OFFSET
1,1
LINKS
Robert Israel, Table of n, a(n) for n = 1..999
FORMULA
a(n) = a(n-1) + 10^(n-1)*(2 + 5*(a(n-1)/2^(n-1) mod 2)), i.e., a(n) ends with a(n-1); if a(n-1) is divisible by 2^n then a(n) begins with a 2, if not then a(n) begins with a 7.
MAPLE
A[1]:= 2:
for n from 2 to 100 do
if A[n-1] mod 2^n = 0 then A[n]:= A[n-1]+2*10^(n-1)
else A[n]:= A[n-1]+7*10^(n-1)
fi
od:
seq(A[i], i=1..100); # Robert Israel, Oct 27 2019
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Henry Bottomley, Mar 06 2000
EXTENSIONS
Formula corrected by Robert Israel, Oct 27 2019
STATUS
approved