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.

A081182 5th binomial transform of (0,1,0,2,0,4,0,8,0,16,...).

Original entry on oeis.org

0, 1, 10, 77, 540, 3629, 23870, 155233, 1003320, 6462841, 41552050, 266875157, 1713054420, 10992415589, 70523904230, 452413483753, 2902085040240, 18615340276081, 119405446835290, 765901642003037, 4912691142818700
Offset: 0

Views

Author

Paul Barry, Mar 12 2003

Keywords

Crossrefs

Binomial transform of A081181.
Cf. A081183.

Programs

  • Magma
    I:=[0, 1]; [n le 2 select I[n] else 10*Self(n-1)-23*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 06 2013
  • Mathematica
    Join[{a=0,b=1},Table[c=10*b-23*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2011 *)
    CoefficientList[Series[x / (1 - 10 x + 23 x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 06 2013 *)
    LinearRecurrence[{10,-23},{0,1},30] (* Harvey P. Dale, Jun 06 2021 *)
  • Sage
    [lucas_number1(n,10,23) for n in range(0, 21)] # Zerinvary Lajos, Apr 26 2009
    

Formula

a(n) = 10a(n-1) - 23a(n-2), a(0)=0, a(1)=1.
G.f.: x/(1 - 10x + 23x^2).
a(n) = ((5 + sqrt(2))^n - (5 - sqrt(2))^n)/(2*sqrt(2)).
a(n) = Sum_{k=0..n} C(n,2k+1) 2^k*5^(n-2k-1).
E.g.f.: exp(5*x)*sinh(sqrt(2)*x)/sqrt(2). - Ilya Gutkovskiy, Aug 12 2017

A081204 Staircase on Pascal's triangle.

Original entry on oeis.org

1, 2, 3, 10, 15, 56, 84, 330, 495, 2002, 3003, 12376, 18564, 77520, 116280, 490314, 735471, 3124550, 4686825, 20030010, 30045015, 129024480, 193536720, 834451800, 1251677700, 5414950296, 8122425444, 35240152720, 52860229080, 229911617056
Offset: 0

Views

Author

Paul Barry, Mar 11 2003

Keywords

Comments

Arrange Pascal's triangle as a square array. This sequence is then a diagonal staircase on the square array.

Crossrefs

Programs

  • Magma
    [Binomial(Ceiling((n)/2) + n, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2013
  • Mathematica
    Table[Binomial[Ceiling[(n)/2] + n, n], {n, 0, 30}] (* Vincenzo Librandi, Aug 07 2013 *)

Formula

a(n) = binomial(ceiling((n)/2) + n, n).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1, k). - Paul Barry, Jul 06 2004
Conjecture: 4*n*(n+1)*(6*n^2 - 15*n + 8)*a(n) + 6*n*(9*n-7)*a(n-1) - 3*(3*n-4)*(3*n-2)*(6*n^2-3*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 07 2014
Conjecture: 8*n^2*(n+1)*a(n) - 12*n*(83*n^2 - 313*n + 232)*a(n-1) + 6*(-9*n^3 - 377*n + 384)*a(n-2) + 9*(3*n-5)*(83*n-64)*(3*n-7)*a(n-3) = 0. - R. J. Mathar, Nov 07 2014

A081205 Staircase on Pascal's triangle.

Original entry on oeis.org

1, 3, 10, 20, 70, 126, 462, 792, 3003, 5005, 19448, 31824, 125970, 203490, 817190, 1307504, 5311735, 8436285, 34597290, 54627300, 225792840, 354817320, 1476337800, 2310789600, 9669554100, 15084504396, 63432274896, 98672427616
Offset: 0

Views

Author

Paul Barry, Mar 11 2003

Keywords

Comments

Arrange Pascal's triangle as a square array. A081204 is then a diagonal staircase on the square array. The steps are (1,3),(10,20),(70,126),(462,792),....

Crossrefs

Programs

  • Magma
    [Binomial(Ceiling((n+1)/2)+(n+1), n): n in [0..30]]; // Vincenzo Librandi Aug 07 2013
  • Mathematica
    Table[Binomial[Ceiling[(n+1)/2] + (n + 1), n], {n, 0, 30}] (* Vincenzo Librandi, Aug 07 2013 *)

Formula

a(n) = binomial(ceiling((n+1)/2)+(n+1), n).
Conjecture: 4*n*(n-1)*(n+4)*(6*n^2-9*n-1)*a(n) +6*(n-1)*(27*n^2+29*n+4)*a(n-1) -3*n*(3*n-1)*(3*n+1)*(6*n^2+3*n-4)*a(n-2)=0. - R. J. Mathar, Nov 19 2014
Showing 1-3 of 3 results.