OFFSET
1,2
COMMENTS
Also solutions to 3^x + 5^x == 2 (mod 11). - Cino Hilliard, May 18 2003
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..5000
Cino Hilliard, solutions to 3^x + 5^x == 2 mod 11/, June 2003.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
FORMULA
From R. J. Mathar, Jan 24 2009: (Start)
G.f.: x^2*(4+x)/((1-x)^2*(1+x)).
a(n) = a(n-2) + 5. (End)
a(n) = 5*n - 6 - a(n-1) (with a(1)=0). - Vincenzo Librandi, Nov 18 2010
a(n+1) = Sum_{k>=0} A030308(n,k)*b(k), with b(0)=4 and b(k) = A020714(k-1) = 5*2^(k-1) for k>0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((5/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = (5*(n-1) + 3*(n-1 mod 2))/2 = (5*(n-1) + A010674(n-1))/2. - G. C. Greubel, Nov 23 2021
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + log(phi)/(2*sqrt(5)) - sqrt(1+2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021
E.g.f.: 1 + ((5*x - 7/2)*exp(x) + (3/2)*exp(-x))/2. - David Lovler, Aug 23 2022
MATHEMATICA
{#, #+4}&/@(5*Range[0, 30])//Flatten (* Harvey P. Dale, Apr 05 2019 *)
PROG
(PARI) forstep(n=0, 200, [4, 1], print1(n", ")) \\ Charles R Greathouse IV, Oct 17 2011
(Magma) [(5*(n-1) + 3*((n-1) mod 2))/2: n in [1..100]]; // G. C. Greubel, Nov 23 2021
(Sage) [(5*(n-1) +3*((n-1)%2))/2 for n in (1..100)] # G. C. Greubel, Nov 23 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved