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A299960 a(n) = (4^(2*n+1) + 1) / 5.

%I A299960 #52 Jan 08 2025 05:51:13
%S A299960 1,13,205,3277,52429,838861,13421773,214748365,3435973837,54975581389,
%T A299960 879609302221,14073748835533,225179981368525,3602879701896397,
%U A299960 57646075230342349,922337203685477581,14757395258967641293,236118324143482260685,3777893186295716170957
%N A299960 a(n) = (4^(2*n+1) + 1) / 5.
%C A299960 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 A299960 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 A299960 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
%C A299960 Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - _M. F. Hasler_, Jan 07 2025
%H A299960 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 A299960 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 A299960 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (17,-16).
%F A299960 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 A299960 O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - _Peter Bala_, Aug 28 2019
%F A299960 a(n) = 17*a(n-1) - 16*a(n-2). - _Wesley Ivan Hurt_, Oct 02 2020
%F A299960 From _Klaus Purath_, Feb 02 2021: (Start)
%F A299960 a(n) = (2^(4*n+2)+1)/5.
%F A299960 a(n) = (A061654(n) + A001025(n))/2.
%F A299960 a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
%F A299960 (End)
%F A299960 E.g.f.: (exp(x) + 4*exp(16*x))/5. - _Stefano Spezia_, Feb 02 2021
%F A299960 We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
%F A299960   (53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
%F A299960   (107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
%F A299960   (173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
%F A299960   (733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
%F A299960   (997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
%F A299960   (41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
%F A299960   (509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
%F A299960   (1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
%F A299960   (773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - _M. F. Hasler_, Jan 07 2025
%e A299960 For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
%e A299960 For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.
%p A299960 A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);
%t A299960 LinearRecurrence[{17, -16}, {1, 13}, 20] (* _Jean-François Alcover_, Feb 22 2018 *)
%o A299960 (PARI) A299960(n)=4^(2*n+1)\5+1
%o A299960 (Python)
%o A299960 def A299960(n): return ((1<<(n<<2)+2)+1)//5 # _Chai Wah Wu_, Jul 29 2022
%Y A299960 Cf. A299959 for the smallest prime factor.
%Y A299960 Cf. A052539, A007583, A095372, A100706.
%K A299960 nonn
%O A299960 0,2
%A A299960 _M. F. Hasler_, Feb 22 2018