A273187 a(n) is the third number in a triple consisting of 3 numbers, which when squared are part of a right diagonal of magic square of squares.
99, 449, 2499, 14449, 84099, 490049, 2856099, 16646449, 97022499, 565488449, 3295908099, 19209960049, 111963852099, 652573152449, 3803475062499, 22168277222449, 129206188272099, 753068852410049, 4389206926188099, 25582172704718449, 149103829302122499
Offset: 0
Examples
a(2)= 449*6 - (99 + 96)= 2499; a(3)= 2499*6 - (449 + 96)= 14449.
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).
Programs
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Mathematica
CoefficientList[Series[(99 - 244 x + 49 x^2)/((1 - x) (1 - 6 x + x^2)), {x, 0, 20}], x] (* Michael De Vlieger, May 20 2016 *) -
PARI
Vec((99-244*x+49*x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Colin Barker, May 18 2016
Formula
a(0)= 99, a(1)= 449, 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.: (99-244*x+49*x^2) / ((1-x)*(1-6*x+x^2)).
(End)
a(n) = 24 + 25*A001541(n+1). - R. J. Mathar, Jun 07 2016