A273187 -id:A273187 - cp's OEIS frontend

A273182 a(n) is the second number in a triple consisting of 3 numbers, which when squared are part of a right diagonal of a magic square of squares.

14, 84, 490, 2856, 16646, 97020, 565474, 3295824, 19209470, 111960996, 652556506, 3803378040, 22167711734, 129202892364, 753049642450, 4389094962336, 25581520131566, 149100025827060, 869018634830794, 5065011783157704, 29521052064115430, 172061300601534876

Offset: 0

Eddie Gutierrez, May 17 2016

The multiplying factor 6 appears to come from the ratio of a(1)/a(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.

			a(2) = 84*6 -14 = 490; a(3) = 490*6 - 84 = 2856; a(4) = 2856*6 - 490 = 16646.

Colin Barker, Table of n, a(n) for n = 0..1000
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
E. Gutierrez, Recursion Methods to Generate New Integer Sequences (Part VIF)
E. Gutierrez, Table of Tuples and Use of Magic Ratio for Tuple Conversion (Part IB)
E. Gutierrez, Table of Tuples for Square of Squares (Part IC)
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A178218, A273187, A273189.

CoefficientList[Series[14/(1 - 6 x + x^2), {x, 0, 21}], x] (* Michael De Vlieger, May 18 2016 *)

Vec(14/(1-6*x+x^2) + O(x^50)) \\ Colin Barker, May 18 2016

a(0)=14, a(1)= 84, a(n+1)= a(n)*6 - a(n-1).
G.f.: 14 / (1-6*x+x^2). - Colin Barker, May 18 2016
E.g.f.: 7*(3*sqrt(2)*sinh(2*sqrt(2)*x) + 4*cosh(2*sqrt(2)*x))*exp(3*x)/2. - Ilya Gutkovskiy, May 18 2016

A273189 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.

Original entry on oeis.org

51, 401, 2451, 14401, 84051, 490001, 2856051, 16646401, 97022451, 565488401, 3295908051, 19209960001, 111963852051, 652573152401, 3803475062451, 22168277222401, 129206188272051, 753068852410001, 4389206926188051, 25582172704718401, 149103829302122451

Offset: 0

Eddie Gutierrez, May 17 2016

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.
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.
The differences between number in the sequence are identical in both of the related sequences.

			a(2)= 401*6 - (51 - 96)= 2451;
a(3)= 2451*6 - (401 - 96)= 14401;
a(4)= 14401*6 - (2451 - 96)= 84051.

Colin Barker, Table of n, a(n) for n = 0..1000
E. Gutierrez, Recursion Methods to Generate New Integer Sequences (Part VIF)
E. Gutierrez, Table of Tuples and Use of Magic Ratio for Tuple Conversion (Part IB)
E. Gutierrez, Table of Tuples for Square of Squares (Part IC)
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A178218, A273182, A273187.

CoefficientList[Series[(51 + 44 x + x^2)/((1 - x) (1 - 6 x + x^2)), {x, 0, 20}], x] (* Michael De Vlieger, May 18 2016 *)
LinearRecurrence[{7,-7,1},{51,401,2451},30] (* Harvey P. Dale, Feb 21 2020 *)

Vec((51+44*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, May 18 2016

a(0)= 51, a(1)= 401, a(n+1)= a(n)*6 - a(n-1) + k where k=96.
From Colin Barker, May 18 2016: (Start)
a(n) = (-24+25/2*(3-2*sqrt(2))^(1+n)+25/2*(3+2*sqrt(2))^(1+n)).
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>2.
G.f.: (51+44*x+x^2) / ((1-x)*(1-6*x+x^2)).
(End)

More terms from Colin Barker, May 18 2016

cp's OEIS Frontend

A273182 a(n) is the second number in a triple consisting of 3 numbers, which when squared are part of a right diagonal of a magic square of squares.

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