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%I #39 Mar 15 2024 23:47:04
%S 0,2,10,60,348,2030,11830,68952,401880,2342330,13652098,79570260,
%T 463769460,2703046502,15754509550,91824010800,535189555248,
%U 3119313320690,18180690368890,105964828892652,617608282987020
%N 2*A084158 (twice Pell triangles).
%C Cross-referenced sequences A116484, A001109, A108475, A090390 are also generated by A*B given in the following FAMP code.
%C Floretion Algebra Multiplication Program, FAMP Code: 1jesleftseq[A*B] with A = - .5'i + .5'j - .5i' + .5j' + 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj' and B = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki'
%C Related to the reciprocals of the differences between successive convergents of the continued fraction of sqrt(2) (i.e., 1, 2, -10, 60, -348, 2030, -11830, 68952, ...). 1/1 + 1/2 - 1/10 + 1/60 - 1/348 + 1/2030 + ... = sqrt(2). 2, 10, 60, ... are products of the denominators of two successive convergents of sqrt(2) (e.g., 11830 = 70*169, cf. A000129 (Pell numbers)). - _Gerald McGarvey_, Feb 28 2006
%C a(n) is half of the even leg (b(n)) of the ordered Pythagorean triple (x(n), y(n)=x(n)+1, z(n)). In fact b(n) = x(n) + (1-(-1)^n)/2: x(0)=0, b(0)=0, a(0)=0; x(1)=3, b(1)=4, a(1)=2. - _George F. Johnson_, Aug 13 2012
%C Given a square shape composed of A001110(n+1) elements, thinking of it graphically as a sum of layers, each layer having an odd number of elements (all layers together being a sum of consecutive odd numbers), a(n) is the number of last layers that we have to subtract from the square to get a square of squares that is made of A002965(2*(n+1))^4 elements. - _Daniel Poveda Parrilla_, Jul 17 2016
%C Also numbers m such that 8*m^2 - 4*m + 1 or 8*m^2 + 4*m + 1 is a perfect square (square roots are then A001653). - _Lamine Ngom_, Jul 25 2023
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,5,-1).
%F G.f.: 2*x/((x+1)*(x^2-6*x+1)).
%F From _George F. Johnson_, Aug 13 2012: (Start)
%F a(n) = ((sqrt(2) + 1)^(2*n+1) - (sqrt(2) - 1)^(2*n+1) - 2*(-1)^n)/8. - corrected by _Ilya Gutkovskiy_, Jul 18 2016
%F 4*a(n)*(2*a(n) + (-1)^n) + 1 = A000129(2*n+1)^2 is a perfect square.
%F For n >= 0, a(n+1) = 3*a(n) + (-1)^n + sqrt(4*a(n)*(2*a(n) + (-1)^n) + 1).
%F For n > 0, a(n-1) = 3*a(n) + (-1)^n - sqrt(4*a(n)*(2*a(n) + (-1)^n) + 1).
%F a(n+1) = 6*a(n) - a(n-1) + 2*(-1)^n.
%F a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
%F For n > 0, a(n+1)*a(n-1) = a(n)*(a(n) + 2*(-1)^n).
%F a(n) = A046729(n)/2. (End)
%F a(n) = A000129(n)*A000129(n+1). - _Philippe Deléham_, Apr 10 2013
%F a(n) = A002965(2*(n+1))*(A002965(2*(n+1)+1) - A002965(2*(n+1))). - _Daniel Poveda Parrilla_, Jul 17 2016
%t Table[Fibonacci[n, 2] Fibonacci[n + 1, 2], {n, 0, 20}] (* or *)
%t LinearRecurrence[{5, 5, -1}, {0, 2, 10}, 21] (* or *)
%t CoefficientList[Series[2 x/((x + 1) (x^2 - 6 x + 1)), {x, 0, 20}], x] (* _Michael De Vlieger_, Jul 17 2016 *)
%Y Cf. A116484, A001109, A108475, A090390.
%Y Cf. A000129.
%K easy,nonn
%O 0,2
%A _Creighton Dement_, Feb 17 2006
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