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a(n) = (3^(6*n) - 2^(6*n))/35.
1

%I #21 Sep 08 2022 08:45:59

%S 0,19,15067,11061667,8068935979,5882573095795,4288416187929211,

%T 3126256706670452803,2279041222725643804363,1661421056715018890883091,

%U 1211175950687522343133931035,882947268073109296732165817059,643668558426698629867350806558827

%N a(n) = (3^(6*n) - 2^(6*n))/35.

%C 3^6 = (3^3)^2 == (-8)^2 (mod 35) = 64 and 2^6 = 64.

%H Vincenzo Librandi, <a href="/A198412/b198412.txt">Table of n, a(n) for n = 0..300</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (793,-46656).

%F a(n) = (3^(6*n) - 2^(6*n))/35.

%F G.f.: 19*x / ( (729*x-1)*(64*x-1) ). - _R. J. Mathar_, Oct 25 2011

%e a(1) = (3^(6*1) - 2^(6*1))/35 = 665/35 = 19.

%p for n from 0 to 30 do:

%p x:= (3^(6*n)- 2^(6*n))/35: printf(`%d, `, x):od:

%t LinearRecurrence[{793,-46656},{0,19},50] (* _Vincenzo Librandi_, Nov 25 2011 *)

%t Table[(3^(6n)-2^(6n))/35,{n,0,20}] (* _Harvey P. Dale_, Aug 14 2019 *)

%o (Magma) [(3^(6*n)- 2^(6*n))/35: n in [0..15]]; // _Vincenzo Librandi_, Nov 25 2011

%o (PARI) a(n)=(3^(6*n)-2^(6*n))/35 \\ _Charles R Greathouse IV_, Jul 06 2017

%K nonn,easy,less

%O 0,2

%A _Michel Lagneau_, Oct 24 2011