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 A082605 Using Euler's 6-term sequence A014556, 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. 5
 2, 3, 5, 11, 17, 41, 65, 161, 257, 641, 1025, 2561, 4097, 10241, 16385, 40961, 65537, 163841, 262145, 655361, 1048577, 2621441, 4194305, 10485761, 16777217, 41943041, 67108865, 167772161, 268435457, 671088641, 1073741825, 2684354561 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Using this definition of a(k) we (formally) work backwards towards a(2)=5 to arrive at the formula for a(k) below. 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. LINKS Index entries for linear recurrences with constant coefficients, signature (1,4,-4). FORMULA (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. a(n) = A056486(n-1)+1. - Ralf Stephan, Mar 19 2004 a(2*n) = 2^n + 1. - Georg Fischer, May 15 2019 G.f.: (2+x-6*x^2+2*x^3-2*x^4)/(1-x-4*x^2+4*x^3). - Georg Fischer, May 15 2019 EXAMPLE a(10) = 1025 MATHEMATICA LinearRecurrence[{1, 4, -4}, {2, 3, 5, 11, 17}, 32] (* Georg Fischer, May 15 2019 *) PROG (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)) CROSSREFS Cf. A014556, A056486. a(0..6) and a(2*n) same as A085613(n+1). Sequence in context: A014556 A062737 A085613 * A007755 A060611 A077497 Adjacent sequences:  A082602 A082603 A082604 * A082606 A082607 A082608 KEYWORD nonn,easy AUTHOR Johan Meyer & Ben de la Rosa (meyerjh.sci(AT)mail.uovs.ac.za), May 23 2003 EXTENSIONS More terms from Ralf Stephan, Mar 19 2004 STATUS approved

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Last modified September 24 17:09 EDT 2020. Contains 337321 sequences. (Running on oeis4.)