A050512 a(n) = (a(n-1)*a(n-3) - a(n-2)^2) / a(n-4), with a(0) = 0, a(1) = a(2) = a(3) = 1, a(4) = -1.

0, 1, 1, 1, -1, -2, -3, -1, 7, 11, 20, -19, -87, -191, -197, 1018, 2681, 8191, -5841, -81289, -261080, -620551, 3033521, 14480129, 69664119, -2664458, -1612539083, -7758440129, -37029252553, 181003520899, 1721180313660, 12437589708389, 19206818781913

Offset: 0

Author

Michael Somos, Dec 28 1999

Keywords

• sign
• easy
• nice

Comments

From Paul Barry, May 31 2010: (Start)

a(n+1) is (-1)^binomial(n,2) times the Hankel transform of the sequence with g.f.

1/(1-x/(1+x^2/(1-x^2/(1-2x^2/(1+(3/4)x^2/(1+(2/9)x^2/(1+21)x^2/(1-... where

-1,1,2,-3/4,-2/9,21,... are the x-coordinates of the multiples of z=(0,0) on the elliptic curve E: y^2 - 2xy - y = x^3-x. (End)

This is a strong elliptic divisibility sequence t_n as given in [Kimberling, p. 16] where x = 1, y = 1, z = -1.

The elliptic curve y^2 + y = x^3 + x^2 (LMFDB label 43.a1) has infinite order point P = (0, 0). The x coordinate of n*P has denominator a(n)^2. - Michael Somos, Feb 14 2023

Examples

			G.f. = x + x^2 + x^3 - x^4 - 2*x^5 - 3*x^6 - x^7 + 7*x^8 + 11*x^9 + 20*x^10 + ...
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..250
• Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
• Paul Barry, Riordan Pseudo-Involutions, Continued Fractions and Somos 4 Sequences, arXiv:1807.05794 [math.CO], 2018.
• Clark Kimberling, Strong divisibility sequences and some conjectures, Fib. Quart., 17 (1979), 13-17.
• LMFDB, Elliptic Curve 43.a1 (Cremona label 43a1)

Crossrefs

Cf. A006769.

Programs

Haskell

a050512 n = a050512_list !! n
a050512_list = 0 : 1 : 1 : 1 : (-1) : zipWith div (zipWith (-) (zipWith (*)
   (drop 4 a050512_list) (drop 2 a050512_list))
     (map (^ 2) (drop 3 a050512_list))) (tail a050512_list)
-- Reinhard Zumkeller, Nov 02 2011

Mathematica

a[n_?OddQ] := a[n] = a[(n-1)/2]^3*a[(n+3)/2] - a[(n-3)/2]*a[(n+1)/2]^3; a[n_?EvenQ] := a[n] = (a[n/2-1]^2*a[n/2+2] - a[n/2-2]*a[n/2+1]^2)*a[n/2]; a[0] = 0; a[1] = a[2] = a[3] = 1; a[4] = -1; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 29 2011 *)
Join[{0},RecurrenceTable[{a[1]==a[2]==a[3]==1,a[4]==-1,a[n]==(a[n-1] a[n-3]-a[n-2]^2)/a[n-4]},a,{n,30}]] (* Harvey P. Dale, Mar 23 2012 *)

PARI

an=vector(200); for(n=1,4,an[ n ]=[ 1,1,1,-1 ][ n ]); for(n=5, length(an),an[ n ]=(an[ n-1 ]*an[ n-3 ]-an[ n-2 ]^2)/an[ n-4 ]); a(n) =sign(n)*an[ abs(n)+(n==0) ]

PARI

{a(n) = my(an); if( n<0, -a(-n), if( n==0, 0, an = vector( max(4, n), i, 1); an[4] = -1; for( k=5, n, an[k] = (an[k-1] * an[k-3] - an[k-2]^2) / an[k-4]); an[n]))}; /* Michael Somos, Jul 07 2014 */

Formula

a(2*n + 1) = a(n + 2) * a(n)^3 - a(n - 1) * a(n + 1)^3 for all n in Z.

a(2*n) = a(n + 2) * a(n) * a(n - 1)^2 - a(n) * a(n - 2) * a(n + 1)^2 for all n in Z.

0 = a(n)*a(n+5) - a(n+1)*a(n+4) - a(n+2)*a(n+3) for all n in Z. - Michael Somos, Jul 07 2014

0 = a(n)*a(n+6) + a(n+1)*a(n+5) - 2*a(n+2)*a(n+4) for all n in Z. - Michael Somos, Jul 07 2014

a(n) = -a(-n) for all n in Z. - Michael Somos, Feb 14 2023