OFFSET
0,5
REFERENCES
James Propp, personal communication.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..18
FORMULA
a(0)=a(1)=a(2)=a(3)=1; for n>3, a(n+2)*a(n-2) = 1 + a(n+1)*a(n)*a(n-1).
a(-n) = a(n+3).
From Vaclav Kotesovec, May 20 2015: (Start)
a(n) ~ c1^(((1+sqrt(13)-sqrt(2*sqrt(13)-2))/4)^n) * c2^(((1+sqrt(13)+sqrt(2*sqrt(13)-2))/4)^n) * (c3^2+c4^2)^((-1)^n * cos(n*arccot(sqrt((2*sqrt(13)-5)/3)))) * exp(2*(-1)^n*arctan(c4/c3) * sin(n*arccot(sqrt((2*sqrt(13)-5)/3)))), where
c1 = 0.0858378165313271469223136812741638183980800626360336156811045938771...
c2 = 1.0479981158737678235689040669973933524451313410375783562899638042343...
c3 = 1.0681060454695696105471945019699938961207077685059613621050203396954...
c4 = 0.0530316436302789163635657674741144158928386126460043035284221194603...
(End)
MATHEMATICA
RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1, a[n]==(1+a[n-1]a[n-2]a[n-3])/ a[n-4]}, a[n], {n, 15}] (* Harvey P. Dale, May 14 2011 *)
PROG
(PARI) {a(n) = if( n<0, n = 3-n); if( n<4, 1, (a(n-1) * a(n-2) * a(n-3) + 1) / a(n-4)) } /* Michael Somos, Oct 16 2006 */
(PARI) a=vector(15); a[1]=a[2]=a[3]=1; a[4]=2; for(n=5, #a, a[n]=(1+a[n-1]*a[n-2]*a[n-3])/a[n-4]); concat(1, a) \\ Altug Alkan, Sep 27 2018
(Haskell)
a051786 n = a051786_list !! n
a051786_list = 1 : 1 : 1 : 1 :
zipWith div (tail $ zipWith3 (\u v w -> 1 + u * v * w)
(drop 2 a051786_list) (tail a051786_list) a051786_list)
a051786_list
-- Reinhard Zumkeller, Jan 07 2014
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
Michael Somos, Dec 09 1999
EXTENSIONS
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 17 2007
STATUS
approved