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 1-6 of 6 results.

A112293 Row sums of number triangle A112292.

Original entry on oeis.org

1, 2, 7, 36, 253, 2278, 25059, 325768, 4886521, 83070858, 1578346303, 33145272364, 762341264373, 19058531609326, 514580353451803, 14922830250102288, 462607737753170929, 15266055345854640658, 534311937104912423031
Offset: 0

Views

Author

Paul Barry, Sep 01 2005

Keywords

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := If[k <= n, (2 n - 1)!! / (2 k - 1)!!, 0];
    a[n_] := Sum[T[n, k], {k, 0, n}];
    Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Jun 13 2019 *)

Formula

a(n) = Sum_{k=0..n} (2n-1)!!/(2k-1)!!.
a(n) = Sum_{k=0..n} 2^(n-k)(n-1/2)!/(k-1/2)!.
a(n) = Sum_{k=0..n} n!C(2n, n)2^(k-n)/(k!C(2k, k)).
a(n) = Sum_{k=0..n} (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)).
Conjecture: a(n) - 2*n*a(n-1) + (2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 28 2014
a(n) = floor((2*n-1)!! * C) for n>0, where C = 1 + sqrt(e*Pi/2)*erf(1/sqrt(2)). - Don Knuth, Mar 25 2018
a(n) = 2^n*(C*Gamma(n + 1/2) + Gamma(n + 1/2, 1/2))*sqrt(e/2) for n >= 0, where C = sqrt(2/(e*Pi)) - erfc(1/sqrt(2)). - Peter Luschny, Mar 25 2018
a(n) = a(n-1) * (2*n-1) + 1 for n > 0 and a(0) = 1; that proves the conjecture of R. J. Mathar from Nov 28 2014; G.f. A(x) satisfies the equation: A(x) = 1/(1-x)^2 + A'(x) * 2*x^2/(1-x), where A' is the first derivative of A. - Werner Schulte, Oct 18 2023

A112294 Diagonal sums of number triangle A112292.

Original entry on oeis.org

1, 1, 4, 18, 121, 1055, 11376, 145852, 2165689, 36532197, 689873284, 14415662558, 330204013569, 8226714806323, 221470234531456, 6406397898751320, 198160750081637521, 6526636388676159977, 228040136335670652324
Offset: 0

Views

Author

Paul Barry, Sep 01 2005

Keywords

Formula

a(n)=sum{k=0..floor(n/2), (2n-2k-1)!!/(2k-1)!!}; a(n)=sum{k=0..floor(n/2), 2^(n-2k)(n-k-1/2)!/(k-1/2)!}; a(n)=sum{k=0..floor(n/2), (n-k)!C(2n-2k, n-k)2^(2k-n)/(k!C(2k, k))}; a(n)=sum{k=0..floor(n/2), (n-k+1)!*C(n-k)2^(2k-n)/((k+1)!*C(k))}.

A094587 Triangle of permutation coefficients arranged with 1's on the diagonal. Also, triangle of permutations on n letters with exactly k+1 cycles and with the first k+1 letters in separate cycles.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
Offset: 0

Views

Author

Paul Barry, May 13 2004

Keywords

Comments

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007
From Tom Copeland, Nov 01 2007: (Start)
T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = Sum_{j=0..n} (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)
6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)
7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
From Peter Bala, Jul 10 2008: (Start)
This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009
T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011
T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011
T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013
For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016
See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016
Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016
The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017
Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017
If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

Examples

			Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}, ...
For n=3 and k=1, T(3,1)=6 since there are exactly 6 permutations of {1,2,3,4} with exactly 2 cycles and with 1 and 2 in separate cycles. The permutations are (1)(2 3 4), (1)(2 4 3), (1 3)(2 4), (1 4)(2 3), (1 3 4)(2), and (1 4 3)(2). - _Dennis P. Walsh_, Jan 24 2011
Triangle begins:
     1,
     1,    1,
     2,    2,    1,
     6,    6,    3,    1,
    24,   24,   12,    4,    1,
   120,  120,   60,   20,    5,    1,
   720,  720,  360,  120,   30,    6,    1,
  5040, 5040, 2520,  840,  210,   42,    7,    1
The production matrix is:
      1,     1,
      1,     1,     1,
      2,     2,     1,    1,
      6,     6,     3,    1,    1,
     24,    24,    12,    4,    1,   1,
    120,   120,    60,   20,    5,   1,   1,
    720,   720,   360,  120,   30,   6,   1,   1,
   5040,  5040,  2520,  840,  210,  42,   7,   1,   1,
  40320, 40320, 20160, 6720, 1680, 336,  56,   8,   1,   1
which is the exponential Riordan array A094587, or [1/(1-x),x], with an extra superdiagonal of 1's.
Inverse begins:
   1,
  -1,  1,
   0, -2,  1,
   0,  0, -3,  1,
   0,  0,  0, -4,  1,
   0,  0,  0,  0, -5,  1,
   0,  0,  0,  0,  0, -6,  1,
   0,  0,  0,  0,  0,  0, -7,  1
		

Crossrefs

Programs

  • Haskell
    a094587 n k = a094587_tabl !! n !! k
    a094587_row n = a094587_tabl !! n
    a094587_tabl = map fst $ iterate f ([1], 1)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
    
  • Maple
    T := proc(n, m): n!/m! end: seq(seq(T(n, m), m=0..n), n=0..9);  # Johannes W. Meijer, Oct 07 2009, revised Nov 25 2012
    # Alternative: Note that if you leave out 'abs' you get A021009.
    T := proc(n, k) option remember; if n = 0 and k = 0 then 1 elif k < 0 or k > n then 0 else abs((n + k)*T(n-1, k) - T(n-1, k-1)) fi end: #  Peter Luschny, Dec 30 2021
  • Mathematica
    Flatten[Table[Table[n!/k!, {k,0,n}], {n,0,10}]] (* Geoffrey Critzer, Dec 11 2011 *)
  • Sage
    def A094587_row(n): return (factorial(n)*exp(x).taylor(x,0,n)).list()
    for n in (0..7): print(A094587_row(n)) # Peter Luschny, Sep 28 2017

Formula

T(n, k) = n!/k! if n >= k >= 0, otherwise 0.
T(n, k) = Sum_{i=k..n} |S1(n+1, i+1)*S2(i, k)| * (-1)^i, with S1, S2 the Stirling numbers.
T(n,k) = (n-k)*T(n-1,k) + T(n-1,k-1). E.g.f.: exp(x*y)/(1-y) = 1 + (1+x)*y + (2+2*x+x^2)*y^2/2! + (6+6*x+3*x^2+x^3)*y^3/3!+ ... . - Peter Bala, Jul 10 2008
A094587 = 1 / ((-1)*A129184 * A127648 + I), I = Identity matrix. - Gary W. Adamson, May 03 2009
From Johannes W. Meijer, Oct 07 2009: (Start)
The o.g.f. of right hand column k is Gf(z;k) = (k-1)!/(1-z)^k, k => 1.
The recurrence relations of the right hand columns lead to Pascal's triangle A007318. (End)
Let f(x) = (1/x)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+1)*R(n,x)-x*R'(n,x). Cf. A132159. - Peter Bala, Oct 28 2011
A padded shifted version of this lower triangular matrix with zeros in the first column and row except for a one in the diagonal position is given by integral(t=0 to t=infinity) exp[-t(I-P)] = 1/(I-P) = I + P^2 + P^3 + ... where P is the infinitesimal generator matrix A218234 and I the identity matrix. The non-padded version is given by P replaced by A132440. - Tom Copeland, Oct 25 2012
From Peter Bala, Aug 28 2013: (Start)
The row polynomials R(n,x) form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x).
Let P(n,x) = Product_{k=0..n-1} (x + k) denote the rising factorial polynomial sequence with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (6, 6, 3, 1) so P(3,x + 1) = (x + 1)*(x + 2)*(x + 3) = 6 + 6*x + 3*x*(x + 1) + x*(x + 1)*(x + 2). (End)
From Tom Copeland, Apr 21 & 26, and Aug 13 2014: (Start)
T-I = M = -A021009*A132440*A021009 with e.g.f. y*exp(x*y)/(1-y). Cf. A132440. Dividing the n-th row of M by n generates the (n-1)th row of T.
T = 1/(I - A132440) = {2*I - exp[(A238385-I)]}^(-1) = unsigned exp[(I-A238385)] = exp[A000670(.)*(A238385-I)] = , umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)/(1-y), so the row polynomials form an Appell sequence with lowering operator d/dx and raising operator x + 1/(1-D).
With L(n,m,x)= Laguerre polynomials of order m, the row polynomials are (-1)^n*n!*L(n,-1-n,x) = (-1)^n*(-1!/(-1-n)!)*K(-n,-1-n+1,x) = n!* K(-n,-n,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-1-n,-:xD:) = (-1)^n*x^(n+1)*:Dx:^n*x^(-1-n) = (-1)^n*x*:xD:^n*x^(-1) = (-1)^n*n!*binomial(xD-1,n) = n!*K(-n,-n,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706 and A132159.
The n-th row of signed M has the coefficients of d[(-:xD:)^n]/d(:Dx:)= f[d/d(-:xD:)](-:xD:)^n with f(y)=y/(y-1), :Dx:^n= n!L(n,0,-:xD:), and (-:xD:)^n = n!L(n,0,:Dx:). M has the coefficients of [D/(1-D)]x^n. (End)
From Tom Copeland, Nov 18 2015: (Start)
Coefficients of the row polynomials of the e.g.f. Sum_{n>=0} P_n(b1,b2,..,bn;t) x^n/n! = e^(P.(..;t) x) = e^(xt) / (1-b.x) = (1 + b1 x + b2 x^2 + b3 x^3 + ...) e^(xt) = 1 + (b1 + t) x + (2 b2 + 2 b1 t + t^2) x^2/2! + (6 b3 + 6 b2 t + 3 b1 t^2 + t^3) x^3/3! + ... , with lowering operator L = d/dt, i.e., L P_n(..;t) = n * P_(n-1)(..;t), and raising operator R = t + d[log(1 + b1 D + b2 D^2 + ...)]/dD = t - Sum_{n>=1} F(n,b1,..,bn) D^(n-1), i.e., R P_n(..,;t) = P_(n+1)(..;t), where D = d/dt and F(n,b1,..,bn) are the Faber polynomials of A263916.
Also P_n(b1,..,bn;t) = CIP_n(t-F(1,b1),-F(2,b1,b2),..,-F(n,b1,..,bn)), the cycle index polynomials A036039.
(End)
The raising operator R = x + 1/(1-D) = x + 1 + D + D^2 + ... in matrix form acting on an o.g.f. (formal power series) is the transpose of the production matrix M below. The linear term x is the diagonal of ones after transposition. The other transposed diagonals come from D^m x^n = n! / (n-m)! x^(n-m). Then P(n,x) = (1,x,x^2,..) M^n (1,0,0,..)^T is a matrix representation of R P(n-1,x) = P(n,x). - Tom Copeland, Aug 17 2016
The row polynomials have e.g.f. e^(xt)/(1-t) = exp(t*q.(x)), umbrally. With p_n(x) the row polynomials of A132013, q_n(x) = v_n(p.(u.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form [T] = [q] = [v]*[p]*[u]. Conversely, p_n(x) = u_n (q.(v.(x))). - Tom Copeland, Nov 10 2016
From the Appell sequence formalism, 1/(1-b.D) t^n = P_n(b1,b2,..,bn;t), the generalized row polynomials noted in the Nov 18 2015 formulas, consistent with the 2007 comments. - Tom Copeland, Nov 22 2016
From Peter Bala, Feb 18 2017: (Start)
G.f.: Sum_{n >= 1} (n*x)^(n-1)/(1 + (n - t)*x)^n = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ....
n-th row polynomial R(n,t) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^k*(x + k - t)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^(n-k)*(x + k + t)^k, for arbitrary x. The particular case of the latter sum when x = 0 and t = 1 is identity 10.35 in Gould, Vol.4. (End)
Rodrigues-type formula for the row polynomials: R(n, x) = -exp(x)*Int(exp(-x)* x^n, x), for n >= 0. Recurrence: R(n, x) = x^n + n*R(n-1, x), for n >= 1, and R(0, x) = 1. d/dx(R(n, x)) = R(n, x) - x^n, for n >= 0 (compare with the formula from Peter Bala, Aug 28 2013). - Wolfdieter Lang, Dec 23 2019
T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * n^i for 0 <= k <= n. - Werner Schulte, Jul 26 2022

Extensions

Edited by Johannes W. Meijer, Oct 07 2009
New description from Dennis P. Walsh, Jan 24 2011

A136214 Triangle U, read by rows, where U(n,k) = Product_{j=k..n-1} (3*j+1) for n > k with U(n,n) = 1.

Original entry on oeis.org

1, 1, 1, 4, 4, 1, 28, 28, 7, 1, 280, 280, 70, 10, 1, 3640, 3640, 910, 130, 13, 1, 58240, 58240, 14560, 2080, 208, 16, 1, 1106560, 1106560, 276640, 39520, 3952, 304, 19, 1, 24344320, 24344320, 6086080, 869440, 86944, 6688, 418, 22, 1
Offset: 0

Views

Author

Paul D. Hanna, Feb 07 2008

Keywords

Comments

Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n, k, p) = G(n-1, n-k, p) then T(n, k, 1) = A094587(n, k), T(n, k, 2) = A112292(n, k) and T(n, k, 3) is this sequence. - Peter Luschny, Jun 01 2009, revised Jun 18 2019

Examples

			Triangle begins:
        1;
        1,       1;
        4,       4,      1;
       28,      28,      7,     1;
      280,     280,     70,    10,    1;
     3640,    3640,    910,   130,   13,   1;
    58240,   58240,  14560,  2080,  208,  16,  1;
  1106560, 1106560, 276640, 39520, 3952, 304, 19, 1; ...
Matrix inverse begins:
   1;
  -1,   1;
   0,  -4,   1;
   0,   0,  -7,   1;
   0,   0,   0, -10,   1;
   0,   0,   0,   0, -13,   1; ...
		

Crossrefs

Programs

  • Magma
    [[n eq 0 select 1 else k eq n select 1 else (&*[3*j+1: j in [k..n-1]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Jun 14 2019
    
  • Maple
    nmax:=8; for n from 0 to nmax do U(n, n):=1 od: for n from 0 to nmax do for k from 0 to n do if n > k then U(n, k) := mul((3*j+1), j = k..n-1) fi: od: od: for n from 0 to nmax do seq(U(n, k), k=0..n) od: seq(seq(U(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jul 04 2011, revised Nov 23 2012
  • Mathematica
    Table[Product[3*j+1, {j,k,n-1}], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 14 2019 *)
  • PARI
    T(n,k)=if(n==k,1,prod(j=k,n-1,3*j+1))
    
  • Sage
    def T(n, k):
        if (k==n): return 1
        else: return product(3*j+1 for j in (k..n-1))
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 14 2019

Formula

Matrix powers: column 0 of U^(k+1) = column k of A136216 for k >= 0; simultaneously, column k = column 0 of A136216^(3k+1) for k >= 0. Element in column 0, row n, of matrix power U^(k+1) = A007559(n)*C(n+k,k), where A007559 are triple factorials found in column 0 of this triangle.

A211417 Integral factorial ratio sequence: a(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!).

Original entry on oeis.org

1, 77636318760, 53837289804317953893960, 43880754270176401422739454033276880, 38113558705192522309151157825210540422513019720, 34255316578084325260482016910137568877961925210286281393760
Offset: 0

Views

Author

Peter Bala, Apr 11 2012

Keywords

Comments

The integrality of this sequence can be used to prove Chebyshev's estimate C(1)*x/log(x) <= #{primes <= x} <= C(2)*x/log(x), for x sufficiently large; the constant C(1) = 0.921292... and C(2) = 1.105550.... Chebyshev's approach used the related step function floor(x) -floor(x/2) -floor(x/3) -floor(x/5) +floor(x/30). See A182067.
This sequence is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin.
The o.g.f. sum {n >= 0} a(n)*z^n is a generalized hypergeometric series of type 8F7 (see Bober, Table 2, Entry 31) and is an algebraic function of degree 483840 over the field of rational functions Q(z) (see Rodriguez-Villegas). Bober remarks that the monodromy group of the differential equation satisfied by the o.g.f. is W(E_8), the Weyl group of the E_8 root system.
See the Bala link for the proof that a(n), n = 0,1,2..., is an integer.
Congruences: a(p^k) == a(p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integer k (write a(n) as C(30*n,15*n)*C(15*n,5*n)/C(6*n,n) and use equation 39 in Mestrovic, p. 12). More generally, the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) may hold for any prime p >= 5 and any positive integers n and k. Cf. A295431. - Peter Bala, Jan 24 2020

Crossrefs

Programs

  • Magma
    [Factorial(30*n)*Factorial(n)/(Factorial(15*n)*Factorial(10*n)*Factorial(6*n)): n in [0..10]]; // Vincenzo Librandi, Oct 03 2015
  • Mathematica
    Table[(30 n)!*n!/((15 n)!*(10 n)!*(6 n)!), {n, 0, 5}] (* Michael De Vlieger, Oct 02 2015 *)
  • PARI
    a(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!);
    vector(10, n, a(n-1)) \\ Altug Alkan, Oct 02 2015
    

Formula

a(n) ~ 2^(14*n-1) * 3^(9*n-1/2) * 5^(5*n-1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 30 2016

A112295 Inverse of a double factorial related triangle.

Original entry on oeis.org

1, -1, 1, 0, -3, 1, 0, 0, -5, 1, 0, 0, 0, -7, 1, 0, 0, 0, 0, -9, 1, 0, 0, 0, 0, 0, -11, 1, 0, 0, 0, 0, 0, 0, -13, 1, 0, 0, 0, 0, 0, 0, 0, -15, 1, 0, 0, 0, 0, 0, 0, 0, 0, -17, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -19, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -21, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -23, 1
Offset: 0

Views

Author

Paul Barry, Sep 01 2005

Keywords

Comments

Inverse of A112292. Similar results can be obtained for higher factorials.

Examples

			Triangle begins
   1;
  -1,  1;
   0, -3,  1;
   0,  0, -5,  1;
   0,  0,  0, -7,  1;
   0,  0,  0,  0, -9,   1;
   0,  0,  0,  0,  0, -11,  1;
		

Crossrefs

Programs

  • Magma
    A112295:= func< n,k | k eq n select 1 else k eq n-1 select 1-2*n else 0 >;
    [A112295(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Feb 17 2021
  • Mathematica
    T[n_, k_]:= If[k==n, 1, If[k==n-1, 1-2*n, 0]];
    Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 17 2021 *)
  • Sage
    def A112295(n,k): return 1 if k==n else 1-2*n if k==n-1 else 0
    flatten([[A112295(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Feb 17 2021
    

Formula

From G. C. Greubel, Feb 17 2021: (Start)
T(n, k) = 1 - 2*n if k = n-1 otherwise 0, with T(n, n) = 1.
Sum_{k=0..n} T(n, k) = 1 - 2*n - [n=0]. (End)
Showing 1-6 of 6 results.