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 1-4 of 4 results.

A058798 a(n) = n*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 5, 18, 85, 492, 3359, 26380, 234061, 2314230, 25222469, 300355398, 3879397705, 54011212472, 806288789375, 12846609417528, 217586071308601, 3903702674137290, 73952764737299909, 1475151592071860890
Offset: 0

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Author

Christian G. Bower, Dec 02 2000

Keywords

Comments

Note that a(n) = (a(n-1) + a(n+1))/(n+1). - T. D. Noe, Oct 12 2012; corrected by Gary Detlefs, Oct 26 2018
a(n) = log_2(A073888(n)) = log_3(A073889(n)).
a(n) equals minus the determinant of M(n+2) where M(n) is the n X n symmetric tridiagonal matrix with entries 1 just above and below its diagonal and diagonal entries 0, 1, 2, .., n-1. Example: M(4)=matrix([[0, 1, 0, 0], [1, 1, 1, 0], [0, 1, 2, 1], [0, 0, 1, 3]]). - Roland Bacher, Jun 19 2001
a(n) = A221913(n,-1), n>=1, is the numerator sequence of the n-th approximation of the continued fraction -(0 + K_{k>=1} (-1/k)) = 1/(1-1/(2-1/(3-1/(4-... The corresponding denominator sequence is A058797(n). - Wolfdieter Lang, Mar 08 2013
The recurrence equation a(n+1) = (A*n + B)*a(n) + C*a(n-1) with the initial conditions a(0) = 0, a(1) = 1 has the solution a(n) = Sum_{k = 0..floor((n-1)/2)} C^k*binomial(n-k-1,k)*( Product_{j = 1..n-2k-1} (k+j)*A + B ). This is the case A = 1, B = 1, C = -1. - Peter Bala, Aug 01 2013

Examples

			Continued fraction approximation 1/(1-1/(2-1/(3-1/4))) = 18/7 = a(4)/A058797(4). - _Wolfdieter Lang_, Mar 08 2013
		

Crossrefs

Column 1 of A007754.
Cf. A073888, A073889, A221913 (alternating row sums).

Programs

  • GAP
    a:=[1,2];; for n in [3..25] do a[n]:=n*a[n-1]-a[n-2]; od; Concatenation([0], a); # Muniru A Asiru, Oct 26 2018
    
  • Magma
    [0] cat [n le 2 select n else n*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 22 2016
    
  • Mathematica
    t = {0, 1}; Do[AppendTo[t, n*t[[-1]] - t[[-2]]], {n, 2, 25}]; t (* T. D. Noe, Oct 12 2012 *)
    nxt[{n_,a_,b_}]:={n+1,b,b*(n+1)-a}; Transpose[NestList[nxt,{1,0,1},20]] [[2]] (* Harvey P. Dale, Nov 30 2015 *)
  • PARI
    m=30; v=concat([1,2], vector(m-2)); for(n=3, m, v[n] = n*v[n-1]-v[n-2]); concat(0, v) \\ G. C. Greubel, Nov 24 2018
  • Sage
    def A058798(n):
        if n < 3: return n
        return hypergeometric([1/2-n/2, 1-n/2],[2, 1-n, -n], -4)*factorial(n)
    [simplify(A058798(n)) for n in (0..20)] # Peter Luschny, Sep 10 2014
    

Formula

a(n) = Sum_{k = 0..floor((n-1)/2)} (-1)^k*binomial(n-k-1,k)*(n-k)!/(k+1)!. - Peter Bala, Aug 01 2013
a(n) = A058797(n+1) + A058799(n-1). - Henry Bottomley, Feb 28 2001
a(n) = Pi*(BesselY(1, 2)*BesselJ(n+1, 2) - BesselJ(1,2)* BesselY(n+1,2)). See the Abramowitz-Stegun reference given under A103921, p. 361 eq. 9.1.27 (first line with Y, J and z=2) and p. 360, eq. 9.1.16 (Wronskian). - Wolfdieter Lang, Mar 05 2013
Limit_{n->oo} a(n)/n! = BesselJ(1,2) = 0.576724807756873... See a comment on asymptotics under A084950.
a(n) = n!*hypergeometric([1/2-n/2, 1-n/2], [2, 1-n, -n], -4) for n >= 2. - Peter Luschny, Sep 10 2014

Extensions

New description from Amarnath Murthy, Aug 17 2002

A093985 a(1) = 1, a(2) = 2; a(n+1) = 2n*a(n) - a(n-1). Symmetrically, a(n) = (a(n-1) + a(n+1))/((n-1) + (n+1)).

Original entry on oeis.org

1, 2, 7, 40, 313, 3090, 36767, 511648, 8149601, 146181170, 2915473799, 63994242408, 1532946343993, 39792610701410, 1112660153295487, 33340011988163200, 1065767723467926913, 36202762585921351842, 1302233685369700739399, 49448677281462706745320
Offset: 1

Views

Author

Amarnath Murthy, May 22 2004

Keywords

Examples

			a(3)=7 because 2*2*a(2) - a(1) = 7.
		

Crossrefs

Programs

  • Magma
    I:=[1,2]; [n le 2 select I[n] else 2*(n-1)*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 15 2017
  • Maple
    a[1]:=1: a[2]:=2: for n from 2 to 21 do a[n+1]:=2*n*a[n]-a[n-1] od: seq(a[n],n=1..21); # Emeric Deutsch, Jul 31 2005
  • Mathematica
    nxt[{n_,a_,b_}]:={n+1,b,2*n*b-a}; NestList[nxt,{2,1,2},20][[All,2]] (* Harvey P. Dale, Jan 09 2021 *)
    a[n_] := (Pi/2)*(BesselY[0, 1]*BesselJ[n, 1.] - BesselJ[0, 1]*BesselY[n, 1.]);
    Table[Round[a[n]], {n, 1, 20}] (* Hugo Pfoertner, Feb 12 2024 *)

Formula

a(n) = Sum_{k = 0..floor((n-1)/2)} (-1)^k*2^(n-2*k-1)*(n-2*k-1)!*(binomial(n-k-1,k))^2. Cf. A058798. - Peter Bala, Aug 01 2013
a(n) = (Pi/2)*(Y[0, 1] * J[n, 1] - J[0, 1] * Y[n, 1]) where Y and J are Bessel functions. - Peter Luschny, Jan 30 2024

Extensions

Corrected and extended by Emeric Deutsch, Jul 31 2005

A093987 a(1) = 1, a(2) = 2, a(n+1) = a(n)^(2n)/ a(n-1). Symmetrically a(n) = {a(n-1)*a(n+1)}^[1/{(n-1) +(n+1)}].

Original entry on oeis.org

1, 2, 16, 8388608, 1532495540865888858358347027150309183618739122183602176
Offset: 1

Views

Author

Amarnath Murthy, May 22 2004

Keywords

Comments

The next term has 535 digits. - Harvey P. Dale, Jan 31 2020

Examples

			a(5) = 8388608^8/16 = 1.53249554e+54.
a(6) =~ 8.5173068734e+534.
		

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[2] = 2; a[n_] := a[n - 1]^(2n - 2)/a[n - 2]; Table[ a[n], {n, 6}] (* Robert G. Wilson v, May 24 2004 *)
    RecurrenceTable[{a[1]==1,a[2]==2,a[n+1]==a[n]^(2n)/a[n-1]},a,{n,5}] (* Harvey P. Dale, Jan 31 2020 *)

Extensions

Edited and extended by Robert G. Wilson v, May 24 2004

A358735 Triangular array read by rows. T(n, k) is the coefficient of x^k in a(n+3) where a(1) = a(2) = a(3) = 1 and a(m+2) = (m*x + 2)*a(m+1) - a(m) for all m in Z.

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 1, 10, 16, 6, 1, 20, 70, 76, 24, 1, 35, 224, 496, 428, 120, 1, 56, 588, 2260, 3808, 2808, 720, 1, 84, 1344, 8140, 23008, 32152, 21096, 5040, 1, 120, 2772, 24772, 107328, 245560, 298688, 178848, 40320
Offset: 0

Views

Author

Michael Somos, Mar 15 2023

Keywords

Comments

This sequence is essentially A204024 except for extra row, alternating signs and reversed rows.
The sequence of polynomials a(m) satisfies a(m)*a(m-2) = a(m-1) * (a(m-1) + x*a(m-2) + a(m-3)) - a(m-2)^2 for all m > 3.

Examples

			a(3) = 1, a(4) = 1 + x, a(5) = 1 + 4*x + 2*x^2.
Triangular array T(n, k) starts:
n\k | 0   1   2   3   4   5
--- + - --- --- --- --- ---
 0  | 1
 1  | 1   1
 2  | 1   4   2
 3  | 1  10  16   6
 4  | 1  20  70  76  24
 5  | 1  35 224 496 428 120
		

Crossrefs

Programs

  • Mathematica
    T[ n_, k_] := If[ n<0, 0, Module[{a = Table[1, n+3], x}, Do[ a[[m]] = a[[m-1]] *(a[[m-1]] + x*a[[m-2]] + a[[m-3]])/a[[m-2]] - a[[m-2]] //Factor//Expand, {m, 4, n+3}]; Coefficient[ a[[n+3]], x, k]]];
  • PARI
    {T(n, k) = if( n<0, 0, my(a = vector(n+3, i, 1)); for(m = 4, n+3, a[m] = a[m-1]*(a[m-1] + 'x*a[m-2] + a[m-3])/a[m-2] - a[m-2]); polcoeff( a[n+3], k))};

Formula

If x=1, then a(n) = A058797(n+2) = Sum_{k=0..n} T(n, k).
If x=2, then a(n) = A093986(n+2).
T(n, n) = n!, T(n, 0) = 1, T(n, 1) = A000292(n). T(n, 2) = 2*A040977(n-2).
Showing 1-4 of 4 results.