OFFSET
1,1
COMMENTS
An algorithm to compute the n-th term of this sequence: Write each of 2n and 2n+1 as products of their divisors in all possible ways and in decreasing order. For each product, equate each divisor in the product to (a1+1)(a2+1)...(ar+1), so that a1 >= a2 >= a3 >= ... >= ar, and solve for the ai. Evaluate A002144(1)^a1 * A002144(2)^a2 * ... * A002144(r)^ar for each set of values determined above, then the smaller of these products is the least integer to have precisely n partitions into a sum of two distinct positive squares. [Ant King, Dec 14 2009; May 26 2010]
LINKS
Ray Chandler, Table of n, a(n) for n = 1..1438 (a(1439) exceeds 1000 digits).
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Lekraj Beedassy, Apr 22 2004
EXTENSIONS
More terms from Ant King, Dec 14 2009 and Feb 07 2010
STATUS
approved