A198410 a(n) = ((3^(n-1) + 1)^3 -1)/3^n.

7, 37, 271, 2269, 19927, 177877, 1596511, 14355469, 129159847, 1162320517, 10460530351, 94143710269, 847290203767, 7625602267957, 68630391713791, 617673439330669, 5559060695695687, 50031545486420197, 450283907053258831, 4052555156505760669, 36472996387631139607

Offset: 2

Author

Michel Lagneau, Oct 24 2011

Keywords

• nonn
• easy

Comments

This sequence is generalizable :

Proposition: p^n divide (p^(n-1) + 1)^ p - 1.

Proof: Let a and p be two integers such that p>=2, and k = gcd(a, p). Then ak divides (a + 1)^p - 1 because (a+1)^p - 1 = [a^p + binomial(p,1)*a^(p-1) + … + binomial(p,p-2)*a^2] + pa == binomial(p,1)*n^(n-1) == 0 (mod p^n) with a = p^(n-1) and k = p.

a(n) is the least k such that k*3^n+1 is a cube. Thus, the cube is given by (3^(n-1)+1)^3. - Derek Orr, Mar 23 2014

Examples

			a(2) = ((3 + 1)^3 - 1)/3^2 = 63/9 = 7.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 2..200
• Index entries for linear recurrences with constant coefficients, signature (13,-39,27).

Crossrefs

Cf. A060073.

Programs

Magma

I:=[7,37,271]; [n le 3 select I[n] else 13*Self(n-1)-39*Self(n-2)+27*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 25 2014

Maple

A198410 := proc(n)
        (3^(n-1)+1)^3 ;
        (%-1)/3^n ;
end proc:
seq(A198410(n), n=2..20) ; # R. J. Mathar, Oct 25 2011

Mathematica

Table[((3^(n - 1) + 1)^3 - 1)/3^n, {n, 2, 20}] (* Wesley Ivan Hurt, Mar 24 2014 *)
CoefficientList[Series[(7 - 54 x + 63 x^2)/((1 - x) (3 x - 1) (9 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 25 2014 *)
LinearRecurrence[{13,-39,27},{7,37,271},30] (* Harvey P. Dale, Mar 04 2015 *)

PARI

a(n)=9^n/27+3^n/3+1 \\ Charles R Greathouse IV, Jun 11 2015

Formula

G.f.: -x^2*(7-54*x+63*x^2) / ( (x-1)*(3*x-1)*(9*x-1) ). - R. J. Mathar, Oct 25 2011

a(n) = 13*a(n-1)-39*a(n-2)+27*a(n-3) for n>2. - Vincenzo Librandi, Mar 25 2014