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.

A129533 Array read by antidiagonals: T(n,k) = binomial(n+1,2)*binomial(n+k,n+1) for 0 <= k <= n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 6, 12, 6, 0, 0, 10, 30, 30, 10, 0, 0, 15, 60, 90, 60, 15, 0, 0, 21, 105, 210, 210, 105, 21, 0, 0, 28, 168, 420, 560, 420, 168, 28, 0, 0, 36, 252, 756, 1260, 1260, 756, 252, 36, 0, 0, 45, 360, 1260, 2520, 3150, 2520, 1260, 360, 45, 0, 0, 55, 495
Offset: 0

Views

Author

Emeric Deutsch, Apr 22 2007

Keywords

Comments

Previous name was: Triangle read by rows: T(n,k)=derivative of the q-binomial coefficient [n,k] evaluated at q=1 (0<=k<=n). - N. J. A. Sloane, Jan 06 2016
For example, T(5,2)=30 because [5,2] = q^6 + q^5 + 2*q^4 + 2*q^3 + 2*q^2 + q + 1 with derivative 6q^5 + 5q^4 + 8q^3 + 6q^2 + 4q + 1, having value 30 at q=1. - Emeric Deutsch, Apr 22 2007
Sum of entries in n-th antidiagonal = n(n-1)2^(n-3) = A001788(n-1).
T(n,k) = A094305(n-2, k-1) for n >= 2, k >= 1.
T(n,k) is total number of pips on a set of generalized linear dominoes with n cells (rather than two) and with the number of pips in each cell running from 0 to k (rather than 6). T(2,6) = 168 gives the total number of pips on a standard set of dominoes. We regard a generalized linear domino with n cells and up to k pips per cell as an ordered n-tuple [i_1, i_2, ..., i_n] with 0 <= i_1 <= i_2 <= ... <= i_n <= k. - Alan Shore and N. J. A. Sloane, Jan 06 2016
T(n,k) can also be written more symmetrically as the trinomial coefficient (n+k; n-1, k-1, 2). - N. J. A. Sloane, Jan 06 2016
As a triangle read by rows, T(n,k) is the total number of inversions over all length n binary words having exactly k 1's. T(n,k) is also the total area above all North East lattice paths from the origin to the point (k,n-k). - Geoffrey Critzer, Mar 22 2018

Examples

			Array begins:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... (A000004)
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, ... (A000217)
0, 3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, ... (A027480)
0, 6, 30, 90, 210, 420, 756, 1260, 1980, 2970, 4290, ... (A033487)
0, 10, 60, 210, 560, 1260, 2520, 4620, 7920, 12870, ... (A266732)
0, 15, 105, 420, 1260, 3150, 6930, 13860, 25740, ... (A240440)
0, 21, 168, 756, 2520, 6930, 16632, 36036, ... (A266733)
...
If regarded as a triangle, this begins:
  0;
  0,  0;
  0,  1,  0;
  0,  3,  3,  0;
  0,  6, 12,  6,  0;
  0, 10, 30, 30, 10,  0;
  0, 15, 60, 90, 60, 15, 0;
  ...

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.

Links

Crossrefs

Cf. A001788.
Rows give A000004, A000217, A027480, A033487, A266732, A240440, A266733.
A128503 and A094305 are very similar sequences.

Programs

  • Maple
    dd:=proc(n,m) if m=0 or n=0 then 0 else (m+n)!/(2*(m-1)!*(n-1)!); fi; end;
f:=n->[seq(dd(n,m),m=0..30)];
for n from 0 to 10 do lprint(f(n)); od: # produces sequence as square array
T:=(n,k)->k*(k+1)*binomial(n,k+1)/2: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    Table[Table[D[Expand[FunctionExpand[QBinomial[n, k, q]]], q] /. q -> 1, {k, 0, n}], {n, 0, 15}] // Grid (* Geoffrey Critzer, Mar 22 2018 *)

Formula

T(n,k) = (1/2)*k*(k+1)*binomial(n,k+1).
G.f.: G(q,z) = qz^2/(1-z-qz)^3.

Extensions

Entry revised by N. J. A. Sloane, Jan 06 2016

A240439 Triangle T(n, k) = Numbers of ways to place k points on a triangular grid of side n so that no three of them are vertices of an equilateral triangle of any orientation. Triangle read by rows.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 6, 15, 15, 3, 1, 10, 45, 105, 114, 39, 3, 1, 15, 105, 420, 969, 1194, 654, 102, 3, 1, 21, 210, 1260, 4773, 11259, 15615, 11412, 3663, 342, 15, 1, 28, 378, 3150, 17415, 64776, 159528, 250233, 234609, 119259, 28395, 2613, 69, 1, 36, 630, 6930
Offset: 1

Views

Author

Heinrich Ludwig, Apr 05 2014

Keywords

Comments

The triangle T(n, k) is irregularly shaped: 0 <= k <= A240114(n). First row corresponds to n = 1.
The maximal number of points that can be placed on a triangular grid of side n so that no three of them form an equilateral triangle is given by A240114(n).

Examples

			The triangle begins:
  1,  1;
  1,  3,   3;
  1,  6,  15,   15,    3;
  1, 10,  45,  105,  114,    39,     3;
  1, 15, 105,  420,  969,  1194,   654,   102,    3;
  1, 21, 210, 1260, 4773, 11259, 15615, 11412, 3663, 342, 15;
There are T(5, 8) = 3 ways to place 8 points (x) on a triangular grid of side 5 under the conditions mentioned above:
          .                x                x
         x x              x .              . x
        x . x            x . .            . . x
       x . . x          x . . .          . . . x
      x . . . x        . x x x x        x x x x .

Links

Crossrefs

Cf. A240114, A240443, A084546,
column 2 is A000217,
column 3 is A050534,
column 4 is A240440,
column 5 is A240441,
column 6 is A240442.

A240441 Number of ways to place 4 points on a triangular grid of side n so that no three of these points are vertices of an equilateral triangle of any orientation.

Original entry on oeis.org

0, 0, 3, 114, 969, 4773, 17415, 52125, 135375, 315675, 676200, 1352085, 2553558, 4595934, 7937874, 13229118, 21369330, 33579450, 51487425, 77229900, 113571975, 164046795, 233117313, 326362179, 450688329, 614572413, 828333870, 1104441975, 1457859900, 1906428300
Offset: 1

Views

Author

Heinrich Ludwig, Apr 05 2014

Keywords

Links

Crossrefs

Cf. A240439, A240440, A240442.

Programs

  • Mathematica
    Table[(n^8+4*n^7-14*n^6-56*n^5+61*n^4+220*n^3-84*n^2-240*n)/384 +If[EvenQ[n],0,(6*n+3)/32],{n,1,20}] (* Vaclav Kotesovec, Apr 05 2014 after Heinrich Ludwig *)
  • PARI
    concat([0,0], Vec(-3*x^3*(x^4+31*x^3+76*x^2+31*x+1)/((x-1)^9*(x+1)^2) + O(x^100))) \\ Colin Barker, Apr 05 2014

Formula

a(n) = (n^8 + 4*n^7 - 14*n^6 - 56*n^5 + 61*n^4 + 220*n^3 - 84*n^2 - 240*n)/384 + IF(MOD(n, 2) = 1)*(6*n + 3)/32.
G.f.: -3*x^3*(x^4+31*x^3+76*x^2+31*x+1) / ((x-1)^9*(x+1)^2). - Colin Barker, Apr 05 2014

A240442 Number of ways to place 5 points on a triangular grid of side n so that no three of these points are vertices of an equilateral triangle of any orientation.

Original entry on oeis.org

0, 39, 1194, 11259, 64776, 275805, 957516, 2859768, 7606821, 18444537, 41458599, 87464157, 174846963, 333687378, 611613150, 1081890447
Offset: 3

Views

Author

Heinrich Ludwig, Apr 08 2014

Keywords

Crossrefs

Cf. A240439, A240440, A240441.

Formula

a(n) = (n^10 + 5*n^9 - 30*n^8 - 150*n^7)/3840 + O(n^6).

Extensions

a(16)-a(18) from Heinrich Ludwig, Apr 25 2014

A243142 Number of inequivalent (mod D_3) ways to place 3 points on a triangular grid of side n so that they are not vertices of an equilateral triangle of any orientation.

Original entry on oeis.org

0, 3, 19, 75, 218, 542, 1178, 2350, 4340, 7585, 12605, 20153, 31094, 46620, 68068, 97212, 136008, 186975, 252855, 337095, 443410, 576378, 740894, 942890, 1188668, 1485757, 1842113, 2267125, 2770670, 3364280, 4060040, 4871928, 5814544, 6904635, 8159643, 9599427
Offset: 2

Views

Author

Heinrich Ludwig, May 30 2014

Keywords

Links

Crossrefs

Cf. A243141, A240440, A001399, A227327, A243143, A243144.

Programs

  • Mathematica
    Drop[CoefficientList[Series[x^3*(2*x^5-5*x^4+x^3-8*x^2-7*x-3) / ((x-1)^7*(x+1)^3), {x, 0, 40}], x],2] (* Vaclav Kotesovec, May 31 2014 after Colin Barker *)
  • PARI
    concat(0, Vec(x^3*(2*x^5-5*x^4+x^3-8*x^2-7*x-3)/((x-1)^7*(x+1)^3) + O(x^100))) \\ Colin Barker, May 30 2014

Formula

a(n) = (n^6 + 3*n^5 - 5*n^4 + 6*n^3 - 68*n^2 + 72*n + IF(MOD(n, 2) = 1)*(27*n^2 - 81*n + 45))/288.
G.f.: x^3*(2*x^5-5*x^4+x^3-8*x^2-7*x-3) / ((x-1)^7*(x+1)^3). - Colin Barker, May 30 2014
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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