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 1-2 of 2 results.

A099597 Array T(n,k) read by antidiagonals: expansion of exp(x+y)/(1-xy).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 9, 4, 1, 1, 5, 19, 19, 5, 1, 1, 6, 33, 82, 33, 6, 1, 1, 7, 51, 229, 229, 51, 7, 1, 1, 8, 73, 496, 1313, 496, 73, 8, 1, 1, 9, 99, 919, 4581, 4581, 919, 99, 9, 1, 1, 10, 129, 1534, 11905, 32826, 11905, 1534, 129, 10, 1, 1, 11, 163, 2377, 25733, 137431, 137431, 25733, 2377, 163, 11, 1
Offset: 0

Views

Author

Ralf Stephan, Oct 28 2004

Keywords

Comments

Rows are polynomials in n whose coefficients are in A099599.
From Peter Bala, Aug 19 2013: (Start)
The k-th superdiagonal sequence of this square array occurs as the sequence of numerators in the convergents to a certain continued fraction representation of the constant BesselI(k,2), where BesselI(k,x) is a modified Bessel function of the first kind:
Let d_k(n) = T(n,n+k) = n! * (n+k)! * Sum_{i=0..n} 1/(i!*(i+k)!) denote the sequence of entries on the k-th superdiagonal. It satisfies the first-order recurrence equation d_k(n) = n*(n+k)*d_k(n-1) + 1 with d_k(0) = 1 and also the second-order recurrence d_k(n) = (n*(n+k)+1)*d_k(n-1) - (n-1)*(n-1+k)*d_k(n-2) with initial conditions d_k(0) = 1 and d_k(1) = k+2. This latter recurrence is also satisfied by the sequence n!*(n+k)!. From this observation we obtain the finite continued fraction expansion d_k(n) = n!*(n+k)!*(1/(k! - k!/((k+2) - (k+1)/((2*k+5) - 2*(k+2)/((3*k+10) - ... - n*(n+k)/(((n+1)*(n+k+1)+1) ))))).
Taking the limit as n -> infinity produces a continued fraction representation for the modified Bessel function value BesselI(k,2) = Sum_{i=0..inf} 1/(i!*(i+k)!) = 1/(k! - k!/((k+2) -(k+1)/((2*k+5) - 2*(k+2)/((3*k+10) - ... - n*(n+k)/(((n+1)*(n+k+1)+1) - ...))))). See A070910 for the case k = 0 and A096789 for the case k = 1. (End)

Examples

			1, 1,  1,   1,    1,     1,
1, 2,  3,   4,    5,     6,
1, 3,  9,  19,   33,    51,
1, 4, 19,  82,  229,   496,
1, 5, 33, 229, 1313,  4581,
1, 6, 51, 496, 4581, 32826,
		

Crossrefs

Rows include A000012, A000027, A058331. Main diagonal is A006040. Antidiagonal sums are in A099598. Cf. A099599.
Cf. A088699. A228229 (main super and subdiagonal).

Programs

  • Maple
    #A099597
    T := proc(n,k) option remember;
    if n = 0 then 1 elif k = 0 then 1
    else n*k*thisproc(n-1,k-1) + 1
    fi
    end:
    # Diplay entries by antidiagonals
    seq(seq(T(n-k,k), k = 0..n), n = 0..10);
    # Peter Bala, Aug 19 2013
  • Mathematica
    T[, 0] = T[0, ] = 1;
    T[n_, k_] := T[n, k] = n k T[n - 1, k - 1] + 1;
    Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

Formula

T(n,k) = Sum_{i=0..min(n,k)} C(n,i)*C(k,i)*i!^2. The LDU factorization of this square array is P * D * transpose(P), where P is Pascal's triangle A007318 and D = diag(0!^2, 1!^2, 2!^2, ... ). Compare with A088699. - Peter Bala, Nov 06 2007
Recurrence equation: T(n,k) = n*k*T(n-1,k-1) + 1 with boundary conditions T(n,0) = T(0,n ) = 1.
Main subdiagonal and main superdiagonal [1, 3, 19, 229, ...] is A228229. - Peter Bala, Aug 19 2013
nth row/column o.g.f.: HypergeometricPFQ[{1,1,-n},{},x/(x-1)]/(1-x) (see comment in A099599). - Natalia L. Skirrow, Jul 18 2025

A089041 Inverse binomial transform of squares of factorial numbers.

Original entry on oeis.org

1, 0, 3, 26, 453, 11844, 439975, 22056222, 1436236809, 117923229512, 11921584264011, 1455483251191650, 211163237294447053, 35913642489947449356, 7077505637217289437423, 1599980633296779087784934, 411293643476907595937924625, 119299057697083019137937718672
Offset: 0

Views

Author

Vladeta Jovovic, Dec 03 2003

Keywords

Comments

a(n) enumerates (ordered) lists of n two-tuples such that all numbers from 1 to n appear as the first as well as the second tuple entry and the j-th list member is not the tuple (j,j), for every j=1,..,n. Called coincidence-free 2-tuple lists of length n. See the Charalambides reference for this combinatorial interpretation.

Examples

			2-tuple combinatorics: a(1)=0 because the only list of 2-tuples with numbers 1 is [(1,1)] and this is a coincidence for j=1.
2-tuple combinatorics: the a(2)=3 coincidence free 2-tuple lists of length n=2 are [(1,2),(2,1)], [(2,1),(1,2)] and [(2,2),(1,1)]. The list [(1,1),(2,2)] has two coincidences (j=1 and j=2).
		

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 187, Exercise 13.(a), for r=2.

Crossrefs

Cf. A001044, A046662 (binomial transform of squares of factorial numbers).
(-1)^n times the polynomials in A099599 evaluated at -1.

Programs

  • Maple
    a:= proc(n) option remember;
           `if`(n<2, 1-n, n^2*a(n-1)+n*(n-1)*a(n-2)+(-1)^n)
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Jun 21 2013
  • Mathematica
    Table[n!Sum[(-1)^k(n-k)!/k!,{k,0,n}],{n,0,15}] (* Geoffrey Critzer, Jun 17 2013 *)

Formula

G.f.: hypergeom([1, 1, 1], [], x/(1+x))/(1+x).
E.g.f.: exp(-x)* hypergeom([1, 1], [], x).
a(n) = n^2*a(n-1) + n*(n-1)*a(n-2) + (-1)^n. - Vladeta Jovovic, Jul 15 2004
a(n) = Sum_{j=0..n} ((-1)^(n-j))*binomial(n,j)*(j!)^2. See the Charalambides reference a(n)=B_{n,2}.
a(n) = (n-1)*(n+1)*a(n-1) + (n-1)*(2*n-1)*a(n-2) + (n-2)*(n-1)*a(n-3). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2*Pi*exp(-2*n)*n^(2*n+1). - Vaclav Kotesovec, Aug 13 2013
G.f.: Sum_{k>=0} (k!)^2*x^k/(1 + x)^(k+1). - Ilya Gutkovskiy, Apr 12 2019

Extensions

Charalambides reference and comments with combinatorial examples from Wolfdieter Lang, Jan 21 2008
Showing 1-2 of 2 results.