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A299960 a(n) = ( 4^(2*n+1) + 1 )/5. 10
1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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
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.
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
EXAMPLE
For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.
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 *)
PROG
(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
CROSSREFS
Cf. A299959 for the smallest prime factor.
Sequence in context: A057807 A057804 A215621 * A194727 A059355 A243783
KEYWORD
nonn
AUTHOR
M. F. Hasler, Feb 22 2018
STATUS
approved

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Last modified May 20 19:00 EDT 2024. Contains 372720 sequences. (Running on oeis4.)