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A299960 a(n) = ( 4^(2*n+1) + 1 )/5. 6
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

LINKS

Table of n, a(n) for n=0..18.

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(2n+1)/5 = A015521(2n+1) = A014985(2n+1) = A007910(4n+1) = A007909(4n+1) = A207262(n+1)/5.

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

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-Fran├žois Alcover, Feb 22 2018 *)

PROG

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

CROSSREFS

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

Sequence in context: A057807 A057804 A215621 * A194727 A059355 A243783

Adjacent sequences:  A299957 A299958 A299959 * A299961 A299962 A299963

KEYWORD

nonn

AUTHOR

M. F. Hasler, Feb 22 2018

STATUS

approved

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Last modified February 16 15:17 EST 2020. Contains 331961 sequences. (Running on oeis4.)