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.

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A291722 Number T(n,k) of permutations p of [n] such that in 0p the sum of all jumps equals k + n; triangle T(n,k), n >= 0, 0 <= k <= n*(n-1)/2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 6, 6, 5, 4, 1, 1, 1, 10, 20, 20, 26, 15, 15, 6, 5, 1, 1, 1, 15, 50, 70, 105, 106, 104, 90, 65, 51, 27, 21, 7, 6, 1, 1, 1, 21, 105, 210, 350, 497, 554, 644, 567, 574, 420, 386, 238, 203, 105, 85, 35, 28, 8, 7, 1, 1
Offset: 0

Views

Author

Alois P. Heinz, Aug 30 2017

Keywords

Comments

An up-jump j occurs at position i in p if p_{i} > p_{i-1} and j is the index of p_i in the increasingly sorted list of those elements in {p_{i}, ..., p_{n}} that are larger than p_{i-1}. A down-jump j occurs at position i in p if p_{i} < p_{i-1} and j is the index of p_i in the decreasingly sorted list of those elements in {p_{i}, ..., p_{n}} that are smaller than p_{i-1}. First index in the lists is 1 here.
From David B. Wilson, Dec 14 2018: (Start)
T(n,k) equals the number of permutations p of [n] such that twice the sum of the leftward-down-jumps of p plus the number of descents of p equals k.
T(n,k) equals the number of cover-inclusive Dyck tilings whose lower boundary is the zig-zag path of order n (UD)^n, and which have k tiles.
A leftward-down-jump j occurs at position i in p if p_{i} > p_{i+1} and there are j positions k for which k p_k > p_{i+1}.
Cover-inclusive Dyck tilings are defined in the Kenyon and Wilson link below. (End)

Examples

			T(4,0) = 1: 1234.
T(4,1) = 6: 1243, 1324, 1342, 2134, 2314, 2341.
T(4,2) = 6: 1432, 2143, 2431, 3214, 3241, 3421.
T(4,3) = 5: 1423, 2413, 3124, 3412, 4321.
T(4,4) = 4: 3142, 4213, 4231, 4312.
T(4,5) = 1: 4123.
T(4,6) = 1: 4132.
T(5,5) = 15: 15234, 25134, 31542, 35124, 41235, 42153, 42531, 43152, 45123, 53214, 53241, 53421, 54213, 54231, 54312.
Triangle T(n,k) begins:
  1;
  1;
  1,  1;
  1,  3,  1,  1;
  1,  6,  6,  5,   4,   1,   1;
  1, 10, 20, 20,  26,  15,  15,  6,  5,  1,  1;
  1, 15, 50, 70, 105, 106, 104, 90, 65, 51, 27, 21, 7, 6, 1, 1;

Links

Crossrefs

Columns k=0-3 give: A000012, A000217(n-1) for n>0, A002415(n-1) for n>0, A291288(n-3) for n>0.
Row sums give A000142.
T(n,n) gives A289489.
Cf. A005990, A008292, A123125, A173018, A258829, A263776, A316292, A316293.

Programs

  • Maple
    b:= proc(u, o) option remember; expand(`if`(u+o=0, 1,
      add(b(u-j, o+j-1)*x^(j-1), j=1..u)+
      add(b(u+j-1, o-j)*x^(j-1), j=1..o)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(0, n)):
seq(T(n), n=0..10);
  • Mathematica
    (* Generating function for tiles for Dyck tilings above the zigzag path of order n *)
(* Computed by looking at descents in the insertion sequence for the Dyck-tiling-ribbon bijection, described in the Kim-Meszaros-Panova-Wilson reference *)
(* Since it's above the zigzag, all insertion positions are even *)
(* When the second argument is specified, refines by position of last insertion *)
tilegen[n_, sn_] := tilegen[n, sn] = If[n == 0 || n == 1, 1,
    Sum[tilegen[n - 1, j] If[j >= sn, t^(j - sn + 1), 1] //
      Expand, {j, 0, 2 (n - 2), 2}]
    ];
tilegen[n_] := tilegen[n + 1, 2 n];
T[n_, k_] := Coefficient[tilegen[n], t, k]; (* David B. Wilson, Dec 14 2018 *)

Formula

Sum_{k>=0} k * T(n,k) = A005990(n).

A098928 Number of cubes that can be formed from the points of a cubical grid of n X n X n points.

Original entry on oeis.org

0, 1, 9, 36, 100, 229, 473, 910, 1648, 2795, 4469, 6818, 10032, 14315, 19907, 27190, 36502, 48233, 62803, 80736, 102550, 128847, 160271, 197516, 241314, 292737, 352591, 421764, 501204, 592257, 696281, 814450, 948112, 1098607, 1267367
Offset: 1

Views

Author

Ignacio Larrosa Cañestro, Oct 19 2004, Sep 29 2009

Keywords

Comments

Skew cubes are allowed.

Examples

			For n = 3 there are 8 cubes of volume 1 and 1 cube of volume 8; thus a(3)=9. - _José María Grau Ribas_, Mar 15 2014
a(6)=229 because we can place 15^2 cubes in a 6 X 6 X 6 cubical grid with their edges parallel to the faces of the grid, plus 4 cubes of edge 3 with a vertex in each face of the lattice and the other two vertices on a diagonal.

Links

Crossrefs

Cf. A103158.
Cf. A000537 (without skew cubes), A002415 (number of squares with corners on an n X n grid), A108279, A102698.

Programs

  • Mathematica
    Needs["Quaternions`"];
(* Initialize variables *)
R = 20;
NN = 1010;
(* Quaternion operations *)
test[q_Quaternion] :=
  Module[{unit, res, a, b, c, u, v, w, p},
   If[Round[Norm[q]] > R, Return[]];
   If[q == Quaternion[0, 0, 0, 0], Return[]];
   unit = Quaternion[0, 1, 0, 0];
   res = q ** unit ** Conjugate[q];
   a = Abs[res[[2]]] + Abs[res[[3]]] + Abs[res[[4]]];
   unit = Quaternion[0, 0, 1, 0];
   res = q ** unit ** Conjugate[q];
   b = Abs[res[[2]]] + Abs[res[[3]]] + Abs[res[[4]]];
   unit = Quaternion[0, 0, 0, 1];
   res = q ** unit ** Conjugate[q];
   c = Abs[res[[2]]] + Abs[res[[3]]] + Abs[res[[4]]];
   For[i = 1, i <= (R - 1)/Max[a, b, c], i++,
    If[SquareFreeQ[i], {u = a*i;
      v = b*i;
      w = c*i;
      p = Max[u, v, w] + 1;
      coe[[p + 1, 4]] += (1);
      coe[[p + 1, 3]] -= (u + v + w);
      coe[[p + 1, 2]] += (u*v + v*w + w*u);
      coe[[p + 1, 1]] -= (u*v*w)}]]];
(* Set up coefficient matrix *)
coe = ConstantArray[0, {NN, 4}];
(* Loop through quaternions *)
rt = Ceiling[Sqrt[R]] + 1;
For[s = -rt, s <= rt, s++,
  For[x = -rt, x <= rt, x++,
   For[y = -rt, y <= rt, y++,
    For[z = -rt, z <= rt, z++, test[Quaternion[s, x, y, z]];
     test[Quaternion[s + 0.5, x + 0.5, y + 0.5, z + 0.5]];]]]];
newCoe = coe;
newCoe[[2 ;; ;; 2]] = coe[[2 ;; ;; 2]]/2;
(* Calculate and output results *)
For[i = 2, i <= R + 1, i++, ans = 0;
  For[j = 4, j >= 1, j--, newCoe[[i, j]] += newCoe[[i - 1, j]];
   ans = ans*(i - 1) + newCoe[[i, j]];
   ];
  Print[i - 1, " ", ans/24];];
(* Haomin Yang, Aug 29 2023 *)

Extensions

Edited by Ray Chandler, Apr 05 2010
Further edited by N. J. A. Sloane, Mar 31 2016

A305653 Expansion of Product_{k>=1} 1/(1 - x^k)^((k+1)*binomial(k+2,3)/2).

Original entry on oeis.org

1, 1, 7, 27, 98, 323, 1085, 3471, 10998, 33874, 102737, 305849, 897899, 2597822, 7423408, 20957775, 58524868, 161741013, 442705279, 1200718351, 3228796864, 8611973548, 22793714865, 59887897679, 156252738062, 404964879419, 1042884107691, 2669317020743, 6792321636929
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 07 2018

Keywords

Comments

Euler transform of A002415, shifted left one place.

Links

Crossrefs

Cf. A000391, A002415, A023871, A279215, A290792.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(a(n-j)*add(d^2*
      (d+2)*(d+1)^2/12, d=numtheory[divisors](j)), j=1..n)/n)
    end:
seq(a(n), n=0..35);  # Alois P. Heinz, Jun 07 2018
  • Mathematica
    nmax = 28; CoefficientList[Series[Product[1/(1 - x^k)^((k + 1) Binomial[k + 2, 3]/2), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 28; CoefficientList[Series[Exp[Sum[x^k (1 + x^k)/(k (1 - x^k)^5), {k, 1, nmax}]], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d^2 (d + 1)^2 (d + 2)/12, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 28}]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^A002415(k+1).
G.f.: exp(Sum_{k>=1} x^k*(1 + x^k)/(k*(1 - x^k)^5)).
a(n) ~ exp(Zeta'(-1)/6 - Zeta(3) / (4*Pi^2) + 149*Zeta(5) / (32*Pi^4) + 15876 * Zeta(3) * Zeta(5)^2 / Pi^12 - 666792 * Zeta(5)^3 / Pi^14 + 108884466432 * Zeta(5)^5 / (5*Pi^24) + Zeta'(-3)/3 + (-7*(7/2)^(1/6) * Pi / (384*sqrt(3)) - 21 * 2^(5/6) * sqrt(3) * 7^(1/6) * Zeta(3) * Zeta(5) / Pi^7 + 3087 * sqrt(3) * (7/2)^(1/6) * Zeta(5)^2 / (2*Pi^9) - 30339036 * 2^(5/6) * sqrt(3) * 7^(1/6) * Zeta(5)^4 / Pi^19) * n^(1/6) + ((7/2)^(1/3) * Zeta(3) / (2*Pi^2) - 21 * (7/2)^(1/3) * Zeta(5) / (2*Pi^4) + 254016 * 2^(2/3) * 7^(1/3) * Zeta(5)^3 / Pi^14) * n^(1/3) + (sqrt(7/6) * Pi / 12 - 756 * sqrt(42) * Zeta(5)^2 / Pi^9) * sqrt(n) + (9 * 2^(1/3) * 7^(2/3) * Zeta(5) / Pi^4) * n^(2/3) + (2 * (2/7)^(1/6) * sqrt(3) * Pi) / 5 * n^(5/6)) * Pi^(1/90) / (2^(247/270) * 3^(34/45) * 7^(23/270) * n^(79/135)). - Vaclav Kotesovec, Jun 08 2018

A108648 a(n) = (n+1)^2*(n+2)^3*(n+3)/24.

Original entry on oeis.org

1, 18, 120, 500, 1575, 4116, 9408, 19440, 37125, 66550, 113256, 184548, 289835, 441000, 652800, 943296, 1334313, 1851930, 2527000, 3395700, 4500111, 5888828, 7617600, 9750000, 12358125, 15523326, 19336968, 23901220, 29329875, 35749200
Offset: 0

Views

Author

Emeric Deutsch, Jun 13 2005

Keywords

Comments

Kekulé numbers for certain benzenoids.

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 24).

Links

Crossrefs

Cf. A108647.

Programs

  • Magma
    [(n+1)^2*(n+2)^3*(n+3)/24: n in [0..30]]; // G. C. Greubel, Oct 28 2022
  • Maple
    a:=(n+1)^2*(n+2)^3*(n+3)/24: seq(a(n),n=0..36);
  • Mathematica
    Table[(n+1)^2*(n+2)^3*(n+3)/24, {n,0,30}] (* G. C. Greubel, Oct 28 2022 *)
  • PARI
    Vec((1 + 11*x + 15*x^2 + 3*x^3) / (1 - x)^7 + O(x^30)) \\ Colin Barker, Apr 22 2020
  • SageMath
    [(n+1)^2*(n+2)^3*(n+3)/24 for n in (0..30)] # G. C. Greubel, Oct 28 2022

Formula

From Colin Barker, Apr 22 2020: (Start)
G.f.: (1 + 11*x + 15*x^2 + 3*x^3) / (1 - x)^7.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>6.
(End)
a(n) = A000217(n+1) * A002415(n+2). - J. M. Bergot, May 21 2022
From Amiram Eldar, May 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 24*zeta(3) + 6*Pi^2 - 87.
Sum_{n>=0} (-1)^n/a(n) = 99 - Pi^2 - 96*log(2) - 18*zeta(3). (End)
E.g.f.: (24 + 408*x + 1020*x^2 + 772*x^3 + 224*x^4 + 26*x^5 + x^6)*exp(x)/4!. - G. C. Greubel, Oct 28 2022

A208950 a(4*n) = n*(16*n^2-1)/3, a(2*n+1) = n*(n+1)*(2*n+1)/6, a(4*n+2) = (4*n+1)*(4*n+2)*(4*n+3)/6.

Original entry on oeis.org

0, 0, 1, 1, 5, 5, 35, 14, 42, 30, 165, 55, 143, 91, 455, 140, 340, 204, 969, 285, 665, 385, 1771, 506, 1150, 650, 2925, 819, 1827, 1015, 4495, 1240, 2728, 1496, 6545, 1785, 3885, 2109, 9139, 2470, 5330, 2870, 12341, 3311, 7095, 3795, 16215, 4324
Offset: 0

Views

Author

Paul Curtz, Mar 03 2012

Keywords

Comments

a(n+2) is divisible by A060819(floor(n/3)).
a(n) is divisible by A176672(floor(n/3)).
Denominator of a(n)/n is of period 24: 1,1,3,4,1,6,1,4,3,1,1,12,1,2,3,4,1,3,1,4,3,2,1,12 (two successive palindromes).
This is the fifth column of the triangle A107711, hence the formula involving gcd(n+2,4) given below follows. - Wolfdieter Lang, Feb 24 2014

Links

Crossrefs

Cf. A107711 (fifth column).
Cf. A000034, A000292, A002415, A051724, A060819, A061037, A107711, A138190, A145979, A176672, A176895.

Programs

  • Magma
    [Binomial(n+1,3)*GCD(n+2,4)/4: n in [0..50]]; // G. C. Greubel, Sep 20 2018
  • Mathematica
    CoefficientList[Series[(x^2 + x^3 + 5 x^4 + 5 x^5 + 31 x^6 + 10 x^7 + 22 x^8 + 10 x^9 + 31 x^10 + 5 x^11 + 5 x^12 + x^13 + x^14)/((1 - x)^4 (1 + x)^4 (1 + 4 x^2 + 6 x^4 + 4 x^6 + x^8)), {x, 0, 47}], x] (* Bruno Berselli, Mar 11 2012 *)
  • Maxima
    A208950(n) := block(
        [a,npr] ,
        if equal(mod(n,4), 0) then (
                a : n/12*(n^2-1)
        ) else if equal(mod(n,2),0) then (
                a : (n-1)*n*(n+1)/6
        ) else (
                npr : (n-1)/2,
                a : npr*(npr+1)*n/6
        ) ,
        return(a)
)$ /* R. J. Mathar, Mar 10 2012 */
  • PARI
    vector(50, n, n--; binomial(n+1,3)*gcd(n+2,4)/4) \\ G. C. Greubel, Sep 20 2018

Formula

a(n) = 4*a(n-4) - 6*a(n-8) + 4*a(n-12) - a(n-16).
a(n+1) = A002415(n+1)/A145979(n-1).
a(n) = A051724(n-1) * A051724(n) * A051724(n+1).
a(n) = A060819(n-1) * A060819(n) * A060819(n+1) / 3.
a(n) * a(n+4) = A061037(n+1) * A061037(n+2) * A061037(n+3) / 9.
a(n) = A138190(n)/A000034(n) for n > 0.
a(n) = A000292(n-1)/A176895(n+2) for n > 0.
a(n)/a(n+4) = n*(n^2-1)/((n+3)*(n+4)*(n+5)).
a(n)/a(n+12) = (n-1)*n*(n+1)/((n+11)*(n+12)*(n+13)).
G.f.: (x^2 + x^3 + 5*x^4 + 5*x^5 + 31*x^6 + 10*x^7 + 22*x^8 + 10*x^9 + 31*x^10 + 5*x^11 + 5*x^12 + x^13 + x^14) / ((1-x)^4*(1+x)^4*(1 + 4*x^2 + 6*x^4 + 4*x^6 + x^8)). - R. J. Mathar, Mar 10 2012
From Wolfdieter Lang, Feb 24 2014: (Start)
G.f.: (1 + x^12 + x*(1+x^10) + 5*x^2*(1+x^8) + 5*x^3*(1+x^7) + 31*x^4*(1+x^4) + 10*x^5*(1+x^2) + 22*x^6)/(1-x^4)^4. This is the preceding g.f. rewritten.
a(n) = binomial(n+1,3)*gcd(n+2,4)/4, n >= 0. From the g.f., see a comment above on A107711. (End)
a(n) = (n*(n-1)*((n+1)*(4+2*(-1)^n + (1+(-1)^n)*(-1)^((2*n+3+(-1)^n)/4))))/48. - Luce ETIENNE, Jan 01 2015
Sum_{n>=2} 1/a(n) = 12 - 27*log(2)/2. - Amiram Eldar, Aug 12 2022

A217476 Coefficient triangle for the square of the monic integer Chebyshev T-polynomials A127672.

Original entry on oeis.org

4, 0, 1, 4, -4, 1, 0, 9, -6, 1, 4, -16, 20, -8, 1, 0, 25, -50, 35, -10, 1, 4, -36, 105, -112, 54, -12, 1, 0, 49, -196, 294, -210, 77, -14, 1, 4, -64, 336, -672, 660, -352, 104, -16, 1, 0, 81, -540, 1386, -1782, 1287, -546, 135, -18, 1, 4, -100, 825, -2640, 4290, -4004, 2275, -800, 170, -20, 1
Offset: 0

Views

Author

Wolfdieter Lang, Oct 17 2012

Keywords

Comments

The monic integer T-polynomials, called R(n,x) (in Abramowitz-Stegun C(n,x)), with their coefficient triangle given in A127672, when squared, become polynomials in y=x^2:
R(n,x)^2 = sum(T(n,k)*y^k,m=0..n).
R(n,x)^2 = 2 + R(2*n,x). From the bisection of the R-(or T-)polynomials, the even part. Directly from the R(m*n,x)=R(m,R(n,x)) property for m=2.
The o.g.f. is G(z,y) := sum((R(n,sqrt(y))^2)*z^n ,n=0..infinity) = (4 + (4 - 3*y)*z + y*z^2)/((1 +(2-y)*z + z^2)*(1-z)). From the bisection.
The o.g.f.s of the columns k>=1 are x^k*(1-x)/(1+x)^(2*k+1),
and for k=0 the o.g.f. is 4/(1-x^2).
Hetmaniok et al. (2015) refer to these as "modified Chebyshev" polynomials. - N. J. A. Sloane, Sep 13 2016

Examples

			The triangle begins:
n\k 0    1    2      3     4      5     6     7    8   9  10
0:  4
1:  0    1
2:  4   -4    1
3:  0    9   -6      1
4:  4  -16   20     -8     1
5:  0   25  -50     35   -10      1
6:  4  -36  105   -112    54    -12     1
7:  0   49 -196    294  -210     77   -14     1
8:  4  -64  336   -672   660   -352   104   -16    1
9:  0   81 -540   1386 -1782   1287  -546   135  -18   1
10: 4 -100  825  -2640  4290  -4004  2275  -800  170 -20   1
...
n=2:  R(2,x) = -2 + y, R(2,x)^2 = 4 -4*y + y^2, with y=x^2.
n=3:  R(3,x) = 3*x - x^3, R(3,x)^2 = 9*y - 6*y^2 +y^3, with y=x^2.
T(4,1) = 8*(-1)^3*binomial(5,3)/5 = -16.
T(4,0) = 2 + 8*(-1)^4*binomial(4,4)/4 = 4.
T(n,1) = (-1)^(n-1)*2*n*(n+1)!/((n-1)!*2!*(n+1)) = -((-1)^n)*n^2 = A162395(n), n >= 1.
T(n,2) = (-1)^n*A002415(n), n >= 0.
T(n,3) = -(-1)^n*A040977(n-3), n >= 3.
T(n,4) = (-1)^n*A053347(n-4), n >= 4.
T(n,5) = -(-1)^n*A054334(n-5), n >= 5.

References

  • E Hetmaniok, P Lorenc, S Damian, et al., Periodic orbits of boundary logistic map and new kind of modified Chebyshev polynomials in R. Witula, D. Slota, W. Holubowski (eds.), Monograph on the Occasion of 100th Birthday Anniversary of Zygmunt Zahorski. Wydawnictwo Politechniki Slaskiej, Gliwice 2015, pp. 325-343.

Crossrefs

Cf. A127672, A158454 (square of S-polynomials), A128495 (sum of square of S-polynomials).

Formula

T(n,k) = [x^(2*k)]R(n,x)^2, with R(n,x) the monic integer version of the Chebyshev T(n,x) polynomial.
T(n,k) = 0 if n=1. ([k=0] means 1 if k=0 else 0).

A047835 a(n) = Product_{i=1..n} ((i+4)*(i+5)*(i+6)*(i+7))/(i*(i+1)*(i+2)*(i+3)).

Original entry on oeis.org

1, 70, 1764, 24696, 232848, 1646568, 9343620, 44537922, 184225041, 677352676, 2254684432, 6892441920, 19571505408, 52101067968, 131018862096, 313203587004, 715536058545, 1569305708586, 3316911815140, 6778924352200, 13435361082000
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of tilings of a <4,n,4> hexagon.
Partial sums of A133708. - Peter Bala, Sep 21 2007

References

  • O. D. Anderson, Find the next sequence, J. Rec. Math., 8 (No. 4, 1975-1976), 241.

Links

Crossrefs

Fourth row of array A103905.
Cf. A002415, A047819, A133708.

Programs

  • Maple
    seq(binomial(n,n-1)*binomial(n+1,n-2)*binomial(n+2,n-3)*binomial(n+3,n-4)/(10*4!), n=4..24); # Zerinvary Lajos, May 29 2007
  • Mathematica
    Table[Product[Times@@((i+Range[4,7])/(i+Range[0,3])),{i,n}],{n,0,20}] (* Harvey P. Dale, Nov 03 2011 *)

Formula

a(n) = C(n,n-1)*C(n+1,n-2)*C(n+2,n-3)*C(n+3,n-4)/(10*4!), n >= 4 . - Zerinvary Lajos, May 29 2007
a(n-4) = (1/3456)*Sum_{1 <= x_1, x_2, x_3, x_4 <= n} (det V(x_1,x_2,x_3,x_4))^2 = (1/3456)*Sum_{1 <= i,j,k,l <= n} ((i-j)(i-k)(i-l)(j-k)(j-l)(k-l))^2, where V(x_1,x_2,x_3,x_4) is the Vandermonde matrix of order 4. - Peter Bala, Sep 21 2007
Empirical g.f.: (x+1)*(x^8 + 52*x^7 + 658*x^6 + 2890*x^5 + 4810*x^4 + 2890*x^3 + 658*x^2 + 52*x + 1)/(1-x)^17. - Colin Barker, Jun 06 2012
Sum_{n>=0} 1/a(n) = 67200*Pi^4 + 5605600*Pi^2 - 185612833/3. - Amiram Eldar, May 29 2022

A121306 Array read by antidiagonals: a(m,n) = a(m,n-1)+a(m-1,n) but with initialization values a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

Original entry on oeis.org

2, 2, 3, 2, 5, 4, 2, 7, 9, 5, 2, 9, 16, 14, 6, 2, 11, 25, 30, 20, 7, 2, 13, 36, 55, 50, 27, 8, 2, 15, 49, 91, 105, 77, 35, 9, 2, 17, 64, 140, 196, 182, 112, 44, 10, 19, 81, 204, 336, 378, 294, 156, 54, 100, 285, 540, 714, 672, 450, 210, 385, 825, 1254, 1386, 1122
Offset: 0

Views

Author

Thomas Wieder, Aug 04 2006, Aug 06 2006

Keywords

Comments

For a(1,0)=1, a(m>1,0)=0 and a(0,n>=0)=0 one gets Pascal's triangle A007318.

Examples

			Array begins
2 2 2 2 2 2 2 2 2 ...
3 5 7 9 11 13 15 17 19 ...
4 9 16 25 36 49 64 81 100 ...
5 14 30 55 91 140 204 285 385 ...
6 20 50 105 196 336 540 825 1210 ...
7 27 77 182 378 714 1254 2079 3289 ...

Crossrefs

The first nine rows are: A006527, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347.
The initial columns are: A000027, A000096, A005581, A005582, A005583, A005584.
Cf. A119800, A007318, A006527, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A000027, A000096, A005581, A005582, A005583, A005584.

Programs

  • Excel
    =Z(-1)S+ZS(-1). The very first row (not included into the table) contains the initialization values: a(0,1)=1, a(0,n>=2)=0. The very first column (not included into the table) contains the initialization values: a(m>=1,0)=1. The value a(0,0)=0 does not enter into the table.

Formula

a(m,n) = a(m,n-1)+a(m-1,n), a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

Extensions

Edited by N. J. A. Sloane, Sep 15 2006

A169937 a(n) = binomial(m+n-1,n)^2 - binomial(m+n,n+1)*binomial(m+n-2,n-1) with m = 14.

Original entry on oeis.org

1, 91, 3185, 63700, 866320, 8836464, 71954064, 488259720, 2848181700, 14620666060, 67255063876, 281248448936, 1081724803600, 3863302870000, 12914469594000, 40680579221100, 121443493851225, 345280521733875, 938920716995625, 2451077240157000, 6162708489537600
Offset: 0

Views

Author

N. J. A. Sloane, Aug 28 2010

Keywords

Comments

13th column (and diagonal) of the triangle A001263. - Bruno Berselli, May 07 2012

References

  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; Prop. 8.4, case n=14.

Links

Crossrefs

The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces the sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
Cf. A002378.

Programs

  • Magma
    [(1/13)*Binomial(n+12,12)^2*(n+13)/(n+1): n in [0..20]]; // Bruno Berselli, Nov 09 2011
  • Maple
    f:=m->[seq( binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1), n=0..20)]; f(14);
  • Mathematica
    Table[Binomial[13+n,n]^2-Binomial[14+n,n+1]Binomial[12+n,n-1],{n,0,20}] (* Harvey P. Dale, Nov 09 2011 *)
  • PARI
    a(n)=binomial(n+12,12)^2*(n+13)/(n+1)/13 \\ Charles R Greathouse IV, Nov 09 2011

Formula

a(n) = (1/13)*A010965(n+12)^2*(n+13)/(n+1). - Bruno Berselli, Nov 09 2011
a(n) = Product_{i=1..12} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 45997360927193/23100 - 201753552*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 16431564019/23100 - 72072*Pi^2. (End)

A185913 Accumulation array of A185912, by antidiagonals.

Original entry on oeis.org

1, 4, 6, 10, 21, 20, 20, 48, 66, 50, 35, 90, 144, 160, 105, 56, 150, 260, 340, 330, 196, 84, 231, 420, 600, 690, 609, 336, 120, 336, 630, 950, 1200, 1260, 1036, 540, 165, 468, 896, 1400, 1875, 2170, 2128, 1656, 825, 220, 630, 1224, 1960, 2730, 3360, 3640, 3384, 2520, 1210, 286, 825, 1620, 2640, 3780, 4851, 5600, 5760, 5130, 3685, 1716, 364, 1056, 2090, 3450, 5040, 6664, 8036, 8820, 8700, 7480, 5214, 2366, 455
Offset: 1

Views

Author

Clark Kimberling, Feb 06 2011

Keywords

Comments

A member of the accumulation chain ... < A185910 < A185911 < A185912 < A185913 < ...
(See A144112 for definitions of weight array and accumulation array.)

Examples

			Northwest corner:
1.....4.....10.....20.....35
6.....21....48.....90.....150
20....66....144....260....420
50....160...340....600....950

Links

Crossrefs

Cf. A144112, A185912.
Row 1 to 2: A000292, 3*A005581.
Column 1: A002415.

Programs

  • Mathematica
    (* The program generates A185912 and its accumulation array A185913 *)
f[n_,k_]:=(k*n/6)(-2+3k+3n+2n^2);
TableForm[Table[f[n,k],{n,1,10},{k,1,15}]]  (* array A185912 *)
Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten
s[n_,k_]:=Sum[f[i,j],{i,1,n},{j,1,k}];
FullSimplify[s[n,k]]  (* formula for A185913 *)
TableForm[Table[s[n,k],{n,1,10},{k,1,15}]] (* array A185913 *)
Table[s[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten

Formula

T(n,k) = C(k+1,2)*C(n+1,2)*(n^2+3*n+2*k)/6, k>=1, n>=1.
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