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 1-5 of 5 results.

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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Author

Keywords

Comments

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

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

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Author

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

Views

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

A098436 Triangle of 3rd central factorial numbers T(n,k).

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 73, 36, 1, 1, 585, 1045, 100, 1, 1, 4681, 28800, 7445, 225, 1, 1, 37449, 782281, 505280, 35570, 441, 1, 1, 299593, 21159036, 33120201, 4951530, 130826, 784, 1, 1, 2396745, 571593565, 2140851900, 652061451, 33209946, 399738, 1296, 1
Offset: 0

Views

Author

Ralf Stephan, Sep 08 2004

Keywords

Examples

			  1;
  1,   1;
  1,   9,    1;
  1,  73,   36,   1;
  1, 585, 1045, 100, 1;
  ...
		

Crossrefs

First column is A023001, first diagonal is A000537.
Row sums are in A098437.
Replace in recurrence (k+1)^3 with k: A008277; with k^2: A008957 (note offsets).

Programs

  • Maple
    A098436 := proc(n,k)
        option remember;
        if k=0 or k = n then
            1;
        else
            (k+1)^3*procname(n-1,k)+procname(n-1,k-1) ;
        end if;
    end proc:
    seq(seq( A098436(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Jan 13 2025
  • Mathematica
    T[n_, n_] = 1;
    T[n_ /; n>=0, k_] /; 0<=k<=n := T[n, k] = (k+1)^3 T[n-1, k]+T[n-1, k-1];
    T[, ] = 0;
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 08 2022 *)

Formula

Recurrence: T(n, k) = (k+1)^3*T(n-1, k) + T(n-1, k-1), T(0, 0)=1.

A303675 Triangle read by rows: coefficients in the sum of odd powers as expressed by Faulhaber's theorem, T(n, k) for n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 6, 1, 120, 30, 1, 5040, 1680, 126, 1, 362880, 151200, 17640, 510, 1, 39916800, 19958400, 3160080, 168960, 2046, 1, 6227020800, 3632428800, 726485760, 57657600, 1561560, 8190, 1, 1307674368000, 871782912000, 210680870400, 22313491200, 988107120, 14217840, 32766, 1
Offset: 1

Views

Author

Kolosov Petro, May 08 2018

Keywords

Comments

T(n,k) are the coefficients in an identity due to Faulhaber: Sum_{j=0..n} j^(2*m-1) = Sum_{k=1..m} T(m,k) binomial(n+k, 2*k). See the Knuth reference, page 10.
More explicitly, Faulhaber's theorem asserts that, given integers n >= 0, m >= 1 and odd, Sum_{k=1..n} k^m = Sum_{k=1..(m+1)/2} C(n+k,n-k)*[(1/k)*Sum_{j=0..k-1} (-1)^j*C(2*k,j)*(k-j)^(m+1)]. The coefficients T(m, k) are indicated by square brackets. Sums similar to this inner part are A304330, A304334, A304336; however, these triangles are (0,0)-based and lead to equivalent but slightly more systematic representations. - Peter Luschny, May 12 2018

Examples

			The triangle begins (see the Knuth reference p. 10):
         1;
         6,          1;
       120,         30,         1;
      5040,       1680,       126,        1;
    362880,     151200,     17640,      510,       1;
  39916800,   19958400,   3160080,   168960,    2046,    1;
6227020800, 3632428800, 726485760, 57657600, 1561560, 8190, 1;
.
Let S(n, m) = Sum_{j=1..n} j^m. Faulhaber's formula gives for m = 7 (m odd!):
F(n, 7) = 5040*C(n+4, 8) + 1680*C(n+3, 6) + 126*C(n+2, 4) + C(n+1, 2).
Faulhaber's theorem asserts that for all n >= 1 S(n, 7) = F(n, 7).
If n = 43 the common value is 1600620805036.
		

References

  • John H. Conway and Richard Guy, The Book of Numbers, Springer (1996), p. 107.

Crossrefs

First column is a bisection of A000142, second column is a bisection of A001720.
Row sums give A100868.

Programs

  • Maple
    T := proc(n,k) local m; m := n-k;
    2*(2*m+1)!*add((-1)^(j+m)*(j+1)^(2*n)/((j+m+2)!*(m-j)!), j=0..m) end:
    seq(seq(T(n, k), k=1..n), n=1..8); # Peter Luschny, May 09 2018
  • Mathematica
    (* After Peter Luschny's above formula. *)
    T[n_, k_] := (1/(n-k+1))*Sum[(-1)^j*Binomial[2*(n-k+1), j]*((n-k+1) - j)^(2*n), {j, 0, n-k+1}]; Column[Table[T[n, k], {n, 1, 10}, {k, 1, n}], Center]
  • Sage
    def A303675(n, k): return factorial(2*(n-k)+1)*A008957(n, k)
    for n in (1..7): print([A303675(n, k) for k in (1..n)]) # Peter Luschny, May 10 2018

Formula

T(n, k) = (2*(n-k)+1)!*A008957(n, k), n >= 1, 1 <= k <= n.
T(n, k) = (1/m)*Sum_{j=0..m} (-1)^j*binomial(2*m,j)*(m-j)^(2*n) where m = n-k+1. - Peter Luschny, May 09 2018

Extensions

New name by Peter Luschny, May 10 2018
Showing 1-5 of 5 results.