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A036969
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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.
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16
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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
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OFFSET
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1,5
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COMMENTS
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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
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)
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REFERENCES
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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.
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LINKS
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FORMULA
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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
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
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)
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EXAMPLE
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Triangle begins:
1;
1, 1;
1, 5, 1;
1, 21, 14, 1;
1, 85, 147, 30, 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}.
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MAPLE
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A036969 := proc(n, k) local j; 2*add(j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!), j=1..k); end;
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MATHEMATICA
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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 *)
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PROG
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(PARI) T(n, k)=if(1<k && k<=n, T(n-1, k-1) + k^2*T(n-1, k), k==1) \\ for illustrative purpose, not efficient ; M. 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
(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]))
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CROSSREFS
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KEYWORD
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nonn,easy,nice,tabl,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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