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%I #26 Oct 25 2023 18:26:45
%S 51,401,2451,14401,84051,490001,2856051,16646401,97022451,565488401,
%T 3295908051,19209960001,111963852051,652573152401,3803475062451,
%U 22168277222401,129206188272051,753068852410001,4389206926188051,25582172704718401,149103829302122451
%N a(n) is the third number in a triple consisting of 3 numbers, which when squared are part of a right diagonal of a magic square of squares.
%C The multiplying factor 6 (in the recursion formulas below) appears to come from the ratio of b(1)/b(0) of the sequence. Each of the lines of tables (V vs VII) or (VI vs VIII) in oddwheel.com/ImaginaryB.html generates this factor.
%C k is obtained from the difference of the offsets of two relate sequences. this one, (II), starting at 51 and a second, (I), at 99 (to be submitted separately). Thus, k =[Ic(n)- IIc(n)]*2. When n=0, Ic(0)=99 and IIc(0)=51 giving the value for k of (99-51)*2=96. Furthermore, k is the same constant number for any value of n.
%C The differences between number in the sequence are identical in both of the related sequences.
%H Colin Barker, <a href="/A273189/b273189.txt">Table of n, a(n) for n = 0..1000</a>
%H E. Gutierrez, <a href="http://www.oddwheel.com/ImaginaryF6.html">Recursion Methods to Generate New Integer Sequences (Part VIF)</a>
%H E. Gutierrez, <a href="http://www.oddwheel.com/ImaginaryB.html">Table of Tuples and Use of Magic Ratio for Tuple Conversion (Part IB)</a>
%H E. Gutierrez, <a href="http://www.oddwheel.com/ImaginaryC.html">Table of Tuples for Square of Squares (Part IC)</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F a(0)= 51, a(1)= 401, a(n+1)= a(n)*6 - a(n-1) + k where k=96.
%F From _Colin Barker_, May 18 2016: (Start)
%F a(n) = (-24+25/2*(3-2*sqrt(2))^(1+n)+25/2*(3+2*sqrt(2))^(1+n)).
%F a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>2.
%F G.f.: (51+44*x+x^2) / ((1-x)*(1-6*x+x^2)).
%F (End)
%e a(2)= 401*6 - (51 - 96)= 2451;
%e a(3)= 2451*6 - (401 - 96)= 14401;
%e a(4)= 14401*6 - (2451 - 96)= 84051.
%t CoefficientList[Series[(51 + 44 x + x^2)/((1 - x) (1 - 6 x + x^2)), {x, 0, 20}], x] (* _Michael De Vlieger_, May 18 2016 *)
%t LinearRecurrence[{7,-7,1},{51,401,2451},30] (* _Harvey P. Dale_, Feb 21 2020 *)
%o (PARI) Vec((51+44*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ _Colin Barker_, May 18 2016
%Y Cf. A178218, A273182, A273187.
%K nonn,easy
%O 0,1
%A _Eddie Gutierrez_, May 17 2016
%E More terms from _Colin Barker_, May 18 2016
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