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.

A096942 Fifth column of (1,5)-Pascal triangle A096940.

Original entry on oeis.org

5, 21, 55, 115, 210, 350, 546, 810, 1155, 1595, 2145, 2821, 3640, 4620, 5780, 7140, 8721, 10545, 12635, 15015, 17710, 20746, 24150, 27950, 32175, 36855, 42021, 47705, 53940, 60760, 68200, 76296, 85085, 94605, 104895, 115995, 127946, 140790, 154570, 169330, 185115
Offset: 0

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Author

Wolfdieter Lang, Jul 16 2004

Keywords

Comments

If Y is a 5-subset of an n-set X then, for n>=8, a(n-8) is the number of 4-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

Crossrefs

Fourth column: A096941; sixth column: A096943.

Programs

  • Magma
    [(n + 20)*Binomial(n + 3, 3) div 4: n in [0..50]]; // Vincenzo Librandi, Oct 01 2013
  • Mathematica
    Table[(n + 20) Binomial[n + 3, 3]/4, {n, 0, 100}]
    CoefficientList[Series[(5 - 4 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Oct 01 2013 *)

Formula

a(n) = (n+20)*binomial(n+3, 3)/4 = 5*b(n) - 4*b(n-1), with b(n) = A000332(n+4) = binomial(n+4, 4).
G.f.: (5-4*x)/(1-x)^5.
a(n) = Sum_{k=1..n} (Sum_{i=1..k} i*(n-k+5)). - Wesley Ivan Hurt, Sep 26 2013

A096944 Seventh column of (1,5)-Pascal triangle A096940.

Original entry on oeis.org

5, 31, 112, 308, 714, 1470, 2772, 4884, 8151, 13013, 20020, 29848, 43316, 61404, 85272, 116280, 156009, 206283, 269192, 347116, 442750, 559130, 699660, 868140, 1068795, 1306305, 1585836, 1913072, 2294248, 2736184, 3246320, 3832752
Offset: 0

Views

Author

Wolfdieter Lang, Jul 16 2004

Keywords

Comments

If Y is a 5-subset of an n-set X then, for n>=10, a(n-10) is the number of 6-subsets of X having at most one element in common with Y. > - Milan Janjic, Dec 08 2007

Crossrefs

Sixth column: A096943; eighth column: A096945.

Formula

G.f.: (5-4*x)/(1-x)^7.
a(n)= (n+30)*binomial(n+5, 5)/6 = 5*b(n)-4*b(n-1), with b(n):=A000579(n+6)=binomial(n+6, 6).

A374452 Iterated rascal triangle R3: T(n,k) = Sum_{m=0..3} binomial(n-k,m)*binomial(k,m).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 69, 56, 28, 8, 1, 1, 9, 36, 84, 121, 121, 84, 36, 9, 1, 1, 10, 45, 120, 195, 226, 195, 120, 45, 10, 1
Offset: 0

Views

Author

Kolosov Petro, Jul 08 2024

Keywords

Comments

Triangle T(n,k) is the third triangle R3 among the rascal-family triangles; A077028 is triangle R1, A374378 is triangle R2.
Triangle T(n,k) equals Pascal's triangle A007318 through row 2i+1, i=2 (i.e., row 7).
Triangle T(n,k) equals Pascal's triangle A007318 through column i, i=2 (i.e., column 3).

Examples

			Triangle begins:
--------------------------------------------------
k=     0   1   2   3    4    5    6   7   8   9 10
--------------------------------------------------
n=0:   1
n=1:   1   1
n=2:   1   2   1
n=3:   1   3   3   1
n=4:   1   4   6   4    1
n=5:   1   5  10  10    5    1
n=6:   1   6  15  20   15    6    1
n=7:   1   7  21  35   35   21    7   1
n=8:   1   8  28  56   69   56   28   8   1
n=9:   1   9  36  84  121  121   84  36   9   1
n=10:  1  10  45 120  195  226  195  120  45  10  1
		

Crossrefs

Programs

  • Mathematica
    t[n_, k_] := Sum[Binomial[n - k, m]*Binomial[k, m], {m, 0, 3}]; Column[Table[t[n, k], {n, 0, 12}, {k, 0, n}], Left]

Formula

T(n,k) = 1 + k*(n-k) + 1/4*(k-1)*k*(n-k-1)*(n-k) + 1/36*(k-2)*(k-1)*k*(n-k-2)*(n-k-1)*(n-k).
Row sums give A008860(n).
Diagonal T(n+1, n) gives A000027(n).
Diagonal T(n+2, n) gives A000217(n).
Diagonal T(n+3, n) gives A000292(n).
Diagonal T(n+4, n) gives A005894(n).
Diagonal T(n+6, n) gives A247608(n).
Column k=4 difference binomial(n+8, 4) - T(n+8, 4) gives C(n+4,4)=A007318(n+4,4).
Column k=5 difference binomial(n+9, 5) - T(n+9, 5) gives sixth column of (1,5)-Pascal triangle A096943.
G.f.: (1 + 4*x^6*y^3 - 3*x*(1 + y) - 6*x^5*y^2*(1 + y) + 2*x^4*y*(2 + 7*y+ 2*y^2) + x^2*(3 + 10*y + 3*y^2) - x^3*(1 + 11*y + 11*y^2 + y^3))/((1 - x)^4*(1 - x*y)^4). - Stefano Spezia, Jul 09 2024
Showing 1-3 of 3 results.