OFFSET
1,1
LINKS
FORMULA
For n > 2, a(n) = 10*A000225(n-3) + 9 = 10*(2^(n-3) - 1) + 9 = 10*2^(n-3) - 1. - Gerald McGarvey, Aug 25 2004
a(1)=1, a(n) = n + Sum_{i=1..n-1} a(i) for n > 1. - Gerald McGarvey, Sep 06 2004
a(n) = 5*2^(n-2) - 1 for n > 1. - Karl V. Keller, Jr., Jun 12 2022
PROG
(Python) print([2]+[(5*2**(n-2) - 1) for n in range(2, 50)]) # Karl V. Keller, Jr., Jun 12 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved