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.

A015013 q-factorial numbers for q=-2.

Original entry on oeis.org

1, 1, -1, -3, 15, 165, -3465, -148995, 12664575, 2165642325, -738484032825, -504384594419475, 688484971382583375, 1880252456845835197125, -10268058666835106011499625, -112158004817839862963610403875, 2450091615245711806440069272649375, 107046952761700394535173066591323843125
Offset: 0

Views

Author

Olivier Gérard

Keywords

Links

Crossrefs

Column k=2 of A384454.
Cf. A001045, A330862.

Programs

  • Magma
    I:=[1]; [n le 1 select I[n] else (((-2)^n - 1) * Self(n-1))/(-3): n in [1..18]]; // Vincenzo Librandi, Oct 26 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==(((-2)^n - 1) * a[n-1])/(-3)}, a, {n, 15}]
Table[QFactorial[n, -2], {n, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
  • PARI
    a(n) = prod(k=1, n, ((-2)^k-1)/(-3)) \\ Michel Marcus, Apr 05 2016

Formula

a(n) = Product_{k=1..n} ((-2)^k - 1) / (-2 - 1).
a(1) = 1, a(n) = (((-2)^n - 1) * a(n-1))/(-3). - Vincenzo Librandi, Oct 26 2012
a(n) = (-1)^(floor((n mod 4)/2)) * Product_{k=1..n} A001045(k). - Altug Alkan, Apr 05 2016
a(n) ~ (-1)^floor(n/2) * c * 2^(n*(n+1)/2) / 3^n, where c = Product_{k>=1} (1 - 1/(-2)^k) = 1.21072413030105918013... (A330862). - Amiram Eldar, Aug 09 2025

Extensions

a(0)=1 prepended by Seiichi Manyama, May 30 2025

A216206 a(n) = Product_{i=1..n} ((-2)^i-1).

Original entry on oeis.org

1, -3, -9, 81, 1215, -40095, -2525985, 325852065, 83092276575, -42626337882975, -43606743654283425, 89350217747626737825, 365889141676531491393375, -2997729737755822508985921375, -49111806293653640164716349886625, 1609344780436736134557590069434814625
Offset: 0

Views

Author

R. J. Mathar, Mar 12 2013

Keywords

Comments

Signed partial products of A062510. This implies that all terms from a(1) on are multiples of 3.

Crossrefs

Cf. A005329, A015013, A015109, A027871, A062510, A330862.

Programs

  • Maple
    A216206 := proc(n)
        mul( (-2)^i-1, i=1..n) ;
end proc:
  • Mathematica
    Table[(-1)^n QPochhammer[-2, -2, n], {n, 0, 15}] (* Bruno Berselli, Mar 13 2013 *)
Table[Product[(-2)^k-1,{k,n}],{n,0,20}] (* Harvey P. Dale, Oct 21 2024 *)

Formula

A015109(n,k) = a(n)/(a(k)*a(n-k)).
a(n) = (-3)^n*A015013(n) for n>0, a(0)=1. - Bruno Berselli and Alonso del Arte, Mar 13 2013
a(n) ~ (-1)^(floor(n/2)+1) * c * 2^(n*(n+1)/2), where c = Product_{k>=1} (1 - 1/(-2)^k) = 1.21072413030105918013... (A330862). - Amiram Eldar, Aug 10 2025

A330863 Decimal expansion of Product_{k>=1} (1 + 1/(-2)^k).

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Apr 28 2020

Keywords

Examples

			(1 - 1/2) * (1 + 1/2^2) * (1 - 1/2^3) * (1 + 1/2^4) * (1 - 1/2^5) * ... = 0.568698946265428505954976737...

Crossrefs

Cf. A000593, A027637, A048651, A062510, A065446, A079555, A081845, A085011, A100221, A132020, A330862, A216206.

Programs

  • Mathematica
    RealDigits[QPochhammer[-1, -1/2]/2, 10, 110] [[1]]
N[3/QPochhammer[-2, 1/4], 120] (* Vaclav Kotesovec, Apr 28 2020 *)
  • PARI
    prodinf(k=1, 1 + 1/(-2)^k) \\ Michel Marcus, Apr 28 2020

Formula

Equals Product_{k>=1} 1/(1 + 1/2^(2*k-1)).
Equals exp(Sum_{k>=1} A000593(k)/(k*(-2)^k)).
From Peter Bala, Dec 15 2020: (Start)
Constant C = (2/3) - (1/3)*Sum_{n >= 0} (-1)^n * 2^(n^2)/( Product_{k = 1..n+1} 4^k - 1 ).
C = Sum_{n >= 0} 1/( Product_{k = 1..n} (-2)^k - 1 ) = 1 - 1/3 - 1/9 + 1/81 + 1/1215 - - + + ... = Sum_{n >= 0} 1/A216206(n).
C = 1 + Sum_{n >= 0} (-1/2)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
3*C = 2 - Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
9*C = 5 - Sum_{n >= 0} (-1/8)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
81*C = 46 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
1215*C = 691 + Sum_{n >= 0} (-1/32)^(n+1)*Product_{k = 1..n} (1 + (-1/2)^k).
The sequence [1, 2, 5, 46, 691, ...] is the sequence of numerators of the partial sums of the series Sum_{n >= 0} 1/A216206(n). (End)
Showing 1-3 of 3 results.

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