A055524 Longest other side of a Pythagorean triangle with n as length of one of the three sides (in fact n is a leg and a(n) the hypotenuse).
5, 5, 13, 10, 25, 17, 41, 26, 61, 37, 85, 50, 113, 65, 145, 82, 181, 101, 221, 122, 265, 145, 313, 170, 365, 197, 421, 226, 481, 257, 545, 290, 613, 325, 685, 362, 761, 401, 841, 442, 925, 485, 1013, 530, 1105, 577, 1201, 626, 1301, 677, 1405, 730, 1513, 785
Offset: 3
Links
- Paolo Xausa, Table of n, a(n) for n = 3..10000
- Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).
Crossrefs
Programs
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Mathematica
A055524[n_] := (3*n^2-(-1)^n*(n^2-2)+6)/8; Array[A055524, 100, 3] (* or *) LinearRecurrence[{0, 3, 0, -3, 0, 1}, {5, 5, 13, 10, 25, 17}, 100] (* Paolo Xausa, Feb 29 2024 *)
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PARI
Vec(-x^3*(2*x^5+x^4-5*x^3-2*x^2+5*x+5)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Sep 15 2014
Formula
a(n) = sqrt(n^2+A055523(n)^2). a(2k) = k^2+1, a(2k+1) = k^2+(k+1)^2.
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x^3*(2*x^5+x^4-5*x^3-2*x^2+5*x+5) / ((x-1)^3*(x+1)^3). - Colin Barker, Sep 15 2014
a(n) = (3*n^2+6-(n^2-2)*(-1)^n)/8. - Luce ETIENNE, Jul 11 2015