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.

A253283 Triangle read by rows: coefficients of the partial fraction decomposition of [d^n/dx^n] (x/(1-x))^n/n!.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 0, 3, 12, 10, 0, 4, 30, 60, 35, 0, 5, 60, 210, 280, 126, 0, 6, 105, 560, 1260, 1260, 462, 0, 7, 168, 1260, 4200, 6930, 5544, 1716, 0, 8, 252, 2520, 11550, 27720, 36036, 24024, 6435, 0, 9, 360, 4620, 27720, 90090, 168168, 180180, 102960, 24310
Offset: 0

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Author

Peter Luschny, Mar 20 2015

Keywords

Comments

The rows give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1) / (n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 31 2022
This is related to the cluster fans of type B (see Fomin and Zelevinsky reference) - F. Chapoton, Nov 17 2022.

Examples

			[1]
[0, 1]
[0, 2,   3]
[0, 3,  12,   10]
[0, 4,  30,   60,   35]
[0, 5,  60,  210,  280,  126]
[0, 6, 105,  560, 1260, 1260,  462]
[0, 7, 168, 1260, 4200, 6930, 5544, 1716]
.
R_0(x) = 1/(x-1)^0.
R_1(x) = 0/(x-1)^1 + 1/(x-1)^2.
R_2(x) = 0/(x-1)^2 + 2/(x-1)^3 + 3/(x-1)^4.
R_3(x) = 0/(x-1)^3 + 3/(x-1)^4 + 12/(x-1)^5 + 10/(x-1)^6.
Then k!*[x^k] R_n(x) is A001286(k+2) and A001754(k+3) for n = 2, 3 respectively.
.
Seen as an array A(n, k) = binomial(n + k, k)*binomial(n + 2*k - 1, n + k):
[0] 1, 1,   3,   10,    35,    126,     462, ...
[1] 0, 2,  12,   60,   280,   1260,    5544, ...
[2] 0, 3,  30,  210,  1260,   6930,   36036, ...
[3] 0, 4,  60,  560,  4200,  27720,  168168, ...
[4] 0, 5, 105, 1260, 11550,  90090,  630630, ...
[5] 0, 6, 168, 2520, 27720, 252252, 2018016, ...
[6] 0, 7, 252, 4620, 60060, 630630, 5717712, ...
		

Crossrefs

T(n, n) = C(2*n-1, n) = A001700(n-1).
T(n, n-1) = A005430(n-1) for n >= 1.
T(n, n-2) = A051133(n-2) for n >= 2.
T(n, 2) = A027480(n-1) for n >= 2.
T(2*n, n) = A208881(n) for n >= 0.
A002002 (row sums).

Programs

  • Maple
    T_row := proc(n) local egf, k, F, t;
    if n=0 then RETURN(1) fi;
    egf := (x/(1-x))^n/n!; t := diff(egf,[x$n]);
    F := convert(t,parfrac,x);
    # print(seq(k!*coeff(series(F,x,20),x,k),k=0..7));
    # gives A000142, A001286, A001754, A001755, A001777, ...
    seq(coeff(F,(x-1)^(-k)),k=n..2*n) end:
    seq(print(T_row(n)),n=0..7);
    # 2nd version by R. J. Mathar, Dec 18 2016:
    A253283 := proc(n,k)
        binomial(n,k)*binomial(n+k-1,k-1) ;
    end proc:
  • Mathematica
    Table[Binomial[n, k] Binomial[n + k - 1, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 22 2017 *)
  • PARI
    T(n,k) = binomial(n,k)*binomial(n+k-1,k-1);
    tabl(nn) = for(n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Apr 29 2018

Formula

The exponential generating functions for the rows of the square array L(n,k) = ((n+k)!/n!)*C(n+k-1,n-1) (associated to the unsigned Lah numbers) are given by R_n(x) = Sum_{k=0..n} T(n,k)/(x-1)^(n+k).
T(n,k) = C(n,k)*C(n+k-1,k-1).
Sum_{k=0..n} T(n,k) = (-1)^n*hypergeom([-n,n],[1],2) = (-1)^n*A182626(n).
Row generating function: Sum_{k>=1} T(n,k)*z^k = z*n* 2F1(1-n,n+1 ; 2; -z). - R. J. Mathar, Dec 18 2016
From Peter Bala, Feb 22 2017: (Start)
G.f.: (1/2)*( 1 + (1 - t)/sqrt(1 - 2*(2*x + 1)*t + t^2) ) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n,x) = (1/2)*(LegendreP(n, 2*x + 1) - LegendreP(n-1, 2*x + 1)) for n >= 1.
The row polynomials are the black diamond product of the polynomials x^n and x^(n+1) (see Dukes and White 2016 for the definition of this product).
exp(Sum_{n >= 1} R(n,x)*t^n/n) = 1 + x*t + x*(1 + 2*x)*t^2 + x*(1 + 5*x + 5*x^2)*t^3 + ... is a g.f. for A033282, but with a different offset.
The polynomials P(n,x) := (-1)^n/n!*x^(2*n)*(d/dx)^n(1 + 1/x)^n begin 1, 3 + 2*x , 10 + 12*x + 3*x^2, ... and are the row polynomials for the row reverse of this triangle. (End)
Let Q(n, x) = Sum_{j=0..n} (-1)^(n - j)*A269944(n, j)*x^(2*j - 1) and P(x, y) = (LegendreP(x, 2*y + 1) - LegendreP(x-1, 2*y + 1)) / 2 (see Peter Bala above). Then n!*(n - 1)!*[y^n] P(x, y) = Q(n, x) for n >= 1. - Peter Luschny, Oct 31 2022
From Peter Bala, Apr 18 2024: (Start)
G.f.: Sum_{n >= 0} binomial(2*n-1, n)*(x*t)^n/(1 - t)^(2*n) = 1 + x*t + (2*x + 3*x^2)*t^2 + (3*x + 12*x^2 + 10*x^3)*t^3 + ....
n-th row polynomial R(n, x) = [t^n] ( (1 - t)/(1 - (1 + x)*t) )^n.
It follows that for integer x, the sequence {R(n, x) : n >= 0} satisfies the Gauss congruences: R(n*p^r, x) == R(n*p^(r-1), x) (mod p^r) for all primes p and positive integers n and r.
R(n, -2) = (-1)^n * A002003(n) for n >= 1.
R(n, 3) = A299507(n). (End)

A208879 Number of words A(n,k), either empty or beginning with the first letter of the cyclic k-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 2, 30, 10, 1, 1, 1, 2, 62, 560, 35, 1, 1, 1, 2, 114, 2830, 11550, 126, 1, 1, 1, 2, 202, 12622, 151686, 252252, 462, 1, 1, 1, 2, 346, 53768, 1754954, 8893482, 5717712, 1716, 1, 1
Offset: 0

Views

Author

Alois P. Heinz, Mar 02 2012

Keywords

Comments

The first and the last letters are considered neighbors in a cyclic alphabet. The words are not considered cyclic here.
A(n,k) is also the number of (n*k-1)-step walks on k-dimensional cubic lattice from (1,0,...,0) to (n,n,...,n) with positive unit steps in all dimensions such that the indices of dimensions used in consecutive steps differ by less than 2 or are in the set {1,k}.

Examples

			A(0,0) = A(n,0) = A(0,k) = 1: the empty word.
A(1,2) = 1 = |{ab}|.
A(1,3) = 2 = |{abc, acb}|.
A(1,4) = 2 = |{abcd, adcb}|.
A(2,2) = 3 = |{aabb, abab, abba}|.
A(2,4) = 62 = |{aabbccdd, aabbcdcd, aabbcddc, aabcbcdd, aabcddcb, aadcbbcd, aadcdcbb, aaddcbbc, aaddcbcb, aaddccbb, ababccdd, ababcdcd, ababcddc, abadcbcd, abadcdcb, abaddcbc, abaddccb, abbadccd, abbadcdc, abbaddcc, abbccdad, abbccdda, abbcdadc, abbcdcda, abcbadcd, abcbaddc, abcbcdad, abcbcdda, abccbadd, abccddab, abcdabcd, abcdadcb, abcdcbad, abcdcdab, abcddabc, abcddcba, adabbccd, adabbcdc, adabcbcd, adabcdcb, adadcbbc, adadcbcb, adadccbb, adcbabcd, adcbadcb, adcbbadc, adcbbcda, adcbcbad, adcbcdab, adccbbad, adccdabb, adcdabbc, adcdabcb, adcdcbab, adcdcbba, addabbcc, addabcbc, addabccb, addcbabc, addcbcba, addccbab, addccbba}|.
Square array A(n,k) begins:
  1,  1,   1,       1,         1,           1,             1, ...
  1,  1,   1,       2,         2,           2,             2, ...
  1,  1,   3,      30,        62,         114,           202, ...
  1,  1,  10,     560,      2830,       12622,         53768, ...
  1,  1,  35,   11550,    151686,     1754954,      19341130, ...
  1,  1, 126,  252252,   8893482,   276049002,    8151741752, ...
  1,  1, 462, 5717712, 552309938, 46957069166, 3795394507240, ...
		

Crossrefs

Columns k=0+1, 2-6 give: A000012, A088218, A208881, A209183, A209184, A209185.
Rows n=0, 2 give: A000012, A208880.
Cf. A208673 (noncyclic alphabet).

Programs

  • Maple
    b:= proc() option remember; local n; n:= nargs;
         `if`(n<2 or {0}={args}, 1,
         `if`(n=2, `if`(args[1]>0, b(args[1]-1, args[2]), 0)+
                   `if`(args[2]>0, b(args[2]-1, args[1]), 0),
         `if`(args[1]>0, b(args[1]-1, seq(args[i], i=2..n)), 0)+
         `if`(args[2]>0, b(args[2]-1, seq(args[i], i=3..n), args[1]), 0)+
         `if`(args[n]>0, b(args[n]-1, seq(args[i], i=1..n-1)), 0)))
        end:
    A:= (n, k)-> `if`(n=0 or k=0, 1, b(n-1, n$(k-1))):
    seq(seq(A(n, d-n), n=0..d), d=0..9);
  • Mathematica
    b[args__] := b[args] = With[{n = Length[{args}]}, If[n<2 || {0} == Union[ {args}], 1, If[n==2, If[{args}[[1]]>0, b[{args}[[1]]-1, {args}[[2]] ], 0] + If[{args}[[2]]>0, b[{args}[[2]]-1, {args}[[1]] ], 0], If[{args}[[1]]>0, b[{args}[[1]]-1, Sequence @@ {args}[[2;;n]] ], 0] + If[{args}[[2]]>0, b[{args}[[2]]-1, Sequence @@ {args}[[3;;n]], {args}[[1]] ], 0]+ If[{args}[[n]]>0, b[{args}[[n]]-1, Sequence @@ Most[{args}]] ],0]] /. Null -> 0];
    a[n_,k_]:= If[n==0 || k==0, 1, b[n-1, Sequence @@ Array[n&, k-1]]];
    Table[Table[a[n, d-n], {n,0,d}], {d,0,9}] // Flatten (* Jean-François Alcover, Dec 13 2013, translated from Maple *)

A361027 Table of generalized de Bruijn's numbers (A006480) read by ascending antidiagonals.

Original entry on oeis.org

2, 30, 3, 560, 20, 20, 11550, 210, 75, 210, 252252, 2772, 504, 504, 2772, 5717712, 42042, 4620, 2352, 4620, 42042, 133024320, 700128, 51480, 15840, 15840, 51480, 700128, 3155170590, 12471030, 656370, 135135, 81675, 135135, 656370, 12471030, 75957810500, 233716340, 9237800
Offset: 0

Views

Author

Peter Bala, Feb 28 2023

Keywords

Comments

The Catalan numbers A000108 are given by the formula Catalan(n) = (2*n)!/(n!*(n + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(r,n) = J(r) * (2*n)!/(n!*(n + r + 1)!), where J(r) = (2*r + 2)!/(2*(r + 1)!) = (2^r)*Product_{j = 0..r} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.
For this table we carry out an analogous construction using the de Bruijn numbers B(n) = (3*n)!/n!^3 = A006480(n) in place of the central binomial numbers. We define the generalized de Bruijn number B(r,n), r = 0, 1, 2, ..., by B(r,n) = F(r) * (3*n)!/(n!*(n + r + 1)!^2), where choosing F(r) = (3*r + 3)!/(3*(r + 1)!) = (3^r)*Product_{j = 0..r} (3*j + 1)*(3*j + 2) appears to produce integer values for these quantities. We have verified this for rows 0, 1, 2 and 3 of the table.
An alternative expression for the generalized de Bruijn numbers is B(r,n) = G(r,n) * B(n+r+1), where G(r) = (1/3)*Product_{j = 0..r} ( (3*j + 1)*(3*j + 2)/((3*n + 3*j + 1)*(3*n + 3*j + 2)) ).
The rows of the square array below are the sequences of generalized de Bruijn numbers {B(0,k)}, {B(1,k)}, {B(2,k)}, ....

Examples

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|       0       1       2        3        4         5         6 ...
  ----------------------------------------------------------------------
   0 |       2       3      20      210     2772     42042    700128 ...
   1 |      30      20      75      504     4620     51480    656370 ...
   2 |     560     210     504     2352    15840    135135   1361360 ...
   3 |   11550    2772    4620    15840    81675    550550   4492488 ...
   4 |  252252   42042   51480   135135   550550   3006003  20271888 ...
   5 | 5717712  700128  656370  1361360  4492488  20271888  ...
  ...
As a triangle:
 Row
  0 |        2
  1 |       30       3
  2 |      560      20     20
  3 |    11550     210     75    210
  4 |   252252    2772    504    504   2772
  5 |  5717712   42042   4620   2352   4620   42042
  ...
		

References

  • N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Crossrefs

A208881 (column 1), A361028(row 0), A361029(row 1), A361030(row 2), A361031(row 3).

Programs

  • Maple
    # as a square array
    T := proc (n,k) (1/3)*27^(n+k+1)*binomial(n+1/3, n+k+1)*binomial(n+2/3,
    n+k+1); end proc:
    for n from 0 to 10 do seq(T(n,k), k = 0..10); end do;
    # as a triangle
    T := proc (n,k) (1/3)*27^(n+k+1)*binomial(n+1/3, n+k+1)*binomial(n+2/3,
    n+k+1); end proc:
    for n from 0 to 10 do seq(T(n-k,k), k = 0..n); end do;

Formula

T(n,k) = (3*n + 3)!/(3*(n + 1)!) * (3*k)!/(k!*(k + n + 1)!^2), n, k >= 0.
T(n,k) = (1/3)*27^(n+1+k)*binomial(n+1/3, n+1+k)*binomial(n+2/3, n+1+k).
T(n,k) = (1/(2*Pi))^2 * 1/27^(n+k+1) * Integral_{x = 0..27} (27 - x)^(n+2/3)*x^(k-2/3) dx * Integral_{x = 0..27} (27 - x)^(n+1/3)*x^(k-1/3) dx.
P-recursive: (n + k + 1)^2*T(n,k) = 3*(3*k - 1)*(3*k - 2)*T(n,k-1) with T(n,0) = 1/(n+1)!^2 * (3*n + 3)!/(3*(n + 1)!).
(n + k + 1)^2*T(n,k) = 3*(3*n + 1)*(3*n + 2)*T(n-1,k) with T(0,k) = 2*(k + 1)*(3*k)!/(k + 1)!^3.
T(n,0) = A208881(n+1).

A199127 Number of n X 2 0..2 arrays with values 0..2 introduced in row major order, the number of instances of each value within one of each other, and no element equal to any horizontal or vertical neighbor.

Original entry on oeis.org

1, 2, 2, 12, 30, 30, 210, 560, 560, 4200, 11550, 11550, 90090, 252252, 252252, 2018016, 5717712, 5717712, 46558512, 133024320, 133024320, 1097450640, 3155170590, 3155170590, 26293088250, 75957810500, 75957810500, 638045608200
Offset: 1

Views

Author

R. H. Hardin, Nov 03 2011

Keywords

Comments

Column 2 of A199133.
a(n) is the last term in row n of triangle in A286030 (see also formulas below). Bob Selcoe, Sep 26 2021

Examples

			Some solutions for n=5:
  0 1   0 1   0 1   0 1   0 1   0 1   0 1   0 1   0 1   0 1
  1 0   1 2   1 2   1 2   1 0   1 2   1 0   1 2   1 2   1 0
  0 2   2 0   0 1   2 0   0 2   2 1   0 2   0 1   2 0   2 1
  2 1   0 2   2 0   0 1   2 1   1 0   2 0   1 2   0 1   1 2
  0 2   2 1   0 2   2 0   1 2   0 2   1 2   2 0   1 2   2 0
		

Crossrefs

Cf. A286030.

Formula

Conjecture: a(3n+2) = a(3n+3) = A208881(n+1). - R. J. Mathar, Nov 01 2015
Conjecture: -(458*n-1205) *(n+2) *(n+1)*a(n) +(-208*n^3+2578*n^2-4613*n-2410) *a(n-1) +9*(-339*n-638) *a(n-2) +27*(n-2) *(458*n^2-289*n-1146) *a(n-3) +54*(n-2) *(n-3) *(104*n-1081) *a(n-4)=0. - R. J. Mathar, Nov 01 2015
Conjecture: (n+2)*(n+1)*a(n) +(5*n^2-2)*a(n-1) +3*(5*n^2-15*n+3) *a(n-2) +3*(n^2 -60*n +81)*a(n-3) +135*(-n^2+3*n-1)*a(n-4) -405*(n-2)*(n-4) *a(n-5) -810*(n-4) *(n-5) *a(n-6)=0. - R. J. Mathar, Nov 01 2015
From Bob Selcoe, Sep 26 2021: (Start)
When n == 0 (mod 3), a(n) = n!/(3*(n/3)!^3);
when n == 1 (mod 3), a(n) = n!/(((n+2)/3)!*((n-1)/3)!^2);
when n == 2 (mod 3), a(n) = n!/(((n-2)/3)!*((n+1)/3)!^2).
(End)
Showing 1-4 of 4 results.