%I #64 Sep 08 2022 08:46:09
%S 1,1,1,1,1,3,13,43,113,253,509,969,1849,3719,8009,18027,40897,91257,
%T 198697,423777,894081,1886011,4007301,8594411,18560081,40181493,
%U 86872293,187197193,402060793,861827743,1846685729,3960390059,8504658049,18283290609,39325827729
%N a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.
%C a(n)/a(n-1) tends to 2.1486... = 1 + 2^(1/5), the real root of the polynomial x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 3.
%C If x^5 = 2 and n >= 0, then there are unique integers a, b, c, d, g such that (1 + x)^n = a + b*x + c*x^2 + d*x^3 + g*x^4. The coefficient a is a(n) (from A052102). - _Alexander Samokrutov_, Jul 11 2015
%C If x=a(n), y=a(n+1), z=a(n+2), s=a(n+3), t=a(n+4) then x, y, z, s, t satisfies Diophantine equation (see link). - _Alexander Samokrutov_, Jul 11 2015
%H G. C. Greubel, <a href="/A247584/b247584.txt">Table of n, a(n) for n = 0..1000</a> (terms 0..48 from Alexander Samokrutov)
%H Alexander Samokrutov, <a href="/A247584/a247584_1.txt">Diophantine equation</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,3).
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5).
%F a(n) = Sum_{k=0...floor(n/5)} (2^k*binomial(n,5*k)). - _Alexander Samokrutov_, Jul 11 2015
%F G.f.: (1-x)^4/(1 -5*x +10*x^2 -10*x^3 +5*x^4 -3*x^5). - _Colin Barker_, Sep 22 2014
%p m:=50; S:=series( (1-x)^4/(1 -5*x +10*x^2 -10*x^3 +5*x^4 -3*x^5), x, m+1):
%p seq(coeff(S, x, j), j=0..m); # _G. C. Greubel_, Apr 15 2021
%t LinearRecurrence[{5,-10,10,-5,3}, {1,1,1,1,1}, 50] (* _Vincenzo Librandi_, Jul 11 2015 *)
%o (PARI) Vec((1-x)^4/(1-5*x+10*x^2-10*x^3+5*x^4-3*x^5) + O(x^100)) \\ _Colin Barker_, Sep 22 2014
%o (Maxima) makelist(sum(2^k*binomial(n,5*k), k, 0, floor(n/5)), n, 0, 50) /* _Alexander Samokrutov_, Jul 11 2015 */
%o (Magma) [n le 5 select 1 else 5*Self(n-1) -10*Self(n-2) +10*Self(n-3) -5*Self(n-4) +3*Self(n-5): n in [1..40]]; // _Vincenzo Librandi_, Jul 11 2015
%o (Sage) [sum(2^j*binomial(n, 5*j) for j in (0..n//5)) for n in (0..50)] # _G. C. Greubel_, Apr 15 2021
%Y The following sequences belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A052101, A052102, A052103, A063727, A083098, A083099, A083100, A084057, A093406, A247344.
%Y Cf. A005531.
%K nonn,easy
%O 0,6
%A _Alexander Samokrutov_, Sep 20 2014