OFFSET
1,1
COMMENTS
Theorem (I. N. Ianakiev): There are infinitely many such numbers. Proof: Any A001110(2n+1), for n>0, is such a number as A001110(2n+1) = (2a+1)^2+(4a^2+4a)(4a^2+4a+1)(1/2), where a = (A002315(n)-1)(1/2). Note: other numbers, not of the form A001110(2n+1), e.g. A001110(6), are also in the sequence (see the example below).
Every term is divisible by its digital root (A010888). - Ivan N. Ianakiev, Oct 17 2013
EXAMPLE
a(3) = A001110(6) = 48024900 = 6918^2 + [576*577*(1/2)].
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Ivan N. Ianakiev, Oct 25 2012
EXTENSIONS
a(8)-a(13) from Donovan Johnson, Nov 02 2012
STATUS
approved