%I #38 Jul 30 2022 08:20:21
%S 1,13,205,3277,52429,838861,13421773,214748365,3435973837,54975581389,
%T 879609302221,14073748835533,225179981368525,3602879701896397,
%U 57646075230342349,922337203685477581,14757395258967641293,236118324143482260685,3777893186295716170957
%N a(n) = ( 4^(2*n+1) + 1 )/5.
%C It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
%C The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - _Peter Bala_, Aug 28 2019
%C Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - _Klaus Purath_, Feb 02 2021
%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
%H H. C. Williams and R. K. Guy, <a href="https://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a> Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (17,-16).
%F a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
%F O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - _Peter Bala_, Aug 28 2019
%F a(n) = 17*a(n-1) - 16*a(n-2). - _Wesley Ivan Hurt_, Oct 02 2020
%F From _Klaus Purath_, Feb 02 2021: (Start)
%F a(n) = (2^(4*n+2)+1)/5.
%F a(n) = (A061654(n) + A001025(n))/2.
%F a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
%F (End)
%F E.g.f.: (exp(x) + 4*exp(16*x))/5. - _Stefano Spezia_, Feb 02 2021
%e For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
%e For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.
%p A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);
%t LinearRecurrence[{17, -16}, {1, 13}, 20] (* _Jean-François Alcover_, Feb 22 2018 *)
%o (PARI) A299960(n)=4^(2*n+1)\5+1
%o (Python)
%o def A299960(n): return ((1<<(n<<2)+2)+1)//5 # _Chai Wah Wu_, Jul 29 2022
%Y Cf. A299959 for the smallest prime factor.
%Y Cf. A052539, A007583, A095372, A100706.
%K nonn
%O 0,2
%A _M. F. Hasler_, Feb 22 2018