)
a(n) ~ GAMMA(1/3) * 3^(n-1) * n^(n+1/6) * (Pi - sqrt(3)*log(2)) / (sqrt(2*Pi) * exp(n)). - Vaclav Kotesovec, Feb 21 2015

Extensions

New name from Peter Bala, Feb 20 2015
a(18)-a(19) from Vincenzo Librandi, Feb 21 2015

A113476 Decimal expansion of (log(2) + Pi/sqrt(3))/3.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Jan 08 2006

Keywords

Comments

This number is transcendental - this follows from a result of Baker (1968) on linear forms of algebraic numbers.

Examples

			0.835648848264721053337... = A073010 + A193535.

References

  • Jolley, Summation of Series, Dover (1961), eq (79) page 16.
  • Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.16

Links

Crossrefs

Cf. A007559, A016777, A024217, A073010, A193534, A193535.

Programs

  • Mathematica
    RealDigits[(Log[2]+\[Pi]/Sqrt[3])/3,10,120][[1]]  (* Harvey P. Dale, Mar 26 2011 *)
  • PARI
    1/3*(log(2)+Pi/sqrt(3))

Formula

Equals Integral_{x = 0..1} dx/(1+x^3) = Sum_{k >= 0} (-1)^k/(3*k+1) = 1 - 1/4 + 1/7 - 1/10 + 1/13 - 1/16 + ... (see A016777). - Benoit Cloitre, Alonso del Arte, Jul 29 2011
Generalized continued fraction: 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 + ... ))))) due to Euler. For a sketch proof see A024217. - Peter Bala, Feb 22 2015
Equals (1/2)*Sum_{n >= 0} n!*(3/2)^n/(Product_{k = 0..n} 3*k + 1) = (1/2)*Sum_{n >= 0} n!*(3/2)^n/A007559(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(3*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Mar 03 2024: (Start)
Equals hypergeom([1/3, 1], [4/3], -1).
Gauss's continued fraction: 1/(1 + 1/(4 + 3^2/(7 + 4^2/(10 + 6^2/(13 + 7^2/(16 + 9^2/(19 + 10^2/(22 + 12^2/(25 + 13^2/(28 + ... )))))))))). (End)
Equals (1/12) * Sum_{n >= 0} (-1/2)^n * (9*n + 7)/((3*n + 2)*(n + 1)*binomial(2*n+1/3, n+1)). - Peter Bala, Mar 05 2025

A024383 a(n) = s(1)*s(2)*...*s(n)*(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....

Original entry on oeis.org

1, 4, 41, 488, 8881, 176556, 4622745, 128838480, 4403082465, 157917434580, 6659489632905, 292097166060600, 14653855170875025, 759940716395000700, 44202442040567948025, 2645857155729629066400, 175060715455871850866625
Offset: 1

Views

Author

Clark Kimberling

Keywords

Crossrefs

Cf. A024199, A024217, A024396, A007696, A181048.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 4 else 4*a(n-1) + (4*n - 3)^2*a(n-2) end if; end:
seq(a(n), n = 0..20);
  • Mathematica
    Table[Product[4*k - 3, {k, 1, n}] * Sum[(-1)^(k+1)/(4*k - 3), {k, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Jan 02 2020 *)
  • PARI
    a(n) = prod(k=1, n, 4*k-3)*sum(k=1, n, (-1)^(k+1)/(4*k-3)); \\ Michel Marcus, Jul 06 2019

Formula

a(n) ~ (Pi^(3/2) + 2*sqrt(Pi)*log(1 + sqrt(2))) * 2^(2*n - 2) * n^(n - 1/4) / (Gamma(1/4) * exp(n)). - Vaclav Kotesovec, Jan 02 2020
From Peter Bala, Mar 21 2024: (Start)
a(n) = Product_{k = 0..n} (4*k + 1) * Sum_{k = 0..n} (-1)^k/(4*k + 1).
a(n) = 4*a(n-1) + (4*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 4.
b(n) := Product_{k = 0..n} (4*k + 1) = A007696(n+1) satisfies the same 3-term recurrence with b(0) = 1 and b(1) = 5, leading to the continued fraction expansion for the constant A181048 = Sum_{k >= 0} (-1)^k/(4*k + 1) = 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler. (End)

Extensions

More terms from Sean A. Irvine, Jul 06 2019
Showing 1-3 of 3 results.

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