%I A082605
%S A082605 2,3,5,11,17,41,65,161,257,641,1025,2561,4097,10241,16385,40961,65537,
%T A082605 163841,262145,655361,1048577,2621441,4194305,10485761,16777217,
%U A082605 41943041,67108865,167772161,268435457,671088641,1073741825,2684354561
%N A082605 Using Euler's 6-term sequence, we define the partial recurrence relation 
               a(0)=2, a(1)=3, a(2)=5; a(k) = 2*a(k-1) - 1 + (-1)^(k-1)*2^(k-2), 
               3 <= k <= 5. Using this definition of a(k) we (formally) work backwards 
               towards a(2)=5 to arrive at the formula for a(k) below.
%C A082605 For k >= 3, a(k) has the simple form a(k) = 2^(k-2)*(4 + 1/2*(1 + (-1)^(k+1)) 
               + 1; and it follows by induction that a(k) is congruent to 17 (mod 
               24) for all k >= 4. Direct calculations show that for k >= 3, the 
               discriminants of the polynomials x^2 + x + a(k), D(k) = 1 - 4*a(k), 
               satisfy the functional equation -D(k) = a(k+2) + 2.
%F A082605 (a(k))_(k>=0) = 2^(k-2)*(4 + sum((-1)^r, r=2..k-1)) + 1, the empty sums 
               corresponding to k=0, 1, 2 of course taken to be zero.
%F A082605 G.f.: [ -2-x+6x^2-2x^3+2x^4]/[8(1-x)(1-2x)(1+2x)]. a(n) = A056486(n-1)+1. 
               - R. Stephan, Mar 19 2004
%e A082605 a(10) = 1025
%p A082605 Note that Maple handles empty sums in an unpredictable way.
%o A082605 (PARI) a(n)=if(n<2,if(n<1,2,3),if(n%2==0,4^(n/2)+1,5/2*4^((n-1)/2)+1))
%Y A082605 Sequence in context: A014556 A062737 A085613 this_sequence A007755 A060611 
               A103598
%Y A082605 Adjacent sequences: A082602 A082603 A082604 this_sequence A082606 A082607 
               A082608
%K A082605 nonn
%O A082605 0,1
%A A082605 Johan Meyer & Ben de la Rosa (meyerjh.sci(AT)mail.uovs.ac.za), May 23 
               2003
%E A082605 More terms from R. Stephan, Mar 19 2004