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%I #50 Nov 05 2024 19:01:00
%S 0,0,2,1,0,0,0,2,0,0,0,0,0,2,2,0,0,2,0,2,0,0,0,2,0,0,0,2,0,0,0,0,0,0,
%T 2,0,0,0,2,0,0,0,0,0,0,2,0,4,0,0,0,0,0,0,2,0,0,0,0,2,0,0,4,1,0,2,0,2,
%U 0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,2,0,0,0
%N Number of ordered pairs (i,j) such that i*j=n and i+j is a square.
%C a(n) = 1 for n > 0 in A141046.
%C a(8820) = 8 and it is the only term in the first 10000 terms that is greater than 6. There are 977 terms in the first 10000 terms that are greater than zero. - _Harvey P. Dale_, Nov 08 2012
%H Reinhard Zumkeller, <a href="/A211996/b211996.txt">Table of n, a(n) for n = 1..10000</a>
%H David Clark, <a href="http://dx.doi.org/10.4153/CMB-1991-029-4">An arithmetical function associated with the rank of elliptic curves</a>, Canad. Math. Bull. Vol. 34 (2), (1991), pp. 181-185.
%H Jean-Marie De Koninck, A. Arthur Bonkli Razafindrasoanaivolala, and Hans Schmidt Ramiliarimanana, <a href="https://doi.org/10.1007/s40993-024-00520-x">Integers with a sum of co-divisors yielding a square</a>, Research in Number Theory, Vol. 10, No. 2 (2024), Article 30; <a href="https://www.jeanmariedekoninck.mat.ulaval.ca/fileadmin/Documents/Publications/2023_integers_with_a_sum_of_co-divisors_yielding_a_square.pdf">author's copy</a>.
%F Sum_{k=1..n} a(k) = c * n^(3/4) + O(sqrt(n)), where c = A377731 (De Koninck et al., 2024). - _Amiram Eldar_, Nov 05 2024
%e For n=3, the pairs (a,b) such that a*b=3 are (1,3) and (3,1). Both pairs add up to a square, so a(3) = 2.
%t nop[n_]:=Module[{divs=Divisors[n]},Count[Thread[{divs,Reverse[divs]}], _?(IntegerQ[Sqrt[Total[#]]]&)]]; Array[nop,90] (* _Harvey P. Dale_, Nov 08 2012 *)
%t a[n_] := DivisorSum[n, 1 &, IntegerQ[Sqrt[# + n/#]] &]; Array[a, 100] (* _Amiram Eldar_, Nov 05 2024 *)
%o (Haskell)
%o a211996 n = length [x | x <- [1..n], let (y, m) = divMod n x,
%o m == 0, a010052 (x + y) == 1]
%o -- _Reinhard Zumkeller_, Oct 28 2012
%o (PARI) a(n) = sumdiv(n, d, issquare(d+n/d)); \\ _Michel Marcus_, Jan 18 2021
%Y Cf. A010052, A141046, A218381, A218382, A377731.
%K nonn
%O 1,3
%A _Michel Marcus_, Oct 25 2012