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 1-4 of 4 results.

A084631 a(n) = floor(C(n+8,8)/C(n+2,2)).

Original entry on oeis.org

1, 3, 7, 16, 33, 61, 107, 178, 286, 442, 663, 969, 1384, 1938, 2664, 3605, 4807, 6325, 8222, 10571, 13455, 16965, 21206, 26295, 32364, 39556, 48032, 57970, 69564, 83028, 98595, 116522, 137085, 160585, 187349, 217730, 252109, 290895, 334529, 383484
Offset: 0

Views

Author

Paul Barry, Jun 01 2003

Keywords

Crossrefs

Cf. A084626.

Programs

  • Magma
    [Floor(Binomial(n+8, 6)/28): n in [0..50]]; // G. C. Greubel, Mar 23 2023
    
  • Mathematica
    Floor[Binomial[Range[0,50] +8, 6]/28] (* G. C. Greubel, Mar 23 2023 *)
  • SageMath
    [binomial(n+8,6)//28 for n in range(51)] # G. C. Greubel, Mar 23 2023

Formula

a(n) = 1 + floor(n*(n+11)*(n^4 +22*n^3 +203*n^2 +902*n +2232)/20160).

A084627 a(n) = floor(C(n+6,6)/C(n+3,3)).

Original entry on oeis.org

1, 1, 2, 4, 6, 8, 11, 14, 18, 22, 28, 34, 40, 48, 57, 66, 77, 88, 101, 115, 130, 146, 163, 182, 203, 224, 248, 272, 299, 327, 357, 388, 421, 456, 494, 533, 574, 617, 662, 709, 759, 810, 864, 921, 980, 1041, 1105, 1171, 1240, 1311, 1386, 1463, 1542, 1625, 1711
Offset: 0

Views

Author

Paul Barry, Jun 01 2003

Keywords

Crossrefs

Programs

  • Magma
    [Floor(Binomial(n+6,3)/20): n in [0..70]]; // G. C. Greubel, Mar 24 2023
    
  • Mathematica
    Table[Floor[Binomial[n+6,6]/Binomial[n+3,3]],{n,0,60}] (* Harvey P. Dale, Feb 07 2015 *)
    Floor[Binomial[Range[6,76],3]/20] (* G. C. Greubel, Mar 24 2023 *)
  • SageMath
    [binomial(n+6,3)//20 for n in range(71)] # G. C. Greubel, Mar 24 2023

Formula

a(n) = 1 + floor(n*(n^2 +15*n +74)/120).
a(n) = floor(binomial(n+6,3)/20). - G. C. Greubel, Mar 24 2023

A084624 a(n) = floor(C(n+5,5)/C(n+2,2)).

Original entry on oeis.org

1, 2, 3, 5, 8, 12, 16, 22, 28, 36, 45, 56, 68, 81, 96, 114, 133, 154, 177, 202, 230, 260, 292, 327, 365, 406, 449, 496, 545, 598, 654, 714, 777, 843, 913, 988, 1066, 1148, 1234, 1324, 1419, 1518, 1621, 1729, 1842, 1960, 2082, 2210, 2342, 2480, 2623, 2772, 2926
Offset: 0

Views

Author

Paul Barry, Jun 01 2003

Keywords

Crossrefs

Programs

  • Magma
    [Floor(Binomial(n+5,3)/10): n in [0..60]]; // G. C. Greubel, Mar 24 2023
    
  • Mathematica
    LinearRecurrence[{3,-3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-3,3,-1},{1,2,3, 5,8,12,16,22,28,36,45,56,68,81,96,114,133,154,177,202,230, 260,292},53] (* Ray Chandler, Jul 17 2015 *)
    Table[Floor[Binomial[n+5,5]/Binomial[n+2,2]],{n,0,60}] (* or *) Table[ Floor[((3+n)(4+n)(5+n))/60],{n,0,60}] (* Harvey P. Dale, Sep 04 2017 *)
    Floor[Binomial[Range[5,65],3]/10] (* G. C. Greubel, Mar 24 2023 *)
  • SageMath
    [(binomial(n+5,3)//10) for n in range(61)] # G. C. Greubel, Mar 24 2023

Formula

a(n) = 1 + floor( n*(n^2 + 12*n + 47)/60 ).
From G. C. Greubel, Mar 24 2023: (Start)
a(n) = floor( binomial(n+5,3)/10 ).
G.f.: (1 -x +x^3 -x^6 +2*x^7 -2*x^8 +2*x^9 -x^10 +x^11 -x^12 +x^14 +x^15 -2*x^16 +x^17)/((1-x)^3*(1-x^20)). (End)

A084625 Binomial transform of A084624.

Original entry on oeis.org

1, 3, 8, 21, 55, 143, 366, 919, 2265, 5491, 13125, 31000, 72485, 168042, 386709, 884161, 2009742, 4543830, 10222264, 22891099, 51041560, 113359224, 250839510, 553173006, 1216070081, 2665518207, 5826533103, 12703217438, 27628250142
Offset: 0

Views

Author

Paul Barry, Jun 01 2003

Keywords

Crossrefs

Programs

  • Magma
    A084625:= func< n | (&+[Binomial(n,j)*Floor(Binomial(j+5,3)/10): j in [0..n]]) >;
    [A084625(n): n in [0..50]]; // G. C. Greubel, Mar 24 2023
    
  • Mathematica
    a[n_]:= a[n]= 2^n +Sum[Binomial[n,j]*Floor[j*(j^2+12*j+47)/60], {j,0, n}];
    Table[a[n], {n,0,50}] (* G. C. Greubel, Mar 24 2023 *)
  • SageMath
    def A084625(n): return sum(binomial(n,j)*(binomial(j+5,3)//10) for j in range(n+1))
    [A084625(n) for n in range(51)] # G. C. Greubel, Mar 24 2023

Formula

a(n) = Sum_{k=0..n} C(n, k)*floor(C(k+5, 5)/C(k+2, 2)).
a(n) = 2^n + Sum_{k=0..n} binomial(n,k)*floor(k*(k^2 +12*k +47)/60). - G. C. Greubel, Mar 24 2023
Showing 1-4 of 4 results.