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 1-10 of 16 results. Next

A060493 A diagonal of A036969.

Original entry on oeis.org

0, 1, 21, 147, 627, 2002, 5278, 12138, 25194, 48279, 86779, 148005, 241605, 380016, 578956, 857956, 1240932, 1756797, 2440113, 3331783, 4479783, 5939934, 7776714, 10064110, 12886510, 16339635, 20531511, 25583481, 31631257, 38826012, 47335512, 57345288, 69059848
Offset: 0

Views

Author

Larry Reeves (larryr(AT)acm.org), Mar 20 2001

Keywords

Crossrefs

Programs

  • PARI
    a(n)=n*(n + 1)*(n + 2)*(2*n + 1)*(2*n + 3)*(5*n - 1)/360

Formula

From Benoit Cloitre, Mar 20 2004: (Start)
a(n) = n*(n + 1)*(n + 2)*(2*n + 1)*(2*n + 3)*(5*n - 1)/360.
a(n) = Sum_{k=1..n} k^2 * Sum_{i=1..k} i^2.
a(n) = Sum_{k=1..n} k^2*A000330(k). (End)
G.f.: -x*(4*x^3+21*x^2+14*x+1) / (x-1)^7. - Colin Barker, Dec 19 2012
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+4) * binomial(2*n, n-j). - Peter Bala, Mar 31 2025

Extensions

Missing a(0)=0 inserted by Alois P. Heinz, Feb 19 2022

A204579 Triangle read by rows: matrix inverse of the central factorial numbers T(2*n, 2*k) (A036969).

Original entry on oeis.org

1, -1, 1, 4, -5, 1, -36, 49, -14, 1, 576, -820, 273, -30, 1, -14400, 21076, -7645, 1023, -55, 1, 518400, -773136, 296296, -44473, 3003, -91, 1, -25401600, 38402064, -15291640, 2475473, -191620, 7462, -140, 1, 1625702400, -2483133696, 1017067024, -173721912, 14739153, -669188, 16422, -204, 1
Offset: 1

Views

Author

M. F. Hasler, Feb 03 2012

Keywords

Comments

This is a signed version of A008955 with rows in reverse order. - Peter Luschny, Feb 04 2012

Examples

			Triangle starts:
  [1]         1;
  [2]        -1,        1;
  [3]         4,       -5,         1;
  [4]       -36,       49,       -14,       1;
  [5]       576,     -820,       273,     -30,       1;
  [6]    -14400,    21076,     -7645,    1023,     -55,    1;
  [7]    518400,  -773136,    296296,  -44473,    3003,  -91,    1;
  [8] -25401600, 38402064, -15291640, 2475473, -191620, 7462, -140, 1;
		

Crossrefs

Cf. A036969, A008955, A008275, A121408, A001044 (column 1), A101686 (alternating row sums), A234324 (central terms).

Programs

  • Maple
    # From Peter Luschny, Feb 29 2024: (Start)
    ogf := n -> local j; z^2*mul(z^2 - j^2, j = 1..n-1):
    Trow := n -> local k; seq(coeff(expand(ogf(n)), z, 2*k), k = 1..n):
    # Alternative:
    f := w -> (w^sqrt(t) + w^(-sqrt(t)))/2: egf := f((x/2 + sqrt(1 + (x/2)^2))^2):
    ser := series(egf, x, 20): cx := n -> coeff(ser, x, 2*n):
    Trow := n -> local k; seq((2*n)!*coeff(cx(n), t, k), k = 1..n):  # (End)
    # Assuming offset 0:
    rowpoly := n -> (-1)^n * pochhammer(1 - sqrt(x), n) * pochhammer(1 + sqrt(x), n):
    row := n -> local k; seq(coeff(expand(rowpoly(n)), x, k), k = 0..n):
    seq(print(row(n)), n = 0..7);  # Peter Luschny, Aug 03 2024
  • Mathematica
    rows = 10;
    t[n_, k_] := 2*Sum[j^(2*n)*(-1)^(k - j)/((k - j)!*(k + j)!), {j, 1, k}];
    T = Table[t[n, k], {n, 1, rows}, {k, 1, rows}] // Inverse;
    Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 14 2018 *)
  • PARI
    select(concat(Vec(matrix(10,10,n,k,T(n,k)/*from A036969*/)~^-1)), x->x)
    
  • Sage
    def A204579(n, k): return (-1)^(n-k)*A008955(n, n-k)
    for n in (0..7): print([A204579(n, k) for k in (0..n)]) # Peter Luschny, Feb 05 2012

Formula

T(n, k) = (-1)^(n-k)*A008955(n, n-k). - Peter Luschny, Feb 05 2012
T(n, k) = Sum_{i=k-n..n-k} (-1)^(n-k+i)*s(n,k+i)*s(n,k-i) = Sum_{i=0..2*k} (-1)^(n+i)*s(n,i)*s(n,2*k-i), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 07 2012
From Peter Bala, Aug 29 2012: (Start)
T(n, k) = T(n-1, k-1) - (n-1)^2*T(n-1, k). (Recurrence equation.)
Let E(x) = cosh(sqrt(2*x)) = Sum_{n >= 0} x^n/{(2*n)!/2^n} and
L(x) = 2*{arcsinh(sqrt(x/2))}^2 = Sum_{n >=1} (-1)^n*(n-1)!^2*x^n/{(2*n)!/2^n}.
L(x) is the compositional inverse of E(x) - 1.
A generating function for the triangle is E(t*L(x)) = 1 + t*x + t*(-1 + t)*x^2/6 + t*(4 - 5*t + t^2)*x^3/90 + ..., where the sequence of denominators [1,1,6,90,...] is given by (2*n)!/2^n. Cf. A008275 with generating function exp(t*log(1+x)).
The e.g.f. is E(t*L(x^2/2)) = cosh(2*sqrt(t)*arcsinh(x/2)) = 1 + t*x^2/2! + t*(t-1)*x^4/4! + t*(t-1)*(t-4)*x^6/6! + .... (End)
From Peter Luschny, Feb 29 2024: (Start)
T(n, k) = [z^(2*k)] z^2*Product_{j=1..n-1} (z^2 - j^2).
T(n, k) = (2*n)! * [t^k] [x^(2*n)] (w^sqrt(t) + w^(-sqrt(t)))/2 where w = (x/2 + sqrt(1 + (x/2)^2))^2. (End)
T(n, k) = [x^k] (-1)^n * Pochhammer(1 - sqrt(x), n) * Pochhammer(1 + sqrt(x), n), assuming offset 0. - Peter Luschny, Aug 03 2024
Integral_{0..oo} x^s / (cosh(x))^(2*n) dx = (2^(2*n - s - 1) * s! * (-1)^(n-1)) / (2*n - 1)!)*Sum_{k=1..n} T(n,k)*DirichletEta(s - 2*k + 2). - Ammar Khatab, Apr 11 2025

Extensions

Typo in data corrected by Peter Luschny, Feb 05 2012

A008955 Triangle of central factorial numbers |t(2n,2n-2k)| read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 4, 1, 14, 49, 36, 1, 30, 273, 820, 576, 1, 55, 1023, 7645, 21076, 14400, 1, 91, 3003, 44473, 296296, 773136, 518400, 1, 140, 7462, 191620, 2475473, 15291640, 38402064, 25401600, 1, 204, 16422, 669188, 14739153, 173721912, 1017067024, 2483133696, 1625702400
Offset: 0

Views

Author

Keywords

Comments

Discussion of Central Factorial Numbers by N. J. A. Sloane, Feb 01 2011: (Start)
Here is Riordan's definition of the central factorial numbers t(n,k) given in Combinatorial Identities, Section 6.5:
For n >= 0, expand the polynomial
x^[n] = x*Product{i=1..n-1} (x+n/2-i) = Sum_{k=0..n} t(n,k)*x^k.
The t(n,k) are not always integers. The cases n even and n odd are best handled separately.
For n=2m, we have:
x^[2m] = Product_{i=0..m-1} (x^2-i^2) = Sum_{k=1..m} t(2m,2k)*x^(2k).
E.g. x^[8] = x^2(x^2-1^2)(x^2-2^2)(x^2-3^2) = x^8-14x^6+49x^4-36x^2,
which corresponds to row 4 of the present triangle.
So the m-th row of the present triangle gives the absolute values of the coefficients in the expansion of Product_{i=0..m-1} (x^2-i^2).
Equivalently, and simpler, the n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first.
For n odd, n=2m+1, we have:
x^[2m+1] = x*Product_{i=0..m-1}(x^2-((2i+1)/2)^2) = Sum_{k=0..m} t(2m+1,2k+1)*x^(2k+1).
E.g. x^[5] = x(x^2-(1/2)^2)(x^2-(3/2)^2) = x^5-10x^3/4+9x/16,
which corresponds to row 2 of the triangle in A008956.
We now rescale to get integers by replacing x by x/2 and multiplying by 2^(2m+1) (getting 1, -10, 9 from the example).
The result is that row m of triangle A008956 gives the coefficients in the expansion of x*Product_{i=0..m} (x^2-(2i+1)^2).
Equivalently, and simpler, the n-th row of A008956 gives the coefficients in the expansion of Product_{i=0..n-1} (x+(2i+1)^2), highest powers first.
Note that the n-th row of A182867 gives the coefficients in the expansion of Product_{i=1..n} (x+(2i)^2), highest powers first.
(End)
Contribution from Johannes W. Meijer, Jun 18 2009: (Start)
We define Beta(n-z,n+z)/Beta(n,n) = Gamma(n-z)*Gamma(n+z)/Gamma(n)^2 = sum(EG2[2m,n]*z^(2m), m = 0..infinity) with Beta(z,w) the Beta function. The EG2[2m,n] coefficients are quite interesting, see A161739. Our definition leads to EG2[2m,1] = 2*eta(2m) and the recurrence relation EG2[2m,n] = EG2[2m,n-1] - EG2[2m-2,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... , with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. We found for the matrix coefficients EG2[2m,n] = sum((-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2,k=1..n) with the central factorial numbers t1(n,m) as defined above, see also the Maple program.
From the EG2 matrix we arrive at the ZG2 matrix, see A161739 for its odd counterpart, which is defined by ZG2[2m,1] = 2*zeta(2m) and the recurrence relation ZG2[2m,n] = ZG2[2m-2,n-1]/(n*(n-1))-(n-1)*ZG2[2m,n-1]/n for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... . We found for the ZG2[2m,n] = Sum_{k=1..n} (-1)^(k+1)*t1(n-1,k-1)* 2* zeta(2*m-2*n+2*k)/((n-1)!*(n)!), and we see that the central factorial numbers t1(n,m) once again play a crucial role.
(End)

Examples

			Triangle begins:
  1;
  1,   1;
  1,   5,   4;
  1,  14,  49,  36;
  1,  30, 273, 820, 576;
  ...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part 1, Springer-Verlag 1985.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

Crossrefs

Cf. A036969.
Columns include A000330, A000596, A000597. Right-hand columns include A001044, A001819, A001820, A001821. Row sums are in A101686.
Appears in A160464 (Eta triangle), A160474 (Zeta triangle), A160479 (ZL(n)), A161739 (RSEG2 triangle), A161742, A161743, A002195, A002196, A162440 (EG1 matrix), A162446 (ZG1 matrix) and A163927. - Johannes W. Meijer, Jun 18 2009, Jul 06 2009 and Aug 17 2009
Cf. A234324 (central terms).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return 1;
        elif k=n then return (Factorial(n))^2;
        else return n^2*T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..8], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Sep 14 2019
  • Haskell
    a008955 n k = a008955_tabl !! n !! k
    a008955_row n = a008955_tabl !! n
    a008955_tabl = [1] : f [1] 1 1 where
       f xs u t = ys : f ys v (t * v) where
         ys = zipWith (+) (xs ++ [t^2]) ([0] ++ map (* u^2) (init xs) ++ [0])
         v = u + 1
    -- Reinhard Zumkeller, Dec 24 2013
    
  • Magma
    T:= func< n,k | Factorial(2*(n+1))*(&+[(-1)^j*Binomial(n,k-j)*(&+[2^(m-2*k)*StirlingFirst(2*(n-k+1)+m, 2*(n-k+1))*Binomial(2*(n-k+1)+2*j-1, 2*(n-k+1)+m-1)/Factorial(2*(n-k+1)+m): m in [0..2*j]]): j in [0..k]]) >;
    [T(n,k): k in [0..n], n in [0..8]]; // G. C. Greubel, Sep 14 2019
    
  • Maple
    nmax:=7: for n from 0 to nmax do t1(n, 0):=1: t1(n, n):=(n!)^2 end do: for n from 1 to nmax do for k from 1 to n-1 do t1(n, k) := t1(n-1, k-1)*n^2 + t1(n-1, k) end do: end do: seq(seq(t1(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jun 18 2009, Revised Sep 16 2012
    t1 := proc(n,k)
            sum((-1)^j*stirling1(n+1,n+1-k+j)*stirling1(n+1,n+1-k-j),j=-k..k) ;
    end proc: # Mircea Merca, Apr 02 2012
    # third Maple program:
    T:= proc(n, k) option remember; `if`(k=0, 1,
          add(T(j-1, k-1)*j^2, j=1..n))
        end:
    seq(seq(T(n, k), k=0..n), n=0..8);  # Alois P. Heinz, Feb 19 2022
  • Mathematica
    t[n_, 0]=1; t[n_, n_]=(n!)^2; t[n_ , k_ ]:=t[n, k] = n^2*t[n-1, k-1] + t[n-1, k]; Flatten[Table[t[n, k], {n,0,8}, {k,0,n}] ][[1 ;; 42]]
    (* Jean-François Alcover, May 30 2011, after recurrence formula *)
  • Maxima
    T(n,m):=(2*(n+1))!*sum((-1)^k*binomial(n,m-k)*sum((2^(i-2*m)*stirling1(2*(n-m+1)+i,2*(n-m+1))*binomial(2*(n-m+1)+2*k-1,2*(n-m+1)+i-1))/(2*(n-m+1)+i)!,i,0,2*k),k,0,m); /* Vladimir Kruchinin, Oct 05 2013 */
    
  • PARI
    T(n,k)=if(k==0,1, if(k==n, (n!)^2, n^2*T(n-1, k-1) + T(n-1, k)));
    for(n=0,8, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 14 2019
    
  • Sage
    # This triangle is (0,0)-based.
    def A008955(n, k) :
        if k==0 : return 1
        if k==n : return factorial(n)^2
        return n^2*A008955(n-1, k-1) + A008955(n-1, k)
    for n in (0..7) : print([A008955(n, k) for k in (0..n)]) # Peter Luschny, Feb 04 2012
    

Formula

The n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first (see Comments section).
The triangle can be obtained from the recurrence t1(n,k) = n^2*t1(n-1,k-1) + t1(n-1,k) with t1(n,0) = 1 and t1(n,n) = (n!)^2.
t1(n,k) = Sum_{j=-k..k} (-1)^j*s(n+1,n+1-k+j)*s(n+1,n+1-k-j) = Sum_{j=0..2*(n+1-k)} (-1)^(n+1-k+j)*s(n+1,j)*s(n+1,2*(n+1-k)-j), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 02 2012
E.g.f.: cosh(2/sqrt(t)*asin(sqrt(t)*z/2)) = 1 + z^2/2! + (1 + t)*z^4/4! + (1 + 5*t + 4*t^2)*z^6/6! + ... (see Berndt, p.263 and p.306). - Peter Bala, Aug 29 2012
T(n,m) = (2*(n+1))!*Sum_{k=0..m} ((-1)^k*binomial(n,m-k)*Sum_{i=0..2*k} ((2^(i-2*m)*stirling1(2*(n-m+1)+i,2*(n-m+1))*binomial(2*(n-m+1)+2*k-1, 2*(n-m+1)+i-1))/(2*(n-m+1)+i)!)). - Vladimir Kruchinin, Oct 05 2013

Extensions

There's an error in the last column of Riordan's table (change 46076 to 21076).
More terms from Vladeta Jovovic, Apr 16 2000
Link added and cross-references edited by Johannes W. Meijer, Aug 17 2009
Discussion of Riordan's definition of central factorial numbers added by N. J. A. Sloane, Feb 01 2011

A241171 Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.

Original entry on oeis.org

1, 1, 6, 1, 30, 90, 1, 126, 1260, 2520, 1, 510, 13230, 75600, 113400, 1, 2046, 126720, 1580040, 6237000, 7484400, 1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400, 1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000, 1, 131070, 96461910, 8203431600, 196556560200, 1882311631200, 8266953895200, 16672848192000, 12504636144000
Offset: 1

Views

Author

N. J. A. Sloane, Apr 22 2014

Keywords

Comments

T(n,k) gives the number of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. An example is given below. Cf. A019538 and A156289. - Peter Bala, Aug 20 2014

Examples

			Triangle begins:
1,
1, 6,
1, 30, 90,
1, 126, 1260, 2520,
1, 510, 13230, 75600, 113400,
1, 2046, 126720, 1580040, 6237000, 7484400,
1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400,
1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000,
...
From _Peter Bala_, Aug 20 2014: (Start)
Row 2: [1,6]
k  Ordered set partitions of {1,2,3,4} into k blocks    Number
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1   {1,2,3,4}                                             1
2   {1,2}{3,4}, {3,4}{1,2}, {1,3}{2,4}, {2,4}{1,3},       6
    {1,4}{2,3}, {2,3}{1,4}
(End)
		

References

  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
  • S. A. Joffe, Calculation of the first thirty-two Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 47 (1914), 103-126.
  • S. A. Joffe, Calculation of eighteen more, fifty in all, Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 48 (1917-1920), 193-271.

Crossrefs

Case m=2 of the polynomials defined in A278073.
Cf. A000680 (diagonal), A094088 (row sums), A000364 (alternating row sums), A281478 (central terms), A327022 (refinement).
Diagonals give A002446, A213455, A241172, A002456.

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->1/(2^(k-1))*Sum([1..k],j->(-1)^(k-j)*Binomial(2*k,k-j)*j^(2*n))))); # Muniru A Asiru, Feb 27 2019
  • Maple
    T := proc(n,k) option remember;
    if k > n then 0
    elif k=0 then k^n
    elif k=1 then 1
    else k*(2*k-1)*T(n-1,k-1)+k^2*T(n-1,k); fi;
    end: # Minor edit to make it also work in the (0,0)-offset case. Peter Luschny, Sep 03 2022
    for n from 1 to 12 do lprint([seq(T(n,k), k=1..n)]); od:
  • Mathematica
    T[n_, k_] /; 1 <= k <= n := T[n, k] = k(2k-1) T[n-1, k-1] + k^2 T[n-1, k]; T[, 1] = 1; T[, ] = 0; Table[T[n, k], {n, 1, 9}, {k, 1, n}] (* _Jean-François Alcover, Jul 03 2019 *)
  • Sage
    @cached_function
    def A241171(n, k):
        if n == 0 and k == 0: return 1
        if k < 0 or k > n: return 0
        return (2*k^2 - k)*A241171(n - 1, k - 1) + k^2*A241171(n - 1, k)
    for n in (1..6): print([A241171(n, k) for k in (1..n)]) # Peter Luschny, Sep 06 2017
    

Formula

T(n,k) = 0 if k <= 0 or k > n, = 1 if k=1, otherwise T(n,k) = k*(2*k-1)*T(n-1,k-1) + k^2*T(n-1,k).
Related to Euler numbers A000364 by A000364(n) = (-1)^n*Sum_{k=1..n} (-1)^k*T(n,k). For example, A000364(3) = 61 = 90 - 30 + 1.
From Peter Bala, Aug 20 2014: (Start)
T(n,k) = 1/(2^(k-1))*Sum_{j = 1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
T(n,k) = k!*A156289(n,k) = k!*(2*k-1)!!*A036969.
E.g.f.: A(t,z) := 1/( 1 - t*(cosh(z) - 1) ) = 1 + t*z^2/2! + (t + 6*t^2)*z^4/4! + (t + 30*t^2 + 90*t^3)*z^6/6! + ... satisfies the partial differential equation d^2/dz^2(A) = D(A), where D = t^2*(2*t + 1)*d^2/dt^2 + t*(5*t + 1)*d/dt + t.
Hence the row polynomials R(n,t) satisfy the differential equation R(n+1,t) = t^2*(2*t + 1)*R''(n,t) + t*(5*t + 1)*R'(n,t) + t*R(n,t) with R(0,t) = 1, where ' indicates differentiation w.r.t. t. This is equivalent to the above recurrence equation.
Recurrence for row polynomials: R(n,t) = t*( Sum_{k = 1..n} binomial(2*n,2*k)*R(n-k,t) ) with R(0,t) := 1.
Row sums equal A094088(n) for n >= 1.
A100872(n) = (1/2)*R(n,2). (End)

A156289 Triangle read by rows: T(n,k) is the number of end rhyme patterns of a poem of an even number of lines (2n) with 1<=k<=n evenly rhymed sounds.

Original entry on oeis.org

1, 1, 3, 1, 15, 15, 1, 63, 210, 105, 1, 255, 2205, 3150, 945, 1, 1023, 21120, 65835, 51975, 10395, 1, 4095, 195195, 1201200, 1891890, 945945, 135135, 1, 16383, 1777230, 20585565, 58108050, 54864810, 18918900, 2027025, 1, 65535, 16076985
Offset: 1

Views

Author

Hartmut F. W. Hoft, Feb 07 2009

Keywords

Comments

T(n,k) is the number of partitions of a set of size 2*n into k blocks of even size [Comtet]. For partitions into odd sized blocks see A136630.
See A241171 for the triangle of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. - Peter Bala, Aug 20 2014
This triangle T(n,k) gives the sum over the M_3 multinomials A036040 for the partitions of 2*n with k even parts, for 1 <= k <= n. See the triangle A257490 with sums over the entries with k parts, and the Hartmut F. W. Hoft program. - Wolfdieter Lang, May 13 2015

Examples

			The triangle begins
  n\k|..1.....2......3......4......5......6
  =========================================
  .1.|..1
  .2.|..1.....3
  .3.|..1....15.....15
  .4.|..1....63....210....105
  .5.|..1...255...2205...3150....945
  .6.|..1..1023..21120..65835..51975..10395
  ..
T(3,3) = 15. The 15 partitions of the set [6] into three even blocks are:
  (12)(34)(56), (12)(35)(46), (12)(36)(45),
  (13)(24)(56), (13)(25)(46), (13)(26)(45),
  (14)(23)(56), (14)(25)(36), (14)(26)(35),
  (15)(23)(46), (15)(24)(36), (15)(26)(34),
  (16)(23)(45), (16)(24)(35), (16)(25)(34).
Examples of recurrence relation
 T(4,3) = 5*T(3,2) + 9*T(3,3) = 5*15 + 9*15 = 210;
 T(6,5) = 9*T(5,4) + 25*T(5,5) = 9*3150 + 25*945 = 51975.
 T(4,2) = 28 + 35 = 63 (M_3 multinomials A036040 for partitions of 8 with 3 even parts, namely (2,6) and (4^2)). - _Wolfdieter Lang_, May 13 2015
		

References

  • L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pages 225-226.

Crossrefs

Diagonal T(n, n) is A001147, subdiagonal T(n+1, n) is A001880.
2nd column variant T(n, 2)/3, for 2<=n, is A002450.
3rd column variant T(n, 3)/15, for 3<=n, is A002451.
Sum of the n-th row is A005046.

Programs

  • Maple
    T := proc(n,k) option remember; `if`(k = 0 and n = 0, 1, `if`(n < 0, 0,
    (2*k-1)*T(n-1, k-1) + k^2*T(n-1, k))) end:
    for n from 1 to 8 do seq(T(n,k), k=1..n) od; # Peter Luschny, Sep 04 2017
  • Mathematica
    T[n_,k_] := Which[n < k, 0, n == 1, 1, True, 2/Factorial2[2 k] Sum[(-1)^(k + j) Binomial[2 k, k + j] j^(2 n), {j, 1, k}]]
    (* alternate computation with function triangle[] defined in A257490 *)
    a[n_]:=Map[Apply[Plus,#]&,triangle[n],{2}]
    (* Hartmut F. W. Hoft, Apr 26 2015 *)

Formula

Recursion: T(n,1)=1 for 1<=n; T(n,k)=0 for 1<=n
Generating function for the k-th column of the triangle T(i+k,k):
G(k,x) = Sum_{i>=0} T(i+k,k)*x^i = Product_{j=1..k} (2*j-1)/(1-j^2*x).
Closed form expression: T(n,k) = (2/(k!*2^k))*Sum_{j=1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
From Peter Bala, Feb 21 2011: (Start)
GENERATING FUNCTION
E.g.f. (including a constant 1):
(1)... F(x,z) = exp(x*(cosh(z)-1))
= Sum_{n>=0} R(n,x)*z^(2*n)/(2*n)!
= 1 + x*z^2/2! + (x + 3*x^2)*z^4/4! + (x + 15*x^2 + 15*x^3)*z^6/6! + ....
ROW POLYNOMIALS
The row polynomials R(n,x) begin
... R(1,x) = x
... R(2,x) = x + 3*x^2
... R(3,x) = x + 15*x^2 + 15*x^3.
The egf F(x,z) satisfies the partial differential equation
(2)... d^2/dz^2(F) = x*F + x*(2*x+1)*F' + x^2*F'',
where ' denotes differentiation with respect to x. Hence the row polynomials satisfy the recurrence relation
(3)... R(n+1,x) = x*{R(n,x) + (2*x+1)*R'(n,x) + x*R''(n,x)}
with R(0,x) = 1. The recurrence relation for T(n,k) given above follows from this.
(4)... T(n,k) = (2*k-1)!!*A036969(n,k).
(End)

A298851 a(n) = [x^n] Product_{k=1..n} 1/(1-k^2*x).

Original entry on oeis.org

1, 1, 21, 1408, 196053, 46587905, 16875270660, 8657594647800, 5974284925007685, 5336898188553325075, 5992171630749371157181, 8260051854943114812198756, 13714895317396748230146099660, 26998129079190909699998105620908, 62173633286588800021263427046090792
Offset: 0

Author

Seiichi Manyama, Feb 01 2018

Keywords

Crossrefs

Programs

  • Maple
    b:= proc(k, n) option remember; `if`(k=0, 1,
          add(b(k-1, j)*j^2, j=1..n))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..14);  # Alois P. Heinz, Feb 19 2022
  • Mathematica
    Table[SeriesCoefficient[Product[1/(1 - k^2*x), {k, 1, n}], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Feb 02 2018 *)
    Join[{1}, Table[2*Sum[(-1)^(n-k) * Binomial[2*n, n-k] * k^(4*n), {k, 0, n}]/(2*n)!, {n, 1, 20}]] (* Vaclav Kotesovec, May 15 2025 *)
  • Maxima
    a(n):=if n<1 then 1 else 2*sum((n-k)^(4*n)/((2*n-k)!*k!*(-1)^k),k,0,n);
    makelist(a(n), n, 0, 20); /* Tani Akinari, Mar 09 2021 */

Formula

From Vaclav Kotesovec, Feb 02 2018, updated May 12 2025: (Start)
a(n) ~ c * d^n * n^(2*n - 1/2), where d = 1.774513671664430848697327843228386312953174421074432567764556466987... and c = 0.617929515483613293691991371141292259390065108300160936187723552669...
In closed form, a(n) ~ n^(2*n - 1/2) * r^(4*n + 1) / (sqrt(Pi*(2 - r^2)) * (r^2 - 1)^n * exp(2*n)), where r = 1.04438203376083348498401390634474776086902815721... is the root of the equation (1-r)/(1+r) = -exp(-4/r). (End)
a(n) = 2*(Sum_{k=0..n} (n-k)^(4*n)/((2*n-k)!*k!*(-1)^k)) for n>0. - Tani Akinari, Mar 09 2021
a(n) = A036969(2n,n) = A269945(2n,n). - Alois P. Heinz, Feb 19 2022
From Seiichi Manyama, May 12 2025: (Start)
a(n) = Sum_{k=0..2*n} (-n)^k * binomial(4*n,k) * Stirling2(4*n-k,2*n).
a(n) = Sum_{k=0..2*n} (-1)^k * Stirling2(k+n,n) * Stirling2(3*n-k,n). (End)

A008958 Triangle of central factorial numbers 4^k T(2n+1, 2n+1-2k).

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 35, 91, 1, 1, 84, 966, 820, 1, 1, 165, 5082, 24970, 7381, 1, 1, 286, 18447, 273988, 631631, 66430, 1, 1, 455, 53053, 1768195, 14057043, 15857205, 597871, 1, 1, 680, 129948, 8187608, 157280838, 704652312, 397027996, 5380840, 1
Offset: 0

Keywords

Examples

			From _Wesley Transue_, Jan 21 2012: (Start)
Triangle begins:
  1;
  1,   1;
  1,  10,      1;
  1,  35,     91,       1;
  1,  84,    966,     820,         1;
  1, 165,   5082,   24970,      7381,         1;
  1, 286,  18447,  273988,    631631,     66430,         1;
  1, 455,  53053, 1768195,  14057043,  15857205,    597871,       1;
  1, 680, 129948, 8187608, 157280838, 704652312, 397027996, 5380840, 1;
(End)
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

Crossrefs

Columns include A000447. Right-hand columns include A002452, A002453.

Programs

  • Mathematica
    Flatten[Table[Sum[(-1)^(q+1) 4^(p-n) (2p+2q-2n-1)^(2n+1)/((2n+1-2p-q)! q!), {q, 0, n-p}], {n, 0, 8}, {p, 0, n}]] (* Wesley Transue, Jan 21 2012 *)

Formula

G.f. of i-th right-hand column is x/Product_{j=1..i+1} (1 - (2j-1)^2*x).

Extensions

More terms from Vladeta Jovovic, Apr 16 2000

A269945 Triangle read by rows. Stirling set numbers of order 2, T(n, n) = 1, T(n, k) = 0 if k < 0 or k > n, otherwise T(n, k) = T(n-1, k-1) + k^2*T(n-1, k), for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 5, 1, 0, 1, 21, 14, 1, 0, 1, 85, 147, 30, 1, 0, 1, 341, 1408, 627, 55, 1, 0, 1, 1365, 13013, 11440, 2002, 91, 1, 0, 1, 5461, 118482, 196053, 61490, 5278, 140, 1, 0, 1, 21845, 1071799, 3255330, 1733303, 251498, 12138, 204, 1
Offset: 0

Author

Peter Luschny, Mar 22 2016

Keywords

Comments

Also known as central factorial numbers T(2*n, 2*k) (cf. A036969).
The analog for the Stirling cycle numbers is A269944.

Examples

			Triangle starts:
  [0] [1]
  [1] [0, 1]
  [2] [0, 1,   1]
  [3] [0, 1,   5,    1]
  [4] [0, 1,  21,   14,   1]
  [5] [0, 1,  85,  147,  30,  1]
  [6] [0, 1, 341, 1408, 627, 55, 1]
		

Crossrefs

Columns k=0..5 give A000007, A000012, A002450(n-1), A002451(n-3), A383838(n-4), A383840(n-5).
Variants are: A008957, A036969.
Cf. A007318 (order 0), A048993 (order 1), A269948 (order 3).
Cf. A000330 (subdiagonal), A002450 (column 2), A135920 (row sums), A269941, A269944 (Stirling cycle), A298851 (central terms).

Programs

  • Maple
    T := proc(n, k) option remember;
        `if`(n=k, 1,
        `if`(k<0 or k>n, 0,
         T(n-1, k-1) + k^2*T(n-1, k))) end:
    for n from 0 to 9 do seq(T(n, k), k=0..n) od;
    # Alternatively with the P-transform (cf. A269941):
    A269945_row := n -> PTrans(n, n->`if`(n=1, 1, 1/(n*(4*n-2))), (n, k)->(-1)^k*(2*n)!/(2*k)!): seq(print(A269945_row(n)), n=0..8);
    # Using the exponential generating function:
    egf := 1 + t^2*(cosh(2*sinh(t*x/2)/t));
    ser := series(egf, x, 20): cx := n -> coeff(ser, x, 2*n):
    Trow := n -> local k; seq((2*n)!*coeff(cx(n), t, 2*(n-k+1)), k = 0..n):
    seq(print(Trow(n)), n = 0..9);  # Peter Luschny, Feb 29 2024
  • Mathematica
    T[n_, n_] = 1; T[n_ /; n >= 0, k_] /; 0 <= k < n := T[n, k] = T[n - 1, k - 1] + k^2*T[n - 1, k]; T[, ] = 0; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Nov 27 2017 *)
  • Sage
    # uses[PtransMatrix from A269941]
    stirset2 = lambda n: 1 if n == 1 else 1/(n*(4*n-2))
    norm = lambda n,k: (-1)^k*factorial(2*n)/factorial(2*k)
    M = PtransMatrix(7, stirset2, norm)
    for m in M: print(m)

Formula

T(n, k) = (-1)^k*((2*n)! / (2*k)!)*P[n, k](s(n)) where P is the P-transform and s(n) = 1/(n*(4*n-2)). The P-transform is defined in the link. Compare also the Sage and Maple implementations below.
T(n, 2) = (4^(n - 1) - 1)/3 for n >= 2 (cf. A002450).
T(n, n-1) = n*(n - 1)*(2*n - 1)/6 for n >= 1 (cf. A000330).
From Fabián Pereyra, Apr 25 2022: (Start)
T(n, k) = (1/(2*k)!)*Sum_{j=0..2*k} (-1)^j*binomial(2*k, j)*(k - j)^(2*n).
T(n, k) = Sum_{j=2*k..2*n} (-k)^(2*n - j)*binomial(2*n, j)*Stirling2(j, 2*k).
T(n, k) = Sum_{j=0..2*n} (-1)^(j - k)*Stirling2(2*n - j, k)*Stirling2(j, k). (End)
T(n, k) = (2*n)! [t^(2*(n-k+1))] [x^(2*n)] (1 + t^2*(cosh(2*sinh(t*x/2)/t))). - Peter Luschny, Feb 29 2024

A002451 Expansion of 1/((1-x)*(1-4*x)*(1-9*x)).

Original entry on oeis.org

1, 14, 147, 1408, 13013, 118482, 1071799, 9668036, 87099705, 784246870, 7059619931, 63542171784, 571901915677, 5147206719578, 46325218390143, 416928397167052, 3752361301126529, 33771274616631006, 303941563175648035, 2735474435084708240, 24619271381777877861
Offset: 0

Keywords

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • T. N. Thiele, Interpolationsrechnung. Teubner, Leipzig, 1909, p. 35.

Crossrefs

A diagonal of A036969.

Programs

Formula

a(n) = 1/24 - 4^(n+2)/15 + 9^(n+2)/40. - Antonio Alberto Olivares, Feb 03 2010
a(n) = 13*a(n-1) - 36*a(n-2) + 1, n >= 2. - Vincenzo Librandi, Mar 23 2011

A008957 Triangle of central factorial numbers T(2*n,2*n-2*k), k >= 0, n >= 1 (in Riordan's notation).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 14, 21, 1, 1, 30, 147, 85, 1, 1, 55, 627, 1408, 341, 1, 1, 91, 2002, 11440, 13013, 1365, 1, 1, 140, 5278, 61490, 196053, 118482, 5461, 1, 1, 204, 12138, 251498, 1733303, 3255330, 1071799, 21845, 1, 1, 285, 25194, 846260, 10787231
Offset: 1

Keywords

Comments

D. E. Knuth [1992] on page 10 gives the formula Sum n^(2m-1) = Sum_{k=1..m} (2k-1)! T(2m,2k) binomial(n+k, 2k) where T(m, k) is the central factorial number of the second kind. - Michael Somos, May 08 2018

Examples

			The triangle starts:
  1;
  1,  1;
  1,  5,    1;
  1, 14,   21,     1;
  1, 30,  147,    85,     1;
  1, 55,  627,  1408,   341,    1;
  1, 91, 2002, 11440, 13013, 1365, 1;
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217, Table 6.2(a).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.

Crossrefs

Row reversed version of A036969. (0,0)-based version: A269945.
Cf. A008955.

Programs

  • Haskell
    a008957 n k = a008957_tabl !! (n-1) (k-1)
    a008957_row n = a008957_tabl !! (n-1)
    a008957_tabl = map reverse a036969_tabl
    -- Reinhard Zumkeller, Feb 18 2013
    
  • Maple
    A036969 := proc(n,k) local j; 2*add(j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!),j=1..k); end; # Gives rows of triangle in reversed order
  • Mathematica
    t[n_, n_] = t[n_, 1] = 1;
    t[n_, k_] := t[n-1, k-1] + k^2 t[n-1, k];
    Flatten[Table[t[n, k], {n, 1, 10}, {k, n, 1, -1}]][[1 ;; 50]] (* Jean-François Alcover, Jun 16 2011 *)
  • PARI
    {T(n, k) = if( n<1 || k>n, 0, n==k || k==1, 1, T(n-1, k-1) + k^2 * T(n-1, k))}; \\ Michael Somos, May 08 2018
    
  • Sage
    def A008957(n, k):
        m = n - k
        return 2*sum((-1)^(j+m)*(j+1)^(2*n)/(factorial(j+m+2)*factorial(m-j)) for j in (0..m))
    for n in (1..7): print([A008957(n, k) for k in (1..n)]) # Peter Luschny, May 10 2018

Formula

From Michael Somos, May 08 2018: (Start)
T(n, k) = T(n-1, k-1) + k^2 * T(n-1, k), where T(n, n) = T(n, 1) = 1.
E.g.f.: x^2 * (cosh(sinh(y*x/2) / (x/2)) - 1) = (1*x^2)*y^2/2! + (1*x^2 + 1*x^4)*y^4/4! + (1*x^2 + 5*x^4 + 1*x^6)*y^6/6! + (1*x^2 + 14*x^4 + 21*x^6 + 1*x^8)*y^8/8! + ... (End)

Extensions

More terms from Vladeta Jovovic, Apr 16 2000
Showing 1-10 of 16 results. Next