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(MAGMAMagma) I:=[4, 100, 3364]; [n le 3 select I[n] else 35*Self(n-1) - 35*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Oct 21 2018
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As T(n) + T(n+1) = (n+1)^2 and T(n+2) + T(n+3) = (n+3)^2, it follows that the equation T(n) + T(n+1) + T(n+2) + T(n+3) = s^2 becomes (n+1)^2 + (n+3)^2 = s^2. Hence the solutions to this equation correspond to those Pythagorean triples with shorter legs that differ by two, such as 6^2 + 8^2 = 10^2.
a(n) Terms are the squares of the hypotenuse for hypotenuses of Pythagorean triangles where other two sides are m and m+2, excepting the initial 4. See A075870. - Richard R. Forberg, Aug 15 2013
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 32.
a(n) = (2 + (3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1))/2.
As the third perfect square that can be expressed as the sum of four consecutive triangular numbers is 3364 = T(39) + T(40) + T(41) + T(42), we have a(3)=3364.
The first term, 4, equals T(-1) + T(0) + T(1) + T(2).
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a(n) = 35a35*a(n-1) -35a35*a(n-2) +a(n-3).
a(n) = 34a34*a(n-1) -a(n-2) -32.
a(n) = 1/2 [(2+(3+2sqrt2*sqrt(2))^(2n2*n+1) + (3-2sqrt2*sqrt(2))^(2n2*n+1)])/2.
a(n) = ceiling((1/2)*(2 + (3+2sqrt2*sqrt(2))^(2n+1))).
G.f.: 4x4*x*(x^2-34x10*x+251)/((1-x)*(x^2-34x34*x+1)).
(PARI) x='x+O('x^50); Vec(4*x*(1-10*x+x^2)/((1-x)*(1-34*x+x^2))) \\ G. C. Greubel, Oct 21 2018
(MAGMA) I:=[4, 100, 3364]; [n le 3 select I[n] else 35*Self(n-1) - 35*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Oct 21 2018
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<a href="/index/Rec#order_03">Index to sequences with entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
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