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 1-8 of 8 results.

A306680 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-1))/((1-x)^k-x^(k+1)).

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 2, 4, 1, 1, 1, 3, 5, 1, 1, 1, 2, 5, 6, 1, 1, 1, 1, 4, 8, 7, 1, 1, 1, 1, 2, 7, 13, 8, 1, 1, 1, 1, 1, 5, 12, 21, 9, 1, 1, 1, 1, 1, 2, 11, 21, 34, 10, 1, 1, 1, 1, 1, 1, 6, 21, 37, 55, 11, 1, 1, 1, 1, 1, 1, 2, 16, 37, 65, 89, 12
Offset: 0

Views

Author

Seiichi Manyama, Mar 05 2019

Keywords

Examples

			A(4,1) = A306713(4,1) = 5, A(4,2) = A306713(8,2) = 4.
Square array begins:
   1,  1,  1,  1,  1,  1, 1, 1, 1, ...
   2,  1,  1,  1,  1,  1, 1, 1, 1, ...
   3,  2,  1,  1,  1,  1, 1, 1, 1, ...
   4,  3,  2,  1,  1,  1, 1, 1, 1, ...
   5,  5,  4,  2,  1,  1, 1, 1, 1, ...
   6,  8,  7,  5,  2,  1, 1, 1, 1, ...
   7, 13, 12, 11,  6,  2, 1, 1, 1, ...
   8, 21, 21, 21, 16,  7, 2, 1, 1, ...
   9, 34, 37, 37, 36, 22, 8, 2, 1, ...
		

Crossrefs

Columns 0-9 give A000027(n+1), A000045(n+1), A005251(n+1), A003522, A005676, A099132, A293169, A306721, A306752, A306753.

Programs

  • Mathematica
    T[n_, k_] := Sum[Binomial[n - j, k*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 11}, {k, 0, n}] // Flatten (* Amiram Eldar, Jun 21 2021 *)

Formula

A(n,k) = Sum_{j=0..n} binomial(n-j,k*j).
A(n,k) = A306713(k*n,k) for k > 0.

A099785 a(n) = Sum_{k=0..floor(n/4)} C(n-k,3*k) * 2^(n-3*k).

Original entry on oeis.org

1, 2, 4, 8, 18, 48, 144, 448, 1380, 4152, 12224, 35456, 102024, 292768, 840416, 2416384, 6959504, 20069280, 57913536, 167158656, 482462752, 1392319488, 4017460224, 11590946816, 33439639616, 96470796672, 278311599616
Offset: 0

Views

Author

Paul Barry, Oct 26 2004

Keywords

Comments

In general a(n) = Sum_{k=0..floor(n/4)} C(n-k,3*k) * u^k * v^(n-4*k) has g.f. (1-v*x)^2/((1-v*x)^3 - u*x^4) and satisfies the recurrence a(n) = 3*v*a(n-1) - 3*v^2*a(n-2) + v^3*a(n-3) + u*a(n-4).

Crossrefs

Programs

  • GAP
    a:=[1,2,4,8];; for n in [5..30] do a[n]:=6*a[n-1] -12*a[n-2] + 8*a[n-3] +2*a[n-4]; od; a; # G. C. Greubel, Sep 04 2019
  • Magma
    I:=[1,2,4,8]; [n le 4 select I[n] else 6*Self(n-1) - 12*Self(n-2) + 8*Self(n-3) + 2*Self(n-4): n in [1..30]]; // G. C. Greubel, Sep 04 2019
    
  • Maple
    seq(coeff(series((1-2*x)^2/((1-2*x)^3 - 2*x^4), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Sep 04 2019
  • Mathematica
    Table[Sum[Binomial[n-k,3k]2^(n-3k),{k,0,Floor[n/4]}],{n,0,30}] (* or *) LinearRecurrence[{6,-12,8,2},{1,2,4,8},30] (* Harvey P. Dale, Apr 01 2012 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-2*x)^2/((1-2*x)^3 - 2*x^4)) \\ G. C. Greubel, Sep 04 2019
    
  • Sage
    def A099785_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P((1-2*x)^2/((1-2*x)^3 - 2*x^4)).list()
    A099785_list(30) # G. C. Greubel, Sep 04 2019
    

Formula

G.f.: (1-2*x)^2/((1-2*x)^3 - 2*x^4).
a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3) + 2*a(n-4).

A099786 a(n) = Sum_{k=0..floor(n/4)} C(n-k,3*k)*3^(n-4*k).

Original entry on oeis.org

1, 3, 9, 27, 82, 255, 819, 2727, 9397, 33312, 120537, 441855, 1631017, 6036879, 22345074, 82589247, 304612975, 1120960983, 4116353265, 15088372416, 55224373105, 201895801851, 737506551321, 2692518758163, 9826402960882
Offset: 0

Views

Author

Paul Barry, Oct 26 2004

Keywords

Comments

In general a(n) = Sum_{k=0..floor(n/4)} C(n-k,3*k) * u^k * v^(n-4*k) has g.f. (1-v*x)^2/((1-v*x)^3 - u*x^4) and satisfies the recurrence a(n) = 3*v*a(n-1) - 3*v^2*a(n-2) + v^3*a(n-3) + u*a(n-4).

Crossrefs

Programs

  • GAP
    a:=[1,3,9,27];; for n in [5..30] do a[n]:=9*a[n-1]-27*a[n-2] + 27*a[n-3] +a[n-4]; od; a; # G. C. Greubel, Sep 04 2019
  • Magma
    I:=[1,3,9,27]; [n le 4 select I[n] else 9*Self(n-1) - 27*Self(n-2) + 27*Self(n-3) +Self(n-4): n in [1..30]]; // G. C. Greubel, Sep 04 2019
    
  • Maple
    seq(coeff(series((1-3*x)^2/((1-3*x)^3 - x^4), x, n+1), x, n), n = 0..30); # G. C. Greubel, Sep 04 2019
  • Mathematica
    LinearRecurrence[{9,-27,27,1},{1,3,9,27},40] (* or *) CoefficientList[ Series[-((1-3 x)^2/(x (x (x (x+27)-27)+9)-1)),{x,0,40}],x] (* Harvey P. Dale, Jun 06 2011 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-3*x)^2/((1-3*x)^3 - x^4)) \\ G. C. Greubel, Sep 04 2019
    
  • Sage
    def A099786_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P((1-3*x)^2/((1-3*x)^3 - x^4)).list()
    A099786_list(30) # G. C. Greubel, Sep 04 2019
    

Formula

G.f.: (1-3*x)^2/((1-3*x)^3 - x^4).
a(n) = 9*a(n-1) - 27*a(n-2) + 27*a(n-3) + a(n-4).

A373906 a(n) = Sum_{k=0..floor(n/4)} binomial(n+2*k,n-4*k).

Original entry on oeis.org

1, 1, 1, 1, 2, 8, 29, 85, 212, 476, 1016, 2172, 4825, 11213, 26763, 64095, 151851, 354737, 820328, 1889968, 4361521, 10106859, 23509678, 54793282, 127709888, 297336790, 691382201, 1606284377, 3731020629, 8668253125, 20146856893, 46840732201, 108918637566, 253262275888
Offset: 0

Views

Author

Seiichi Manyama, Jun 22 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(n+2*k, n-4*k));

Formula

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6).
G.f.: 1/(1 - x - x^4/(1 - x)^5).

A096750 Expansion of (1-x+x^2)/(1-2x+2x^2-x^3-x^4).

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 6, 7, 8, 12, 21, 33, 44, 55, 76, 119, 185, 263, 351, 480, 706, 1066, 1551, 2156, 2982, 4269, 6281, 9162, 13013, 18252, 25921, 37513, 54449, 78045, 110626, 157124, 225490, 325403, 467576, 666960, 949661, 1358381, 1951976, 2803811
Offset: 0

Views

Author

Paul Barry, Jul 07 2004

Keywords

Crossrefs

Cf. A003522.

Programs

  • Mathematica
    CoefficientList[Series[(1-x+x^2)/(1-2x+2x^2-x^3-x^4),{x,0,50}],x] (* or *) LinearRecurrence[ {2,-2,1,1},{1,1,1,1},50] (* Harvey P. Dale, Jun 16 2024 *)

Formula

a(n) = 2*a(n-1)-2*a(n-2)+a(n-3)+a(n-4).

A138653 a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+a(n-4).

Original entry on oeis.org

1, 2, 4, 8, 15, 27, 48, 86, 156, 285, 521, 950, 1728, 3140, 5707, 10379, 18884, 34362, 62520, 113737, 206897, 376362, 684652, 1245504, 2265815, 4121947, 7498552, 13641134, 24815508, 45143621, 82124025, 149397854, 271780616, 494415932, 899427827, 1636214155
Offset: 0

Views

Author

Paul Curtz, May 15 2008

Keywords

Comments

Sequence is identical to its shifted third differences.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3, -3, 1, 1},{1, 2, 4, 8},14] (* Ray Chandler, Sep 23 2015 *)
  • PARI
    Vec((1-x+x^2+x^3)/(1-3*x+3*x^2-x^3-x^4) + O(x^50)) \\ Colin Barker, Oct 18 2016

Formula

G.f.: (1-x+x^2+x^3) / (1-3*x+3*x^2-x^3-x^4). - Colin Barker, Oct 18 2016

A370722 a(n) = Sum_{k=0..floor(n/7)} binomial(n-4*k,3*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 5, 11, 21, 36, 57, 85, 122, 173, 249, 371, 575, 918, 1485, 2398, 3830, 6030, 9369, 14422, 22107, 33909, 52226, 80888, 125925, 196706, 307653, 480873, 750275, 1168085, 1815191, 2817518, 4371772, 6785606, 10539893, 16384908, 25488736
Offset: 0

Views

Author

Seiichi Manyama, Feb 28 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3, -3, 1, 0, 0, 0, 1}, Table[1, 7], 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\7, binomial(n-4*k, 3*k));
    
  • PARI
    my(N=50, x='x+O('x^N)); Vec((1-x)^2/((1-x)^3-x^7))

Formula

G.f.: (1-x)^2/((1-x)^3 - x^7).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-7).

A375169 Expansion of (1 - x) / ((1 - x)^3 - x^4).

Original entry on oeis.org

1, 2, 3, 4, 6, 11, 22, 43, 80, 144, 257, 462, 839, 1532, 2798, 5099, 9274, 16855, 30640, 55728, 101393, 184490, 335659, 610628, 1110790, 2020635, 3675822, 6686979, 12164896, 22130208, 40258737, 73237462, 133231279, 242370396, 440913550, 802098203, 1459155634
Offset: 0

Views

Author

Seiichi Manyama, Aug 05 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=40, x='x+O('x^N)); Vec((1-x)/((1-x)^3-x^4))

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-4).
a(n) = Sum_{k=0..floor(n/4)} binomial(n+1-k,n-4*k).
a(n) = (n + 1)*hypergeom([(1-n)/4, (2-n)/4, (3-n)/4, -n/4], [2/3, 4/3, -1-n], -4^4/3^3). - Stefano Spezia, Jun 18 2025
Showing 1-8 of 8 results.