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A036969 Triangle read by rows: T(n,k) = T(n-1,k-1) + k^2*T(n-1,k), 1 < k <= n, T(n,1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 14, 1, 1, 85, 147, 30, 1, 1, 341, 1408, 627, 55, 1, 1, 1365, 13013, 11440, 2002, 91, 1, 1, 5461, 118482, 196053, 61490, 5278, 140, 1, 1, 21845, 1071799, 3255330, 1733303, 251498, 12138, 204, 1, 1, 87381, 9668036, 53157079, 46587905
Offset: 1

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Or, triangle of central factorial numbers T(2n,2k) (in Riordan's notation).
Can be used to calculate the Bernoulli numbers via the formula B_2n = (1/2)*Sum_{k = 1..n} (-1)^(k+1)*(k-1)!*k!*T(n,k)/(2*k+1). E.g., n = 1: B_2 = (1/2)*1/3 = 1/6. n = 2: B_4 = (1/2)*(1/3 - 2/5) = -1/30. n = 3: B_6 = (1/2)*(1/3 - 2*5/5 + 2*6/7) = 1/42. - Philippe Deléham, Nov 13 2003
From Peter Bala, Sep 27 2012: (Start)
Generalized Stirling numbers of the second kind. T(n,k) is equal to the number of partitions of the set {1,1',2,2',...,n,n'} into k disjoint nonempty subsets V1,...,Vk such that, for each 1 <= j <= k, if i is the least integer such that either i or i' belongs to Vj then {i,i'} is a subset of Vj. An example is given below.
Thus T(n,k) may be thought of as a two-colored Stirling number of the second kind. See Matsumoto and Novak, who also give another combinatorial interpretation of these numbers. (End)

Examples

			Triangle begins:
  1;
  1,    1;
  1,    5,      1;
  1,   21,     14,      1;
  1,   85,    147,     30,     1;
  1,  341,   1408,    627,    55,    1;
  1, 1365,  13013,  11440,  2002,   91,   1;
  1, 5461, 118482, 196053, 61490, 5278, 140, 1;
  ...
T(3,2) = 5: The five set partitions into two sets are {1,1',2,2'}{3,3'}, {1,1',3,3'}{2,2'}, {1,1'}{2,2',3,3'}, {1,1',3}{2,2',3'} and {1,1',3'}{2,2',3}.
		

References

  • L. Carlitz, A conjecture concerning Genocchi numbers. Norske Vid. Selsk. Skr. (Trondheim) 1971, no. 9, 4 pp. [The triangle appears on page 2.]
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.

Crossrefs

Columns are A002450, A002451.
Diagonals are A000330 and A060493.
Transpose of A008957.
(0,0)-based version: A269945.
Cf. A008955, A008956, A156289, A135920 (row sums), A204579 (inverse), A000290.

Programs

  • Haskell
    a036969 n k = a036969_tabl !! (n-1) (k-1)
    a036969_row n = a036969_tabl !! (n-1)
    a036969_tabl = iterate f [1] where
       f row = zipWith (+)
         ([0] ++ row) (zipWith (*) (tail a000290_list) (row ++ [0]))
    -- 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;
  • Mathematica
    t[n_, k_] := 2*Sum[j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!), {j, 1, k}]; Flatten[ Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, Oct 11 2011 *)
    t1[n_, k_] := (1/(2 k)!) * Sum[Binomial[2 k, j]*(-1)^j*(k - j)^(2 n), {j, 0, 2 k}]; Column[Table[t1[n, k], {n, 1, 10}, {k, 1, n}]] (* Kolosov Petro ,Jul 26 2023 *)
  • PARI
    T(n,k)=if(1M. F. Hasler, Feb 03 2012
    
  • PARI
    T(n,k)=2*sum(j=1,k,(-1)^(k-j)*j^(2*n)/(k-j)!/(k+j)!)  \\ M. F. Hasler, Feb 03 2012
    
  • Sage
    def A036969(n,k) : return (2/factorial(2*k))*add((-1)^j*binomial(2*k,j)*(k-j)^(2*n) for j in (0..k))
    for n in (1..7) : print([A036969(n,k) for k in (1..n)]) # Peter Luschny, Feb 03 2012
    

Formula

T(n,k) = A156289(n,k)/A001147(k). - Peter Bala, Feb 21 2011
From Peter Bala, Oct 14 2011: (Start)
O.g.f.: Sum_{n >= 1} x^n*t^n/Product_{k = 1..n} (1 - k^2*t^2) = x*t + (x + x^2)*t^2 + (x + 5*x^2 + x^3)*t^3 + ....
Define polynomials x^[2*n] = Product_{k = 0..n-1} (x^2 - k^2). This triangle gives the coefficients in the expansion of the monomials x^(2*n) as a linear combination of x^[2*m], 1 <= m <= n. For example, row 4 gives x^8 = x^[2] + 21*x^[4] + 14*x^[6] + x^[8].
A008955 is a signed version of the inverse.
The n-th row sum = A135920(n). (End)
T(n,k) = (2/(2*k)!)*Sum_{j=0..k-1} (-1)^(j+k+1) * binomial(2*k,j+k+1) * (j+1)^(2*n). This formula is valid for n >= 0 and 0 <= k <= n. - Peter Luschny, Feb 03 2012
From Peter Bala, Sep 27 2012: (Start)
Let E(x) = cosh(sqrt(2*x)) = Sum_{n >= 0} x^n/((2*n)!/2^n). A generating function for the triangle is E(t*(E(x)-1)) = 1 + t*x + t*(1 + t)*x^2/6 + t*(1 + 5*t + t^2)*x^3/90 + ..., where the sequence of denominators [1, 1, 6, 90, ...] is given by (2*n)!/2^n. Cf. A008277 which has generating function exp(t*(exp(x)-1)). An e.g.f. is E(t*(E(x^2/2)-1)) = 1 + t*x^2/2! + t*(1 + t)*x^4/4! + t*(1 + 5*t + t^2)*x^6/6! + ....
Put c(n) := (2*n)!/2^n. The column k generating function is (1/c(k))*(E(x)-1)^k = Sum_{n >= k} T(n,k)*x^n/c(n). The inverse array is A204579.
The production array begins:
1, 1;
0, 4, 1;
0, 0, 9, 1;
0, 0, 0, 16, 1;
... (End)
x^n = Sum_{k=1..n} T(n,k)*Product_{i=0..k-1} (x-i^2), see Stanley link. - Michel Marcus, Nov 19 2014; corrected by Kolosov Petro, Jul 26 2023
From Kolosov Petro, Jul 26 2023: (Start)
T(n,k) = (1/(2*k)!) * Sum_{j=0..2k} binomial(2k, j)*(-1)^j*(k - j)^(2n).
T(n,k) = (1/(k*(2k-1)!)) * Sum_{j=0..k} (-1)^(k-j)*binomial(2k, k-j)*j^(2n). (End)

Extensions

More terms from Vladeta Jovovic, Apr 16 2000