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A046177
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Squares (A000290) which are also hexagonal numbers (A000384).
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8
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1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481
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OFFSET
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1,2
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COMMENTS
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Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007
Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011
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REFERENCES
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M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.
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LINKS
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FORMULA
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a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
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MATHEMATICA
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LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)
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PROG
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(PARI) Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015
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CROSSREFS
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Cf. A001110 (Numbers that are both triangular and square).
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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