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.

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

Offset: 0

Author

Eddie Gutierrez, May 17 2016

Keywords

• nonn
• easy

Comments

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.

Examples

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

Links

• 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).

Crossrefs

Cf. A178218, A273182, A273187.

Programs

Mathematica

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 *)

PARI

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

Formula

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)

Extensions

More terms from Colin Barker, May 18 2016