OFFSET
1,4
COMMENTS
What accounts for the high proportion of semiprimes in this sequence? Primes: 3, 7, 31, 1093, 846721, 393723877, ... Semiprimes: 85 = 5 * 17 393 = 3 * 131 5071 = 11 * 461 14119 = 7 * 2017 65523 = 3 * 21841 182449 = 43 * 4243 5087942581 = 11113 * 457837 849655943137 = 17 * 49979761361 3943144558081 = 31 * 127198211551 - Jonathan Vos Post, Feb 04 2005
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..1803
P. Heideman and E. Hogan, A new family of Somos-like recurrences.
P. Heideman and E. Hogan, A new family of Somos-like recurrences, El. J. Combin. 15 (2008) #R54. [From R. J. Mathar, Dec 04 2008]
Index entries for linear recurrences with constant coefficients, signature (0, 14, 0, -14, 0, 1).
FORMULA
Both sequences u=(a(2n-1))_{n>0} and u=(a(2n))_{n>0} satisfy the order 3 linear recursion : u(n)=14u(n-1)-14u(n-2)+u(n-3).
a(2*n-1) = ceiling((1/11)*sqrt(1002/5-78*sqrt(33/5))*(sqrt(15)/2+sqrt(11)/ 2)^(2*n-1)).
a(2*n) = ceiling((1/11)*(13-sqrt(165))*(sqrt(15)/2+sqrt(11)/2)^(2*n)).
G.f.: x*(1+x-13*x^2-11*x^3+7*x^4+3*x^5)/(1-14*x^2+14*x^4-x^6). - Jaume Oliver Lafont, Sep 25 2009
a(n) = (4-(-1)^n)*a(n-1)-a(n-2)-1. - Bruno Langlois, Aug 21 2016
Sequences u=(a(2n)) and v=(a(2n-1)) satisfy order 2 linear recursions : u(n)=13*u(n-1)-u(n-2)-5 and v(n)=13*v(n-1)-v(n-2)-7. - Bruno Langlois, Aug 21 2016
MATHEMATICA
LinearRecurrence[{0, 14, 0, -14, 0, 1}, {1, 1, 1, 3, 7, 31}, 26] (* Ray Chandler, Jul 24 2016 *)
nxt[{a_, b_, c_}]:={b, c, (c*b+c+b)/a}; NestList[nxt, {1, 1, 1}, 30][[All, 1]] (* Harvey P. Dale, Mar 11 2019 *)
PROG
(PARI) a(k=3, n) = {K = (k-1)/2; vds = vector(n); for (i=1, 2*K+1, vds[i] = 1; ); for (i=2*K+2, n, vds[i] = (vds[i-1]*vds[i-2*K]+vds[i-K]+vds[i-K-1])/vds[i-2*K-1]; ); for (i=1, n, print1(vds[i], ", "); ); } \\ Michel Marcus, Oct 28 2012
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Jul 28 2002, revised Feb 03 2005
STATUS
approved