A247584 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.

1, 1, 1, 1, 1, 3, 13, 43, 113, 253, 509, 969, 1849, 3719, 8009, 18027, 40897, 91257, 198697, 423777, 894081, 1886011, 4007301, 8594411, 18560081, 40181493, 86872293, 187197193, 402060793, 861827743, 1846685729, 3960390059, 8504658049, 18283290609, 39325827729

Offset: 0

Author

Alexander Samokrutov, Sep 20 2014

Keywords

• nonn
• easy

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).

Crossrefs

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.

Cf. A005531.

Programs

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

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 *)

Maxima

makelist(sum(2^k*binomial(n,5*k), k, 0, floor(n/5)), n, 0, 50); /* Alexander Samokrutov, Jul 11 2015 */

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

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

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