cp's OEIS Frontend

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.

Showing 1-2 of 2 results.

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.

Original entry on oeis.org

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

Views

Author

Alexander Samokrutov, Sep 20 2014

Keywords

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

Crossrefs

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

A097081 a(n) = Sum_{k=0..n} C(n,4k)*2^k.

Original entry on oeis.org

1, 1, 1, 1, 3, 11, 31, 71, 145, 289, 601, 1321, 2979, 6683, 14743, 32111, 69697, 151777, 332113, 728689, 1598883, 3503627, 7668079, 16774775, 36704017, 80343361, 175916521, 385196761, 843365379, 1846290395, 4041672871, 8847607391
Offset: 0

Views

Author

Paul Barry, Jul 23 2004

Keywords

Crossrefs

Cf. A093406.

Programs

  • Mathematica
    Table[Sum[Binomial[n,4k]2^k,{k,0,n}],{n,0,40}] (* or *) LinearRecurrence[ {4,-6,4,1},{1,1,1,1},40] (* Harvey P. Dale, Feb 26 2012 *)

Formula

G.f.: (1-x)^3/((1-x)^4-2*x^4);
a(n) = Sum_{k=0..floor(n/2)} binomial(n,4*k)*2^k;
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)+a(n-4).
Showing 1-2 of 2 results.