login
A133895
Numbers m such that binomial(m+5,m) mod 5 = 0.
0
20, 21, 22, 23, 24, 45, 46, 47, 48, 49, 70, 71, 72, 73, 74, 95, 96, 97, 98, 99, 120, 121, 122, 123, 124, 145, 146, 147, 148, 149, 170, 171, 172, 173, 174, 195, 196, 197, 198, 199, 220, 221, 222, 223, 224, 245, 246, 247, 248, 249, 270, 271, 272, 273, 274, 295
OFFSET
0,1
COMMENTS
Also numbers m such that floor(1+(m/5)) mod 5 = 0.
Partial sums of the sequence 20,1,1,1,1,21,1,1,1,1, 21, ... which has period 5.
FORMULA
a(n)=5n+20-4*(n mod 5).
G.f.: g(x)=(20+x+x^2+x^3+x^4+x^5)/((1-x^5)(1-x)).
G.f.: g(x)=(20-19x-x^6) /((1-x^5)(1-x)^2).
KEYWORD
nonn,easy
AUTHOR
Hieronymus Fischer, Oct 20 2007
STATUS
approved