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 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 b-a=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=(b-a)*(b+a) gives a=31 and b=32; 3*21=(b-a)*(b+a) gives a=9 and b=12; 7*9=(b-a)*(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 (PARI) a(n)=my(b, N=2*n+1); sum(a=0, n, if(issquare(N+a^2, &b), a^2+b^2)) \\ Charles R Greathouse IV, Apr 28 2014 CROSSREFS Cf. A237626. Sequence in context: A147272 A147427 A147408 * A249513 A147090 A097117 Adjacent sequences:  A241654 A241655 A241656 * A241658 A241659 A241660 KEYWORD nonn AUTHOR J. M. Bergot, Apr 26 2014 EXTENSIONS a(4) corrected by Charles R Greathouse IV, Apr 28 2014 STATUS approved

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Last modified February 18 02:57 EST 2020. Contains 332006 sequences. (Running on oeis4.)