OFFSET
0,2
COMMENTS
mod(A092899(n),4)=1,3,3,3,... = sum{k=0..n, mod(2^k,4)} Partial sums of 1,2,4,8,4,8,4,8....
LINKS
FORMULA
a(n)=4floor((n+1)/2)+4n-5+6*0^n; a(n)=sum{k=0...n, mod(A078008(k), 4)}+sum{k=0..n, 2*mod(A001045(k), 4)}.
For n > 0, a(n) = 6*n - 4 - (-1)^n; a(n+3) = a(n+2) + a(n+1) - a(n) - Warut Roonguthai, Oct 19 2005
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 12 2004
STATUS
approved