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LinearRecurrence[{35, -35, 1}, {0, 1, 9, 289}, 25] (* Paolo Xausa, Jul 22 2024 *)
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From Colin Barker, Sep 15 2014: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n > 3. G.f.: -x*(9*x^2-26*x+1) / ((x-1)*(x^2-34*x+1)). - _Colin Barker_, Sep 15 2014
G.f.: -x*(9*x^2 - 26*x + 1) / ((x-1)*(x^2 - 34*x + 1)). (End)
a(n) = (4 + 2*(17 + 12*sqrt(2))^(1-n) + (34 - 24*sqrt(2))*(17 + 12*sqrt(2))^n)/8 for n > 0. - Colin Barker, Mar 02 2016
a(2) = 9 because 9 = 3^2 = 1001 base _2 and 100 base _2 = 4 = 2^2.
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Giovanni Lucca, <a href="https://ijgeometry.com/product/giovanni-lucca-circle-chains-inside-the-arbelos-and-integer-sequences/">Circle chains inside the arbelos and integer sequences</a>, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
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