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 1-5 of 5 results.

A090447 Triangle of partial products of binomials.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 9, 9, 1, 4, 24, 96, 96, 1, 5, 50, 500, 2500, 2500, 1, 6, 90, 1800, 27000, 162000, 162000, 1, 7, 147, 5145, 180075, 3781575, 26471025, 26471025, 1, 8, 224, 12544, 878080, 49172480, 1376829440, 11014635520, 11014635520, 1, 9, 324
Offset: 0

Views

Author

Wolfdieter Lang, Dec 23 2003

Keywords

Examples

			[1]; [1,1]; [1,2,2]; [1,3,9,9]; ...
		

Crossrefs

Column sequences: A000027 (natural numbers), A006002, A090448-9.
Cf. A090450 (row sums), A090451 (alternating row sums).
Cf. A008949 (partial row sums in Pascal's triangle).

Programs

  • Mathematica
    a[n_, m_] := Product[Binomial[n, p], {p, 0, m}]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Sep 01 2016 *)

Formula

a(n, m) = Product_{p=0..m} binomial(n, p), n>=m>=0, else 0. Partial row products in Pascal's triangle A007318.
a(n, m) = (Product_{p=0..m} fallfac(n, m-p))/superfac(m), n>=m>=0, else 0; with fallfac(n, m) := A008279(n, m) (falling factorials) and superfac(m) = A000178(m) (superfactorials).
a(n, m) = (Product_{p=0..m} (n-p)^(m-p))/superfac(m), n>=m>=0, with 0^0:=0, else 0.

A189346 Number of sets of four points on an n X n grid (or geoboard), exactly three of which are collinear.

Original entry on oeis.org

0, 0, 48, 532, 3088, 11340, 33824, 83288, 183344, 364304, 681872, 1194100, 1992976, 3182332, 4941360, 7420640, 10874720, 15539952, 21812720, 30011924, 40650368, 54187196, 71463440, 92990296, 119675712, 152314920, 192393872, 240690060
Offset: 1

Views

Author

Martin Renner, Apr 20 2011

Keywords

Comments

The four points build a triangle on an n X n grid, with one of them located on a side of the triangle.
The number of sets of four points with the three collinear points in a horizontal or vertical line is 2*n^2*(n-1)*binomial(n,3) = 4*A090448(n). The number of sets of four points with the three collinear points in a diagonal line of slope 1 is 2*n*(n-1)*binomial(n,3) + 4*Sum_{k=3..n-1}(n^2-k)*binomial(k,3). The sum of these two values is a lower bound for this sequence. - Nathaniel Johnston, Apr 23 2011

Crossrefs

Programs

  • Maple
    A189346 := proc(n)local a,b,j,k,l,m,s,slopes,num,den,tot: tot := 0: slopes := {}: for b from 1 to ceil(n/2)-1 do for a from 0 to b do slopes := slopes union {a/b}: od: od: for s from 1 to nops(slopes) do num := numer(slopes[s]): den := denom(slopes[s]): if(num = 0)then tot := tot + 2*n^2*(n-1)*binomial(n,3): elif(num = den)then tot := tot + 2*(2*add(binomial(k,3)*(n^2-k), k=3..n) - binomial(n,3)*(n^2 - n)): else for j from 1 to n - 2*den do for k from 1 to n - 2*num do tot := tot + 4*(n^2 - 3): for l from 1 to n do for m from 1 to n do if((not l = j or not m = k) and (not l = j + den or not m = k + num) and (not l = j + 2*den or not m = k + 2*num) and (m - k)*den = num*(l - j))then tot := tot - 4: fi: od: od: od: od: fi: od: return tot: end:
    seq(A189346(n),n=1..15); # Nathaniel Johnston, Apr 23 2011

Extensions

a(6)-a(28) from Nathaniel Johnston, Apr 23 2011

A120409 a(n) = n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6/(1!*2!*3!*4!*5!*6!).

Original entry on oeis.org

162000, 26471025, 1376829440, 36294822144, 600112800000, 7031325609000, 63117561830400, 457937132487120, 2790771598030416, 14702257341646875, 68449036271616000, 286552568263270400, 1093771338292039680, 3849852478998931776, 12612749124441600000
Offset: 1

Views

Author

Zerinvary Lajos, Jul 05 2006

Keywords

Crossrefs

Programs

  • Maple
    [seq(n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6/(1!*2!*3!*4!*5!*6!),n=1..27)];
  • Mathematica
    Table[(Times@@Table[(n+k)^(k+1),{k,0,5}])/Times@@(Range[6]!),{n,15}] (* Harvey P. Dale, Jun 07 2022 *)
  • Sage
    [binomial(n,1)*binomial(n,3)*binomial(n,5)*binomial(n,2)*binomial(n,4)*binomial(n,6) for n in range(6, 19)] # Zerinvary Lajos, May 17 2009

Formula

Sum_{n>=1} 1/a(n) = 789878089*Pi^2/18000 + 64687*Pi^4/150 - 16*Pi^6/21 + 6603436*zeta(3)/25 + 80136*zeta(5) - 56698539425671/64800000. - Amiram Eldar, Sep 08 2022

Extensions

Offset changed from 0 to 1 by Georg Fischer, May 08 2021

A120408 a(n) = n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5/(1!*2!*3!*4!*5!).

Original entry on oeis.org

2500, 162000, 3781575, 49172480, 432081216, 2857680000, 15219319500, 68309049600, 266863130820, 929327871472, 2937513954375, 8547581952000, 23153892070400, 58918947333120, 141893427649968, 325406324160000, 714327643354500, 1507601438758800, 3070631112865855
Offset: 1

Views

Author

Zerinvary Lajos, Jul 05 2006

Keywords

Crossrefs

Programs

  • Maple
    [seq(n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5/(1!*2!*3!*4!*5!),n=1..37)];
  • Mathematica
    Table[n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5/(1!*2!*3!*4!*5!),{n,19}] (* James C. McMahon, Oct 05 2024 *)

Extensions

Offset changed from 0 to 1 by Georg Fischer, May 08 2021
a(17)-a(19) from James C. McMahon, Oct 05 2024

A120410 a(n) = n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6*(n+6)^7/(1!*2!*3!*4!*5!*6!*7!).

Original entry on oeis.org

0, 26471025, 11014635520, 1306613597184, 72013536000000, 2320337450970000, 49989108969676800, 785820119347897920, 9577928124440387712, 94609025993497640625, 783056974947287040000, 5572874347584082739200, 34808179069805870776320, 193986366711798174329088
Offset: 0

Views

Author

Zerinvary Lajos, Jul 05 2006

Keywords

Crossrefs

Programs

  • Maple
    [seq(n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6*(n+6)^7/(1!*2!*3!*4!*5!*6!*7!),n=1..17)];
  • Mathematica
    Table[n*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6*(n+6)^7/(1!*2!*3!*4!*5!*6!*7!), {n, 0, 10}] (* Amiram Eldar, Sep 08 2022 *)

Formula

Sum_{n>=1} 1/a(n) = 422971791896349857/972000000 - 845737633741*Pi^2/22500 - 230834541*Pi^4/500 - 58492*Pi^6/15 - 18320341039*zeta(3)/1800 - 15501934*zeta(5)/5 - 5040*zeta(7). - Amiram Eldar, Sep 08 2022

Extensions

a(0) prepended by Amiram Eldar, Sep 08 2022
Showing 1-5 of 5 results.