A119626 Erroneous version of A178674.
6, 12, 30, 84, 246, 732, 2790, 8364, 25086, 75252, 225750, 677264
Offset: 1
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a(1)=3*5-8=7; a(2)=3*7-8=13; a(3)=3*13-8=31.
[3^n+4: n in [0..30]]; // Vincenzo Librandi, May 13 2014
Table[3^n + 4, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, May 19 2011 *) CoefficientList[Series[(5 - 13 x)/((1 - x) (1 - 3 x)), {x, 0, 30}], x] (* Vincenzo Librandi, May 13 2014 *) LinearRecurrence[{4,-3},{5,7},30] (* Harvey P. Dale, Mar 11 2023 *)
List([0..30], n -> 4^n + 4); # G. C. Greubel, Jan 27 2019
[4^n+4: n in [0..35]];
I:=[5, 8]; [n le 2 select I[n] else 5*Self(n-1)-4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 18 2013
Table[4^n +4, {n, 0, 40}] (* or *) CoefficientList[Series[(5-17x)/((4x - 1)(x-1)), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 18 2013 *) LinearRecurrence[{5,-4},{5,8},30] (* Harvey P. Dale, Sep 12 2023 *)
vector(40, n, n--; 4^n+4) \\ G. C. Greubel, Jan 27 2019
[4^n+4 for n in range(40)] # G. C. Greubel, Jan 27 2019
All values of 3^k+3 are multiples of 2, so 3^0+3 = 4 is the least value of k which is a multiple of 2. a(10) = 5 and A301918(10) = 41 so 3^5+3 = 246 is the first multiple of 41 which can be written in the form 3^k+3.
For Pn=5, a(1)=-2/2=-1, for Pn=7, a(2)=0/2=0, for Pn=11,a(3)=4/2=2
a(n)=(prime(n)-7)/2 \\ Charles R Greathouse IV, Jun 15 2013
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