OFFSET
0,6
COMMENTS
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.
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
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
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..48 from Alexander Samokrutov)
Alexander Samokrutov, Diophantine equation
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,3).
FORMULA
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5).
a(n) = Sum_{k=0...floor(n/5)} (2^k*binomial(n,5*k)). - Alexander Samokrutov, Jul 11 2015
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
MAPLE
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):
seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Apr 15 2021
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 3}, {1, 1, 1, 1, 1}, 50] (* Vincenzo Librandi, Jul 11 2015 *)
PROG
(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
(Maxima) makelist(sum(2^k*binomial(n, 5*k), k, 0, floor(n/5)), n, 0, 50) /* Alexander Samokrutov, Jul 11 2015 */
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alexander Samokrutov, Sep 20 2014
STATUS
approved