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.

A045896 Denominator of n/((n+1)*(n+2)) = A026741/A045896.

Original entry on oeis.org

1, 6, 6, 20, 15, 42, 28, 72, 45, 110, 66, 156, 91, 210, 120, 272, 153, 342, 190, 420, 231, 506, 276, 600, 325, 702, 378, 812, 435, 930, 496, 1056, 561, 1190, 630, 1332, 703, 1482, 780, 1640, 861, 1806, 946, 1980, 1035, 2162, 1128, 2352, 1225, 2550, 1326, 2756, 1431
Offset: 0

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Keywords

Comments

Also period length divided by 2 of pairs (a,b), where a has period 2*n-2 and b has period n.
From Paul Curtz, Apr 17 2014: (Start)
Difference table of A026741/A045896:
0, 1/6, 1/6, 3/20, 2/15, 5/42, ...
1/6, 0, -1/60, -1/60, -1/70, -1/84, ... = 1/6, -A051712/A051713
-1/6, -1/60, 0, 1/420, 1/420, 1/504, ...
3/20, 1/60, 1/420, 0, -1/2520, -1/2520, ...
-2/15, -1/70, -1/420, -1/2520, 0, 1/13860, ...
5/42, 1/84, 1/504, 1/2520, -1/13860, 0, ...
Autosequence of the first kind. The main diagonal is A000004. The first two upper diagonals are equal. Their denominators are A000911. (End)

Crossrefs

Programs

  • Haskell
    import Data.Ratio ((%), denominator)
    a045896 n = denominator $ n % ((n + 1) * (n + 2))
    -- Reinhard Zumkeller, Dec 12 2011
    
  • Maple
    seq((n+1)*(n+2)*(3-(-1)^n)/4, n=0..20); # C. Ronaldo
    with(combinat): seq(lcm(n+1,binomial(n+2,n)), n=0..50); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[LCM[2*n + 2, n + 2]/2, {n, 0, 40}] (* corrected by Amiram Eldar, Sep 14 2022 *)
    Denominator[#[[1]]/(#[[2]]#[[3]])&/@Partition[Range[0,60],3,1]] (* Harvey P. Dale, Aug 15 2013 *)
  • PARI
    Vec((2*x^3+3*x^2+6*x+1)/(1-x^2)^3+O(x^99)) \\ Charles R Greathouse IV, Mar 23 2016

Formula

G.f.: (2*x^3+3*x^2+6*x+1)/(1-x^2)^3.
a(n) = (n+1)*(n+2) if n odd; or (n+1)*(n+2)/2 if n even = (n+1)*(n+2)*(3-(-1)^n)/4. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 16 2004
a(2*n) = A000384(n+1); a(2*n+1) = A026741(n+1). - Reinhard Zumkeller, Dec 12 2011
Sum_{n>=0} 1/a(n) = 1 + log(2). - Amiram Eldar, Sep 11 2022
From Amiram Eldar, Sep 14 2022: (Start)
a(n) = lcm(2*n+2, n+2)/2.
a(n) = A045895(n+2)/2. (End)
E.g.f.: (2 + 8*x + x^2)*cosh(x)/2 + (2 + 2*x + x^2)*sinh(x). - Stefano Spezia, Apr 24 2024

A204556 Left edge of the triangle A045975.

Original entry on oeis.org

1, 2, 9, 24, 45, 90, 133, 224, 297, 450, 561, 792, 949, 1274, 1485, 1920, 2193, 2754, 3097, 3800, 4221, 5082, 5589, 6624, 7225, 8450, 9153, 10584, 11397, 13050, 13981, 15872, 16929, 19074, 20265, 22680, 24013, 26714, 28197, 31200, 32841, 36162, 37969, 41624
Offset: 1

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Author

Reinhard Zumkeller, Jan 18 2012

Keywords

Programs

  • Haskell
    a204556 = head . a045975_row
    
  • Magma
    [n*(2*n^2-3*n+(-1)^n*(n-3)+3)/4: n in [1..50]]; // G. C. Greubel, Jun 15 2018
  • Mathematica
    Table[n*(2*n^2 - 3*n + (-1)^n*(n - 3) + 3)/4, {n, 1, 50}] (* G. C. Greubel, Jun 15 2018 *)
  • PARI
    Vec(x*(1+x+4*x^2+12*x^3+3*x^4+3*x^5)/((1+x)^3*(x-1)^4) + O(x^99)) \\ Charles R Greathouse IV, Jun 12 2015
    
  • PARI
    for(n=1, 50, print1(n*(2*n^2-3*n+(-1)^n*(n-3)+3)/4, ", ")) \\ G. C. Greubel, Jun 15 2018
    

Formula

a(n) = A045975(n,1);
a(n) = A031940(n-1) * n for n > 1;
a(n) = A204557(n) - A045895(n).
G.f.: x*(1+x+4*x^2+12*x^3+3*x^4+3*x^5) / ((1+x)^3*(x-1)^4). - R. J. Mathar, Aug 13 2012
From Colin Barker, Jan 28 2016: (Start)
a(n) = n*(2*n^2-3*n+(-1)^n*(n-3)+3)/4.
a(n) = (n^3-n^2)/2 for n even.
a(n) = (n^3-2*n^2+3*n)/2 for n odd.
(End)

A204557 Right edge of the triangle A045975.

Original entry on oeis.org

1, 4, 21, 36, 85, 120, 217, 280, 441, 540, 781, 924, 1261, 1456, 1905, 2160, 2737, 3060, 3781, 4180, 5061, 5544, 6601, 7176, 8425, 9100, 10557, 11340, 13021, 13920, 15841, 16864, 19041, 20196, 22645, 23940, 26677, 28120, 31161, 32760, 36121, 37884, 41581
Offset: 1

Views

Author

Reinhard Zumkeller, Jan 18 2012

Keywords

Programs

  • Haskell
    a204557 = last . a045975_row
    
  • Magma
    [n*(2*n^2+(3-(-1)^n)*n-(-1)^n-3)/4: n in [1..50]]; // G. C. Greubel, Jun 15 2018
  • Mathematica
    Table[n*(2*n^2+(3-(-1)^n)*n-(-1)^n-3)/4, {n, 1, 50}] (* G. C. Greubel, Jun 15 2018 *)
    LinearRecurrence[{1,3,-3,-3,3,1,-1},{1,4,21,36,85,120,217},50] (* Harvey P. Dale, Feb 20 2021 *)
  • PARI
    Vec(-x*(-1-3*x-14*x^2-6*x^3-x^4+x^5)/((1+x)^3*(x-1)^4) + O(x^100)) \\ Colin Barker, Jan 28 2016
    

Formula

a(n) = A045975(n,n);
a(n) = A079326(n+1) * n;
a(n) = A204556(n) + A045895(n).
G.f.: -x*(-1-3*x-14*x^2-6*x^3-x^4+x^5) / ((1+x)^3*(x-1)^4). - R. J. Mathar, Aug 13 2012
From Colin Barker, Jan 28 2016: (Start)
a(n) = n*(2*n^2+(3-(-1)^n)*n-(-1)^n-3)/4.
a(n) = (n^3+n^2-2*n)/2 for n even.
a(n) = (n^3+2*n^2-n)/2 for n odd.
(End)
Showing 1-3 of 3 results.