|
Search: id:A047208
|
|
|
| A047208 |
|
Numbers that are congruent to {0, 4} mod 5. |
|
+0
4
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
Also solutions to 3^x + 5^x == 2 mod 11. - Cino Hilliard (hillcino368(AT)gmail.com), May 18 2003
|
|
LINKS
|
Cino Hilliard, solutions to 3^x + 5^x == 2 mod 11/
|
|
FORMULA
|
G.f.: x^2(4+x)/((1-x)^2(1+x)). a(n)=a(n-2)+5. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jan 24 2009]
a(n)=(1/4)*[3-3*(-1)^n+10*n], with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Feb 10 2009]
a(n)=5*n-a(n-1)-6 (with a(1)=0) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 22 2009]
|
|
EXAMPLE
|
For n=2, a(2)=5*2-0-6=4; n=3, a(3)=5*3-4-6=5: n=4, a(4)=5*4-5-6=9 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 22 2009]
|
|
CROSSREFS
|
Cf.: A010685 (first differences). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jan 24 2009]
Sequence in context: A064473 A001983 A143575 this_sequence A032381 A050036 A059582
Adjacent sequences: A047205 A047206 A047207 this_sequence A047209 A047210 A047211
|
|
KEYWORD
|
nonn,new
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com).
|
|
|
Search completed in 0.002 seconds
|
