

A241657


The sum of a^2 + b^2 for all nonnegative integers a,b such that b^2  a^2 = 2*n+1.


0



1, 5, 13, 25, 50, 61, 85, 130, 145, 181, 250, 265, 338, 410, 421, 481, 610, 650, 685, 850, 841, 925, 1183, 1105, 1250, 1450, 1405, 1586, 1810, 1741, 1861, 2275, 2210, 2245, 2650, 2521, 2665, 3255, 3050, 3121, 3731, 3445, 3770, 4210, 3961, 4250, 4810
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

A sample of 54 terms found none with last digit 2,4,7, or 9, and both ending digit 0 and 5 had 17; 15 had final digit 1.
The square of 29 = a(20); a(47)a(48)=1, probably the only time this will occur.
Eleven primes all ending in 1 were found.


LINKS



FORMULA

For each pair of divisors d and d' of 2*n+1 with d*d'=2*n+1 and d<=d', find a and b to satisfy ba=d and b+a=d', then compute a^2 + b^2. Find the sum of all these results.
If 2*n+1 is not a square, a(n)=sum[d(2*n+1)^2 {d(2*n+1) a divisor of 2*n+1}].
If 2*n+1 is a square, a(n)=(sum[d(2*n+1)^2 {d(2*n+1) a divisor of 2*n+1}] +
2*n+1)/2.


EXAMPLE

For n=31, 2*31+1=63=3^2*7, with divisors 1,3,7,9,21,63.
Grouping in pairs 1*63=(ba)*(b+a) gives a=31 and b=32; 3*21=(ba)*(b+a) gives a=9 and b=12; 7*9=(ba)*(b+a) gives a=1 and b=8.
The sum 1^2 + 8^2 + 9^2 + 12^2 + 31^2 + 32^2 = 2275 = a(31).


PROG



CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



