OFFSET
1,4
COMMENTS
Proof. By A127276 and A001788, we have odd part(A176126(n))=odd part(|A127276(n)|) = odd part(n*(n+1)-4), {odd part(A176126(n-1)), n>=3}={odd part((n+1)*(n+2)-4), n>=1}.
Let n=2^b*k, where k=k(n) is odd.
Then {odd part(A176126(n-1)), n>=3}={odd part((2^b*k+1)*(2^b*k+2)-4)}={odd part(2^(2*b)*k^2+3*2^b*k-2)}. Hence, if b>0, then {odd part(A176126(n-1), n>=3)= {odd part(2^(2*b-1)*k^2+3*2^(b-1)*k-1)}.
On the other hand, in this case odd part(a(n))=odd part(2^(b-1)*k*(2^b*k+3)-1)=odd part(2^(2*b-1)*k^2+3*2^(b-1)*k-1). It is left to consider the case of odd n. Setting n=2*m-1, m>=1, we easily find that for both expressions the odd part equals odd part(2*m^2+m-2).
The smallest prime divisor of a(n) is more than or equal to 13.
LINKS
Peter J. C. Moses, Table of n, a(n) for n = 1..1000
FORMULA
MATHEMATICA
Map[#/2^IntegerExponent[#, 2]&[(# (#+3)/2-1)]&, Range[100]] (* Peter J. C. Moses, Feb 02 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Shevelev, Feb 02 2014
STATUS
approved