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 1-9 of 9 results.

A086764 Triangle T(n, k), read by row, related to Euler's difference table A068106 (divide column k of A068106 by k!).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 9, 11, 7, 3, 1, 44, 53, 32, 13, 4, 1, 265, 309, 181, 71, 21, 5, 1, 1854, 2119, 1214, 465, 134, 31, 6, 1, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 1, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1
Offset: 0

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Author

Philippe Deléham, Aug 02 2003

Keywords

Comments

The k-th column sequence, k >= 0, without leading zeros, enumerates the ways to distribute n beads, n >= 1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k+1 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords each contribute a factor 1, hence for n=0 one has 1. See A000255 for the description of a fixed cord with beads. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

Examples

			Formatted as a square array:
      1      3     7    13   21   31  43 57 ... A002061;
      2     11    32    71  134  227 356    ... A094792;
      9     53   181   465 1001 1909        ... A094793;
     44    309  1214  3539 8544             ... A094794;
    265   2119  9403 30637                  ... A023043;
   1854  16687 82508                        ... A023044;
  14833 148329                              ... A023045;
Formatted as a triangular array (mirror of A076731):
       1;
       0      1;
       1      1     1;
       2      3     2     1;
       9     11     7     3    1;
      44     53    32    13    4    1;
     265    309   181    71   21    5    1;
    1854   2119  1214   465  134   31    6   1;
   14833  16687  9403  3539 1001  227   43   7   1;
  133496 148329 82508 30637 8544 1909  356  57   8   1;
		

Crossrefs

Programs

  • Magma
    A086764:= func< n,k | (&+[(-1)^j*Binomial(n-k,j)*Factorial(n-j): j in [0..n]])/Factorial(k) >;
    [A086764(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Mathematica
    T[n_,k_]:=(1/k!)*Sum[(-1)^j*Binomial[n-k,j]*(n-j)!,{j,0,n}];Flatten[Table[T[n,k],{n,0,11},{k,0,n}]] (* Indranil Ghosh, Feb 20 2017 *)
    T[n_, k_] := (n!/k!) HypergeometricPFQ[{k-n},{-n},-1];
    Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)
  • SageMath
    def A086764(n,k): return sum((-1)^j*binomial(n-k,j)*factorial(n-j) for j in range(n+1))//factorial(k)
    flatten([[A086764(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, n) = 1; T(n+1, n) = n.
T(n+2, n) = A002061(n+1) = n^2 + n + 1; T(n+3, n) = n^3 + 3*n^2 + 5*n + 2.
T(n, k) = (k + 1)*T(n, k + 1) - T(n-1, k); T(n, n) = 1; T(n, k) = 0, if k > n.
T(n, k) = (n-1)*T(n-1, k) + (n-k-1)*T(n-2, k).
k!*T(n, k) = A068106(n, k). [corrected by Georg Fischer, Aug 13 2022]
Sum_{k>=0} T(n, k) = A003470(n+1).
T(n, k) = (1/k!) * Sum_{j>=0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, Jun 13 2005
From Peter Bala, Aug 14 2008: (Start)
The following remarks all relate to the array read as a square array: e.g.f for column k: exp(-y)/(1-y)^(k+1); e.g.f. for array: exp(-y)/(1-x-y) = (1 + x + x^2 + x^3 + ...) + (x + 2*x^2 + 3*x^3 + 4*x^4 + ...)*y + (1 + 3*x + 7*x^2 + 13*x^3 + ...)*y^2/2! + ... .
This table is closely connected to the constant e. The row, column and diagonal entries of this table occur in series formulas for e.
Row n for n >= 2: e = n!*(1/T(n,0) + (-1)^n*[1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) + 1/(3!*T(n,2)*T(n,3)) + ...]). For example, row 3 gives e = 6*(1/2 - 1/(1!*2*11) - 1/(2!*11*32) - 1/(3!*32*71) - ...). See A095000.
Column 0: e = 2 + Sum_{n>=2} (-1)^n*n!/(T(n,0)*T(n+1,0)) = 2 + 2!/(1*2) - 3 !/(2*9) + 4!/(9*44) - ... .
Column k, k >= 1: e = (1 + 1/1! + 1/2! + ... + 1/k!) + 1/k!*Sum_{n >= 0} (-1)^n*n!/(T(n,k)*T(n+1,k)). For example, column 3 gives e = 8/3 + 1/6*(1/(1*3) - 1/(3*13) + 2/(13*71) - 6/(71*465) + ...).
Main diagonal: e = 1 + 2*(1/(1*1) - 1/(1*7) + 1/(7*71) - 1/(71*1001) + ...).
First subdiagonal: e = 8/3 + 5/(3*32) - 7/(32*465) + 9/(465*8544) - ... .
Second subdiagonal: e = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...). See A143413.
Third subdiagonal: e = 3 - (2*3*5)/(2*53) + (3*4*7)/(53*1214) - (4*5*9)/(1214*30637) + ... .
For the corresponding results for the constants 1/e, sqrt(e) and 1/sqrt(e) see A143409, A143410 and A143411 respectively. For other arrays similarly related to constants see A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)). (End)
G.f. for column k is hypergeom([1,k+1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
T(n, k) = (n!/k!)*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017

Extensions

More terms from David Wasserman, Mar 28 2005
Additional comments from Zerinvary Lajos, Mar 30 2006
Edited by N. J. A. Sloane, Sep 24 2011

A003470 a(n) = n*a(n-1) - a(n-2) + 1 + (-1)^n.

Original entry on oeis.org

1, 1, 3, 8, 31, 147, 853, 5824, 45741, 405845, 4012711, 43733976, 520795003, 6726601063, 93651619881, 1398047697152, 22275111534553, 377278848390249, 6768744159489931, 128228860181918440, 2557808459478878871, 53585748788874537851, 1176328664895760953853
Offset: 0

Views

Author

Keywords

Comments

Row sums of A086764. - Philippe Deléham, Apr 27 2004
a(n+2m) == a(n) (mod m) for all n and m. - Robert Israel, Dec 06 2016

Examples

			G.f. = 1 + x + 3*x^2 + 8*x^3 + 31*x^4 + 147*x^5 + 853*x^6 + 5824*x^7 + ...
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({a(n) -(n-1)*a(n-1)-(n-2)*a(n-2)+a(n-3)-2=0,a(0)=1,a(1)=1,a(2)=3},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Dec 06 2016
  • Mathematica
    t = {1, 1}; Do[AppendTo[t, n*t[[-1]] - t[[-2]] + 1 + (-1)^n], {n, 2, 20}] (* T. D. Noe, Oct 07 2013 *)
    T[n_, k_] := HypergeometricPFQ[{k+1, k-n},{},-1];
    Table[Sum[(-1)^k T[n,k], {k,0,n}], {n,0,22}] (* Peter Luschny, Oct 05 2017 *)

Formula

Diagonal sums of reverse of permutation triangle A008279. a(n) = Sum_{k=0..floor(n/2)} (n-k)!/k!. - Paul Barry, May 12 2004
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*(n-2k)!. - Paul Barry Dec 15 2010
G.f.: 1/(1-x^2-x/(1-x/(1-x^2-2x/(1-2x/(1-x^2-3x/(1-3x/(1-x^2-4x/(1-4x/(1-.... (continued fraction);
G.f.: 1/(1-x-x^2-x^2/(1-3x-x^2-4x^2/(1-5x-x^2-9x^2/(1-7x-x^2-16x^2/(1-... (continued fraction). - Paul Barry, Dec 15 2010
G.f.: hypergeom([1,1],[],x/(1-x^2))/(1-x^2). - Mark van Hoeij, Nov 08 2011
G.f.: 1/Q(0), where Q(k)= 1 - x^2 - x*(k+1)/(1-x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 20 2013
From Robert Israel, Dec 06 2016: (Start)
a(2m) = hypergeom([1,-m,m+1],[],-1).
a(2m+1) = hypergeom([1,-m,m+2],[],-1)*(m+1).
a(2m-1) + a(2m+1) = (2m+1) a(2m). (End)
0 = a(n)*(-a(n+2) - a(n+3)) + a(n+1)*(-2 + a(n+1) - 2*a(n+3) + a(n+4)) + a(n+2)*(-2*a(n+3) + a(n+4)) + a(n+3)*(+2 - a(n+3)) if n >= 0. - Michael Somos, Dec 06 2016
0 = a(n)*(-a(n+2) + a(n+4)) + a(n+1)*(+a(n+1) - a(n+2) - a(n+3) + 3*a(n+4) - a(n+5)) + a(n+2)*(-a(n+3) + a(n+4)) + a(n+3)*(-a(n+4) + a(n+5)) + a(n+4)*(-a(n+4)) if n >= 0. - Michael Somos, Dec 06 2016
a(n) = Sum_{k=0..n} (-1)^k*hypergeom([k+1, k-n], [], -1). - Peter Luschny, Oct 05 2017
D-finite with recurrence: a(n) -n*a(n-1) +(n-2)*a(n-3) -a(n-4)=0. - R. J. Mathar, Apr 29 2020
a(n) ~ n! * (1 + 1/n + 1/(2*n^2) + 2/(3*n^3) + 25/(24*n^4) + 77/(40*n^5) + 2971/(720*n^6) + 6287/(630*n^7) + 1074809/(40320*n^8) + 28160749/(362880*n^9) + ...). - Vaclav Kotesovec, Nov 25 2022

Extensions

More terms from Gabriel Cunningham (gcasey(AT)mit.edu), Oct 25 2004

A145905 Square array read by antidiagonals: Hilbert transform of triangle A060187.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 27, 25, 7, 1, 1, 81, 125, 49, 9, 1, 1, 243, 625, 343, 81, 11, 1, 1, 729, 3125, 2401, 729, 121, 13, 1, 1, 2187, 15625, 16807, 6561, 1331, 169, 15, 1, 1, 6561, 78125, 117649, 59049, 14641, 2197, 225, 17, 1, 1, 19683, 390625, 823543
Offset: 0

Views

Author

Peter Bala, Oct 27 2008

Keywords

Comments

Definition of the Hilbert transform of a triangular array:
For many square arrays in the database the entries in a row are polynomial in the column index, of degree d say and hence the row generating function has the form P(x)/(1-x)^(d+1), where P is some polynomial function. Often the array whose rows are formed from the coefficients of these P polynomials is of independent interest. This suggests the following definition.
Let [L(n,k)]n,k>=0 be a lower triangular array and let R(n,x) := sum {k = 0 .. n} L(n,k)*x^k, denote the n-th row generating polynomial of L. Then we define the Hilbert transform of L, denoted Hilb(L), to be the square array whose n-th row, n >= 0, has the generating function R(n,x)/(1-x)^(n+1).
In this particular case, L is the array A060187, the array of Eulerian numbers of type B, whose row polynomials are the h-polynomials for permutohedra of type B. The Hilbert transform is an infinite Vandermonde matrix V(1,3,5,...).
We illustrate the Hilbert transform with a few examples:
(1) The Delannoy number array A008288 is the Hilbert transform of Pascal's triangle A007318 (view as the array of coefficients of h-polynomials of n-dimensional cross polytopes).
(2) The transpose of the array of nexus numbers A047969 is the Hilbert transform of the triangle of Eulerian numbers A008292 (best viewed in this context as the coefficients of h-polynomials of n-dimensional permutohedra of type A).
(3) The sequence of Eulerian polynomials begins [1, x, x + x^2, x + 4*x^2 + x^3, ...]. The coefficients of these polynomials are recorded in triangle A123125, whose Hilbert transform is A004248 read as square array.
(4) A108625, the array of crystal ball sequences for the A_n lattices, is the Hilbert transform of A008459 (viewed as the triangle of coefficients of h-polynomials of n-dimensional associahedra of type B).
(5) A142992, the array of crystal ball sequences for the C_n lattices, is the Hilbert transform of A086645, the array of h-vectors for type C root polytopes.
(6) A108553, the array of crystal ball sequences for the D_n lattices, is the Hilbert transform of A108558, the array of h-vectors for type D root polytopes.
(7) A086764, read as a square array, is the Hilbert transform of the rencontres numbers A008290.
(8) A143409 is the Hilbert transform of triangle A073107.

Examples

			Triangle A060187 (with an offset of 0) begins
1;
1, 1;
1, 6, 1;
so the entries in the first three rows of the Hilbert transform of
A060187 come from the expansions:
Row 0: 1/(1-x) = 1 + x + x^2 + x^3 + ...;
Row 1: (1+x)/(1-x)^2 = 1 + 3*x + 5*x^2 + 7*x^3 + ...;
Row 2: (1+6*x+x^2)/(1-x)^3 = 1 + 9*x + 25*x^2 + 49*x^3 + ...;
The array begins
n\k|..0....1.....2.....3......4
================================
0..|..1....1.....1.....1......1
1..|..1....3.....5.....7......9
2..|..1....9....25....49.....81
3..|..1...27...125...343....729
4..|..1...81...625..2401...6561
5..|..1..243..3125.16807..59049
...
		

Crossrefs

Cf. A008292, A039755, A052750 (first superdiagonal), A060187, A114172, A145901.

Programs

  • Maple
    T:=(n,k) -> (2*k + 1)^n: seq(seq(T(n-k,k),k = 0..n),n = 0..10);

Formula

T(n,k) = (2*k + 1)^n, (see equation 4.10 in [Franssens]). This array is the infinite Vandermonde matrix V(1,3,5,7, ....) having a LDU factorization equal to A039755 * diag(2^n*n!) * transpose(A007318).

A271705 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(n,j)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 15, 9, 1, 1, 64, 66, 16, 1, 1, 325, 490, 190, 25, 1, 1, 1956, 3915, 2120, 435, 36, 1, 1, 13699, 34251, 23975, 6755, 861, 49, 1, 1, 109600, 328804, 283136, 101990, 17696, 1540, 64, 1, 1, 986409, 3452436, 3534636, 1554966, 342846, 40404, 2556, 81, 1
Offset: 0

Views

Author

Peter Luschny, Apr 14 2016

Keywords

Comments

This is the Sheffer (aka exponential Riordan) matrix T = P*L = A007318*A271703 = (exp(x), x/(1-x)). Note that P = A007318 is Sheffer (exp(t), t) (of the Appell type). The Sheffer a-sequence is [1,1,repeat(0)] and the z-sequence has e.g.f. (x/(1+x))*(1 - exp(-x/(1+x)) given in A288869 / A000027. Because the column k=0 has only entries 1, the z-sequence gives fractional representations of 1. See A288869. - Wolfdieter Lang, Jun 20 2017

Examples

			Triangle starts:
  1;
  1,    1;
  1,    4,    1;
  1,   15,    9,    1;
  1,   64,   66,   16,   1;
  1,  325,  490,  190,  25,  1;
  1, 1956, 3915, 2120, 435, 36, 1;
  ...
Recurrence: T(3, 2) = (3/2)*4 + 3*1 = 9. - _Wolfdieter Lang_, Jun 20 2017
		

Crossrefs

Cf. A000290 (diag n, n-1), A062392 (diag n, n-2).
Cf. A007526 (col. 1), A134432 (col. 2).
Cf. A052844 (row sums), A059110 (matrix inverse).

Programs

  • Magma
    B:=Binomial;
    A271705:= func< n,k | k eq 0 select 1 else (&+[B(n, j+k)*B(j+k, k)*B(j+k-1, k-1)*Factorial(j): j in [0..n-k]]) >;
    [A271705(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 09 2022
    
  • Maple
    L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*binomial(n,n-k)*binomial(n-1,n-k)):
    T := (n,k) -> add(L(j,k)*binomial(-j-1,-n-1)*(-1)^(n-j), j=0..n):
    seq(seq(T(n,k), k=0..n), n=0..9);
  • Mathematica
    T[n_, k_]:= If[k==0, 1, Sum[((k*j!)/(j+k))*Binomial[n, j+k]*Binomial[j+k, k]^2, {j,0,n-k}]];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 09 2022 *)
  • Sage
    b=binomial
    def A271705(n,k): return 1 if (k==0) else sum(factorial(j-k)*b(n, j)*b(j, k)*b(j-1, k-1) for j in (k..n))
    flatten([[A271705(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jan 09 2022

Formula

From Wolfdieter Lang, Jun 20 2017: (Start)
T(n, k) = Sum_{m=k..n} A007318(n, m)*A271703(m, k), n >= k >= 0, and 0 for k < m. See also the name.
E.g.f. of column k: exp(x)*(x/(1-x))^k/k! (Sheffer property), k >= 0.
E.g.f. of triangle (or row polynomials in x): exp(z)*exp(x*z/(1-z)).
Recurrence for T(n, k), k >= 1, with T(n, 0) = 1, T(n, k) = 0 if n < k: T(n, k) = (n/k)*T(n-1, k-1) + n*T(n-1, k), n >= 1, k = 1..n. (From the a-sequence with column k=0 as input.) (End)
T(n, k) = Sum_{j=0..n-k} j!*binomial(n, j+k)*binomial(j+k, k)*binomial(j+k-1, k-1) with T(n, 0) = 1. - G. C. Greubel, Jan 09 2022
From Natalia L. Skirrow, Jun 11 2025: (Start)
T(n, k) = C(n, k)*hypergeom([k-n, k], [], -1), which equals C(n, k)*A143409(n-k, k-1) for k>0.
By the saddle point method upon the e.g.f., n-th row polynomial converges with n (for all y) to n^n*exp(2*sqrt(n*y) - n - y/2 + 1)/sqrt(2*sqrt(n/y)); as such, the n-th row's expectation is ~ sqrt(n)-1/4 and the n-th row's variance is ~ (sqrt(n)-1)/2. (End)

A143411 Square array, read by antidiagonals: form the Euler-Seidel matrix for the sequence {2^k*k!} and then divide column k by 2^k*k!.

Original entry on oeis.org

1, 3, 1, 13, 5, 1, 79, 33, 7, 1, 633, 277, 61, 9, 1, 6331, 2849, 643, 97, 11, 1, 75973, 34821, 7993, 1225, 141, 13, 1, 1063623, 493825, 114751, 17793, 2071, 193, 15, 1, 17017969, 7977173, 1870837, 292681, 34361, 3229, 253, 17, 1
Offset: 0

Views

Author

Peter Bala, Aug 19 2008

Keywords

Comments

This table is closely connected to the constant 1/sqrt(e). The row, column and diagonal entries of this table occur in series acceleration formulas for 1/sqrt(e). For a similar table based on the differences of the sequence {2^k*k!} and related to the constant sqrt(e), see A143410. For other arrays similarly related to constants see A086764 (for e), A143409 (for 1/e), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

Examples

			The Euler-Seidel matrix for the sequence {2^k*k!} begins
  ========================================
  n\k|     0     1     2     3     4     5
  ========================================
  0  |     1     2     8    48   384  3840
  1  |     3    10    56   432  4224
  2  |    13    66   488  4656
  3  |    79   554  5144
  4  |   633  5698
  5  |  6331
  ...
.
  Dividing the k-th column by 2^k*k! gives
  ========================================
  n\k|     0     1     2     3     4     5
  ========================================
  0  |     1     1     1     1     1     1
  1  |     3     5     7     9    11
  2  |    13    33    61    97
  3  |    79   277   643
  4  |   633  2849
  5  |  6331
  ...
		

Crossrefs

Programs

  • Magma
    A:= func< n,k | (&+[Binomial(n,j)*Factorial(k+j)*2^j/Factorial(k): j in [0..n]]) >; // Array
    A143411:= func< n,k | A(n-k,k) >; // antidiagonal triangle
    [A143411(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    with combinat: T := (n, k) -> 1/k!*add(2^j*binomial(n,j)*(k+j)!, j = 0..n): for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
  • Mathematica
    A[n_, k_]:= (1/k!)*Sum[Binomial[n,j]*(k+j)!*2^j, {j,0,n}]; (* array *)
    A143411[n_, k_]:= A[n-k,k]; (* antidiagonals *)
    Table[A143411[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 05 2023 *)
  • SageMath
    def A(n,k): return sum(binomial(n,j)*factorial(j+k)*2^j/factorial(k) for j in range(n+1)) # array
    def A143411(n,k): return A(n-k,k) # antidiagonal triangle
    flatten([[A143411(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n,k) = (1/k!)*Sum_{j = 0..n} 2^j*binomial(n,j)*(k+j)!.
Relation with Poisson-Charlier polynomials c_n(x,a):
T(n,k) = (-1)^n*c_n(-(k+1),1/2).
Recurrence relations:
T(n,k) = 2*n*T(n-1,k) + T(n,k-1);
T(n,k) = 2*(n+k)*T(n-1,k) + T(n-1,k-1);
T(n,k) = 2*(k+1)*T(n-1,k+1) + T(n-1,k).
Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k-1)*T(n,k-1) + T(n,k-2).
E.g.f. for column k: exp(y)/(1 - 2*y)^(k+1).
E.g.f. for array: exp(y)/(1 - x - 2*y) = (1 + x + x^2 + ...) + (3 + 5*x + 7*x^2 + ...)*y + (13 + 37*x + 61*x^2 + ...)*y^2/2! + ... .
Series acceleration formulas for 1/sqrt(e):
Row n: 1/sqrt(e) = 2^n*n!*(1/T(n,0) - 1/(2*1!*T(n,0)*T(n,1)) + 1/(2^2*2!*T(n,1)*T(n,2)) - 1/(2^3*3!*T(n,2)*T(n,3)) + ...). For example, row 3 gives 1/sqrt(e) = 48*(1/79 - 1/(2*79*277) + 1/(8*277*643) - 1/(48*643*1225) + ...).
Column k: 1/sqrt(e) = (1 - (1/2)/1! + (1/2)^2/2! - ... + (-1/2)^k/k!) + (-1)^(k+1)/(2^k*k!)*( Sum_{n = 0..inf} 2^n*n!/(T(n,k)*T(n+1,k)) ). For example, column 3 gives 1/sqrt(e) = 29/48 + 1/48*( 1/(1*9) + 2/(9*97) + 8/(97*1225) + 48/(1225*17793) + ... ).
Main diagonal: 1/sqrt(e) = 1 - 2*( 1/(1*5) - 1/(5*61) + 1/(61*1225) - ... ). See A065919.

A326659 T(n,k) = [0=0]; triangle T(n,k), n >= 0, 0 <= k <= n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 1, 15, 18, 6, 1, 64, 132, 96, 24, 1, 325, 980, 1140, 600, 120, 1, 1956, 7830, 12720, 10440, 4320, 720, 1, 13699, 68502, 143850, 162120, 103320, 35280, 5040, 1, 109600, 657608, 1698816, 2447760, 2123520, 1108800, 322560, 40320
Offset: 0

Views

Author

Alois P. Heinz, Sep 12 2019

Keywords

Comments

[] is an Iverson bracket.

Examples

			Triangle T(n,k) begins:
  1;
  1,     1;
  1,     4,     2;
  1,    15,    18,      6;
  1,    64,   132,     96,     24;
  1,   325,   980,   1140,    600,    120;
  1,  1956,  7830,  12720,  10440,   4320,   720;
  1, 13699, 68502, 143850, 162120, 103320, 35280, 5040;
  ...
		

Crossrefs

Columns k=0-2 give: A000012, A007526, 2*A134432(n-1).
Main diagonal gives A000142.
Row sums give A308876.

Programs

  • Maple
    T:= proc(n, k) option remember;
          `if`(0=0, 1, 0)
        end:
    seq(seq(T(n, k), k=0..n), n=0..10);
  • Mathematica
    T[n_ /; n >= 0, k_ /; k >= 0] := T[n, k] = Boole[0 < k <= n]*n*(T[n-1, k-1] + T[n-1, k]) + Boole[k == 0 && n >= 0];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 09 2021 *)

Formula

E.g.f. of column k: exp(x)*(x/(1-x))^k.
T(n,k) = k! * A271705(n,k).
T(n,k) = n * A073474(n-1,k-1) for n,k >= 1.
T(n,1) = n * A000522(n-1) for n >= 1.
T(n,2) = n * A093964(n-1) for n >= 1.
Sum_{k=1..n} k * T(n,k) = A327606(n).

A143410 Form the difference table of the sequence {2^k*k!}, then divide k-th column entries by 2^k*k!. Array read by ascending antidiagonals, T(n, k) for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 5, 3, 1, 29, 17, 5, 1, 233, 131, 37, 7, 1, 2329, 1281, 353, 65, 9, 1, 27949, 15139, 4105, 743, 101, 11, 1, 391285, 209617, 56189, 10049, 1349, 145, 13, 1, 6260561, 3325923, 883885, 156679, 20841, 2219, 197, 15, 1, 112690097, 59475329, 15700313
Offset: 0

Views

Author

Peter Bala, Aug 19 2008

Keywords

Comments

This table is closely connected to the constant sqrt(e). The row, column and diagonal entries of this table occur in series acceleration formulas for sqrt(e). For a similar table based on the Euler-Seidel matrix of the sequence {2^k*k!} and related to the constant 1/sqrt(e), see A143411. For other arrays similarly related to constants see A086764 (for e), A143409 (for 1/e), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

Examples

			Table of differences of {2^k*k!}
  =====================================================
  Column                0     1     2     3     4     5
  =====================================================
  Sequence 2^k*k!       1     2     8    48   384  3840
  First differences     1     6    40   336  3456
  Second differences    5    34   296  3120
  Third differences    29   262  2824
  Fourth differences  233  2562
  ...
Remove the common factor 2^k*k! from k-th column entries:
  ====================================
  n\k|   0      1      2      3      4
  ====================================
  0  |   1      1      1      1      1
  1  |   1      3      5      7      9
  2  |   5     17     37     65    101
  3  |  29    131    353    743   1349
  4  | 233   1281   4105  10049  20841
		

Crossrefs

Programs

  • Maple
    T := (n, k) -> (-1)^n/k!*add((-2)^j*binomial(n,j)*(k+j)!, j = 0..n):
    for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;

Formula

T(n,k) = ((-1)^n/k!)*Sum {j = 0..n} (-2)^j*C(n,j)*(k+j)!.
Relation with Poisson-Charlier polynomials c_n(x,a): T(n,k) = c_n(-(k+1),-1/2).
Recurrence relations: T(n,k) = 2*n*T(n-1,k) + T(n,k-1); T(n,k) = 2*(n+k)*T(n-1,k) - T(n-1,k-1); T(n,k) = 2*(k+1)*T(n-1,k+1) - T(n-1,k);
Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k+1)*T(n,k-1) - T(n,k-2).
E.g.f. for column k: exp(-y)/(1-2*y)^(k+1).
E.g.f. for array: exp(-y)/(1-x-2*y) = (1 + x + x^2 + ...) + (1 + 3*x + 5*x^2 + ...)*y + (5 + 17*x + 37*x^2 + ...)*y^2/2! + ... .
Series acceleration formulas for sqrt(e):
Row n: sqrt(e) = 2^n*n!*(1/T(n,0) + (-1)^n*(1/(2*1!*T(n,0)*T(n,1)) + 1/(2^2*2!*T(n,1)*T(n,2)) + 1/(2^3*3!*T(n,2)*T(n,3)) + ...)). For example, row 3 gives sqrt(e) = 48*(1/29 - 1/(2*29*131) - 1/(8*131*353) - 1/(48*353*743) - ...).
Column k: sqrt(e) = (1 + (1/2)/1! + (1/2)^2 / 2! + ... + (1/2)^k/k!) + 1/(2^k*k!) * Sum_{n>= 0} ((-2)^n *n!/(T(n,k)*T(n+1,k))). For example, column 3 gives sqrt(e) = 79/48 + (1/48)*(1/(1*7) - 2/(7*65) + 8/(65*743) - 48/(743*10049) + ...).
Main diagonal: sqrt(e) = 1 + 2*(1/(1*3) - 1/(3*37) + 1/(37*743) - ...). See A143412.
T(n, k) = (-1)^n*(-1/2)^(k + 1)*KummerU(k + 1, k + n + 2, -1/2). - Peter Luschny, Jan 02 2020

A273596 For n >= 2, a(n) is the number of slim rectangular diagrams of length n.

Original entry on oeis.org

1, 3, 9, 32, 139, 729, 4515, 32336, 263205, 2401183, 24275037, 269426592, 3257394143, 42615550453, 599875100487, 9040742057760, 145251748024649, 2478320458476795, 44755020000606961, 852823700470009056, 17101229029400788083, 359978633317886558801, 7936631162022905081707
Offset: 2

Views

Author

Tamas Dekany, May 26 2016

Keywords

Examples

			The initial term is the diagram of the four element diamond shape lattice.
		

Crossrefs

Programs

  • Maple
    A273596 := proc (n) option remember; `if`(n = 2, 1, `if`(n = 3, 3, (n-2)*procname(n-1) + procname(n-2) + 2)) end: seq(A273596(n), n = 2..20); # Peter Bala, Jan 08 2017
  • Mathematica
    x = 15;
    SRectD = Table[0, {x}];
    For[n = 2, n < x, n++,
    For[a = 1, a < n, a++,
       For[b = 1, b <= n - a, b++,
        SRectD[[n]] +=
          Binomial[n - a - 1, b - 1]*
           Binomial[n - b - 1, a - 1]*(n - a - b)!;
        ]
       ]
      Print[n, " ", SRectD[[n]]]
    ]
    (* Alternatively: *)
    T[n_, k_] := HypergeometricPFQ[{k+1, k-n}, {}, -1];
    Table[Sum[T[n,k], {k,0,n}], {n,0,22}] (* Peter Luschny, Oct 05 2017 *)
  • PARI
    a(n)= sum(rps=1, n, sum(r=1, n, s = rps-r; binomial(n-r-1, s-1) * binomial(n-s-1, r-1) * (n-r-s)!)); \\ Michel Marcus, Jun 12 2016

Formula

a(n) = Sum_{1<=r,s; r+s<=n} binomial(n-r-1, s-1) * binomial(n-s-1, r-1) * (n-r-s)!.
a(n) ~ exp(2) * n! / n^2. - Vaclav Kotesovec, Jun 29 2016
a(n) = Sum_{k=0..n} hypergeom([k+1, k-n], [], -1). - Peter Luschny, Oct 05 2017
From Peter Bala, Jan 08 2018: (Start)
a(n) = Sum_{k = 0..n-2} k!*binomial(n+k-1, 2*k+1).
a(n) = (n - 2)*a(n-1) + a(n-2) + 2, with a(2) = 1, a(3) = 3.
a(n+2) = 1/n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)* A000522(n)^2.
Row sums of array A143409 read as a triangle.
O.g.f.: Sum_{n >= 0} n!*x^(n+2)/(1 - x)^(2*n+2). Cf. A000179, A000271, A000904 and A127548.
O.g.f. with offset 0: 1/(1 - x) o 1/(1 - x) = 1 + 3*x + 9*x^2 + 32*x^3 + ..., where o denotes the white diamond multiplication of power series. See the Bala link for details. (End)

A271704 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j,-n)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 11, 8, 1, 0, 49, 57, 15, 1, 0, 261, 424, 174, 24, 1, 0, 1631, 3425, 1930, 410, 35, 1, 0, 11743, 30336, 21855, 6320, 825, 48, 1, 0, 95901, 294553, 259161, 95235, 16835, 1491, 63, 1, 0, 876809, 3123632, 3251500, 1452976, 325150, 38864, 2492, 80, 1
Offset: 0

Views

Author

Peter Luschny, Apr 14 2016

Keywords

Examples

			Triangle starts:
  [1]
  [0, 1]
  [0, 3,     1]
  [0, 11,    8,     1]
  [0, 49,    57,    15,    1]
  [0, 261,   424,   174,   24,   1]
  [0, 1631,  3425,  1930,  410,  35,  1]
  [0, 11743, 30336, 21855, 6320, 825, 48, 1]
		

Crossrefs

A001339 (col. 1), A005563 (diag. n,n-1).

Programs

  • Maple
    L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*binomial(n,n-k)*binomial(n-1,n-k)):
    T := (n,k) -> add(L(j,k)*binomial(-j,-n)*(-1)^(n-j), j=0..n):
    seq(seq(T(n,k), k=0..n), n=0..9);

Formula

From Natalia L. Skirrow, Jun 12 2025: (Start)
Definition can also be written Sum_{j=0..n} C(n-1, j-1)*L(j, k), where C(n, -1) = (1 if n = -1 else 0) over integer n.
T(n, k) = C(n-1, k-1)*A143409(n-k, k) = k*(k+1)*A271705(n+1, k+1)/(n*(n+1)) for n > 0.
E.g.f. for sequence b(n, k) = T(n+1, k+1): F/(1-x)^2, where F = exp(x + y*x/(1-x)) is e.g.f. of A271705.
E.g.f. for kth column of b: exp(x)*x^(k-1)/(1-x)^(k+1)/(k-1)!. (These cannot be integrated with x to give e.g.f.s for T(n, k) using standard functions.)
T(n, k) = Sum_{i=0..n-k} (n-1)_i*(i+1)*A271705(n-1-i, k-1), where (n)_i = n!/(n-i)! is the falling factorial. (End)
Showing 1-9 of 9 results.