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%I #31 May 03 2018 18:53:49
%S 3,30,85,168,279,418,585,780,1003,1254,1533,1840,2175,2538,2929,3348,
%T 3795,4270,4773,5304,5863,6450,7065,7708,8379,9078,9805,10560,11343,
%U 12154,12993,13860,14755,15678,16629,17608,18615,19650,20713,21804,22923
%N a(n) = n*(14*n + 13) + 3.
%C Sequence found by reading the line from 3, in the direction 3, 30,..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the semi-diagonal parallel to A195024 and also parallel to A195028 in the same square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].
%C 56*a(n) + 1 is a perfect square. - _Bruno Berselli_, Feb 14 2017
%H Matthew House, <a href="/A195029/b195029.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 14*n^2 + 13*n + 3 = A195028(n) + 3 = (2*n + 1)*(7*n + 3).
%F From _Colin Barker_, Apr 09 2012: (Start)
%F G.f.: (3 + 21*x + 4*x^2)/(1 - x)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%t Table[n (14 n + 13) + 3, {n, 0, 40}] (* _Bruno Berselli_, Feb 14 2017 *)
%t LinearRecurrence[{3,-3,1},{3,30,85},50] (* _Harvey P. Dale_, May 03 2018 *)
%o (PARI) a(n)=n*(14*n+13)+3 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Bisection of A022264.
%Y Cf. A144555, A152760, A195019, A195020, A195021, A195023, A195024, A195025, A195026, A195027, A195028, A195320.
%Y Cf. A185019, A193053, A198017.
%K nonn,easy
%O 0,1
%A _Omar E. Pol_, Sep 07 2011
%E Edited by _Bruno Berselli_, Feb 14 2017
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