A047208 Numbers that are congruent to {0, 4} mod 5.

0, 4, 5, 9, 10, 14, 15, 19, 20, 24, 25, 29, 30, 34, 35, 39, 40, 44, 45, 49, 50, 54, 55, 59, 60, 64, 65, 69, 70, 74, 75, 79, 80, 84, 85, 89, 90, 94, 95, 99, 100, 104, 105, 109, 110, 114, 115, 119, 120, 124, 125, 129, 130, 134, 135, 139, 140, 144, 145, 149

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A001622, A010674, A010685 (first differences), A274406.

Programs

Magma

[(5*(n-1) + 3*((n-1) mod 2))/2: n in [1..100]]; // G. C. Greubel, Nov 23 2021

Mathematica

{#,#+4}&/@(5*Range[0,30])//Flatten (* Harvey P. Dale, Apr 05 2019 *)

PARI

forstep(n=0,200,[4,1],print1(n", ")) \\ Charles R Greathouse IV, Oct 17 2011

Sage

[(5*(n-1) +3*((n-1)%2))/2 for n in (1..100)] # G. C. Greubel, Nov 23 2021

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