%I #76 Sep 08 2022 08:46:04
%S 0,2,30,420,5852,81510,1135290,15812552,220240440,3067553610,
%T 42725510102,595089587820,8288528719380,115444312483502,
%U 1607931846049650,22395601532211600,311930489604912752,4344631252936566930,60512907051507024270,842836067468161772852
%N Numbers m such that 16*m*(3*m+1)+1 is a square.
%C If S(h) is the sum of the squares of the integers from 1 to h, then S(h)/h = (h+1)*(2*h+1)/6. If S(h)/h is a square, then h+1 is even and 2*h+1 is a multiple of 3, so that (h+1)/2 and (2*h+1)/3 must be odd squares. The numbers h that satisfy this condition are in A084231. Let h = 1+24*k+72*k^2; for any natural k we have (h+1)/2 = (6*k+1)^2, but only if 16*k*(3*k+1)+1 is a square also (2*h+1)/3 is a square. Searching what integer numbers k satisfy this condition leads to this sequence.
%H Bruno Berselli, <a href="/A217855/b217855.txt">Table of n, a(n) for n = 0..500</a>
%H Forum Rudi Mathematici, <a href="http://www.trekportal.it/coelestis/showpost.php?p=625864&postcount=923">Un po' di calcoli...un po' di logica</a>, post #923 (in Italian)
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F a(n) = ((2 + sqrt(3))^(2*n + 1) + (2 - sqrt(3))^(2*n + 1) - 4)/24.
%F a(n) = (cosh((2*n + 1)*log(2 + sqrt(3))) - 2)/12.
%F a(n) = a(-n-1) = 15*a(n-1) - 15*a(n-2) + a(n-3).
%F a(n) = 14*a(n-1) - a(n-2) + 2.
%F a(n) = 2*A076139(n).
%F 72*a(n)^2 + 24*a(n)+1 = A084231(n+1).
%F 16*a(n)*(3*a(n) + 1) + 1 = A028230(n+1)^2.
%F G.f.: 2*x/((1 - x)*(1 - 14*x + x^2)). - _Bruno Berselli_, Oct 16 2012
%F a(n) = 2*A076139(n) = 1/2*A123480(n) = 1/4*A045899(n) = 2/3*A076140(n). - _Peter Bala_, Dec 31 2012
%e a(0) = ((2+sqrt(3))^1+(2-sqrt(3))^1-4)/24 = 0,
%e a(1) = ((2+sqrt(3))^3+(2-sqrt(3))^3-4)/24 = 2,
%e a(2) = ((2+sqrt(3))^5+(2-sqrt(3))^5-4)/24 = 30;
%e a(2) = 14*a(1)-a(0)+2 = 30,
%e a(3) = 14*a(2)-a(1)+2 = 420,
%e a(4) = 14*a(3)-a(2)+2 = 5852;
%e a(3) = 15*a(2)-15*a(1)+a(0) = 420,
%e a(4) = 15*a(3)-15*a(2)+a(1) = 5852.
%e Regarding to the comment, a(3) = 420 and so 72*a(3)^2+24*a(3)+1 = A084231(4) = 12710881, therefore Sum_{i=1..12710881} i^2/12710881 = 7338631^2 = A084232(3)^2. - _Bruno Berselli_, Oct 17 2012
%t LinearRecurrence[{15, -15, 1}, {0, 2, 30}, 20] (* _Bruno Berselli_, Oct 16 2012 *)
%o (Maxima) makelist(expand(((2+sqrt(3))^(2*n+1)+(2-sqrt(3))^(2*n+1)-4)/24), n, 0, 19); /* _Bruno Berselli_, Oct 16 2012 */
%o (Magma) m:=19; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2/((1-x)*(1-14*x+x^2)))); // _Bruno Berselli_, Oct 16 2012
%o (PARI) a=vector(20); a[1]=0; a[2]=2; a[3]=30; for(i=4, #a, a[i]=15*a[i-1]-15*a[i-2]+a[i-3]); a \\ _Bruno Berselli_, Oct 16 2012
%Y Cf. A028230, A076139, A084231, A084232. A045899, A076140, A123480.
%K nonn,easy
%O 0,2
%A _Arnaldo Vicentini_, Oct 12 2012