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

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957

Offset: 0

Author

M. F. Hasler, Feb 22 2018

Keywords

• nonn

Comments

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.

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

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

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

Examples

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.
		

Links

• H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
• H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
• Index entries for linear recurrences with constant coefficients, signature (17,-16).

Crossrefs

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

Programs

Maple

A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);

Mathematica

LinearRecurrence[{17, -16}, {1, 13}, 20] (* Jean-François Alcover, Feb 22 2018 *)

PARI

A299960(n)=4^(2*n+1)\5+1

Python

def A299960(n): return ((1<<(n<<2)+2)+1)//5 # Chai Wah Wu, Jul 29 2022

Formula

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.

O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019

a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020

From Klaus Purath, Feb 02 2021: (Start)

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

a(n) = (A061654(n) + A001025(n))/2.

a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.

(End)

E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021

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),

(53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),

(107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),

(173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),

(733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),

(997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),

(41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),

(509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),

(1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),

(773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025