This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A245323 #41 Oct 22 2016 09:35:06
%S A245323 4,170,7320,328380,15124186,704915600,33014404692,1549142827050,
%T A245323 72743819556328,3416820019114700,160507201018772634,
%U A245323 7540231471940495520,354226959651753624100,16641065639596669234730,781774759322033582085816,36726752905662141638238300
%N A245323 a(n) = F(6*n-3)*(L(2*n-1)+1), where F = A000045 are the Fibonacci and L = A000032 are the Lucas numbers.
%C A245323 Let n = 2*m-1 where m is 1,2,3,...; then a(m) = F(3*n)*(L(n)+1) with F(n) a Fibonacci number and L(n) a Lucas number (A000032); also a(n) = F(n)*(L(n)^3+L(n)^2+L(n)+1) which is a repdigit in base L(n), made of four digits F(n).
%C A245323 For n = 1, unary representation must be used to give a repdigit, then 4(10) = 1111(1).
%C A245323 For n = 3, 170(10) = 2222(4).
%C A245323 For n = 5, 7320(10) = 5555(11).
%C A245323 Starting from n = 5, the repdigit is the 4th term of a Fibonacci-type sequence of 5 palindromes.
%C A245323 For example this sequence for n=5 in base 11 is: 1331, 2112, 3443, 5555, 8998 which are the 5 2-digit Fibonacci numbers in base 10 concatenated with their reversed forms.
%C A245323 If the repdigit F(3*n)*(L(n)+1) is the most noticeable result in base L(n), there are other recurrences; naming f the digit F(n) and g the digit F(n)*2, we find
%C A245323 F(0*n)*(L(n)+1) = 0
%C A245323 F(1*n)*(L(n)+1) = ff
%C A245323 F(2*n)*(L(n)+1) = ff0
%C A245323 F(3*n)*(L(n)+1) = ffff
%C A245323 F(4*n)*(L(n)+1) = ffgg0
%C A245323 The base L(n) gives a visual aspect to the formula
%C A245323 F(n*k) = L(n)*F(n*(k-1)) + F(n*(k-2)) with n odd, which is a particular case of the general formula for any integers n,k,m:
%C A245323 F(n*(k)+m) = L(n) * F(n*(k-1)+m) - (-1)^n * F(n*(k-2)+m)
%H A245323 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (72,-1304,6066,-1304,72,-1).
%F A245323 a((n-1)/2) = F(3*n)*(L(n)+1) for any positive odd n. [Corrected by _M. F. Hasler_, Oct 20 2016]
%F A245323 a(n) = F(n)*(L(n)^3+L(n)^2+L(n)+1).
%F A245323 G.f.: 2*x*(51*x^4-622*x^3+148*x^2-59*x+2) / ((x^2-47*x+1)*(x^2-18*x+1)*(x^2-7*x+1)). - _Colin Barker_, Jul 18 2014
%e A245323 Example: for n = 5, F(15) = 610, L(5) = 11, then a(5) = 610*12 = 7320 which is 5555 in base 11; F(5) = 5.
%t A245323 LinearRecurrence[{72,-1304,6066,-1304,72,-1}, {4,170,7320,328380,15124186,704915600},30] (* _Harvey P. Dale_, Aug 26 2014 *)
%t A245323 Table[Fibonacci[6 n - 3] (LucasL[2 n - 1] + 1), {n, 16}] (* _Michael De Vlieger_, Oct 21 2016 *)
%o A245323 (PARI) vector(50, m, fibonacci(6*m-3)*(lucas(2*m-1)+1)) \\ _Colin Barker_, Jul 18 2014
%Y A245323 Cf. A000045, A000032, A000042.
%K A245323 nonn,base
%O A245323 1,1
%A A245323 _Rémi Schulz_, Jul 18 2014
%E A245323 More terms from _Colin Barker_, Jul 18 2014
%E A245323 Edited and partially corrected by _M. F. Hasler_, Oct 09 and Oct 20 2016