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A239450
Numbers m such that T(m)^2 + T(m^2) is a perfect square, where T = A000217.
1
0, 3, 47, 659, 9183, 127907, 1781519, 24813363, 345605567, 4813664579, 67045698543, 933826115027, 13006519911839, 181157452650723, 2523197817198287, 35143611988125299, 489487370016555903, 6817679568243657347, 94958026585394646959, 1322594692627281400083
OFFSET
1,2
COMMENTS
The sequence of numbers k such that k^2 is representable as T(m)^2 + T(m^2) begins: 0, 9, 1927, 376289, 73032399, 14168386297, 2748600555479.
After 0, numbers k such that 3*k^2 + 2*k + 3 is a square. - Bruno Berselli, Aug 31 2017
FORMULA
a(n) = 15*(a(n-1) - a(n-2)) + a(n-3) for n > 4. - Giovanni Resta, Mar 19 2014
From Bruno Berselli, Mar 19 2014: (Start)
G.f.: x^2*(1 + x)*(3 - x)/((1 - x)*(1 - 14*x + x^2)).
a(n) = 4*A046174(n-1) - 1 for n > 1. (End)
a(n) = (-1 + (7 + 4*sqrt(3))^n*(-19 + 11*sqrt(3)) - (7 - 4*sqrt(3))^n*(19 + 11*sqrt(3)))/3 for n > 1. - Colin Barker, Mar 05 2016
EXAMPLE
T(3)^2 + T(3^2) = 6^2 + 9*10/2 = 36+45 = 81. Because 81 is a perfect square, 3 is in the sequence.
MATHEMATICA
a[1]=0; a[2]=3; a[3]=47; a[4]=659; a[n_] := a[n] = 15*(a[n-1] - a[n-2]) + a[n-3]; Array[a, 50] (* Giovanni Resta, Mar 19 2014 *)
Join[{0}, LinearRecurrence[{15, -15, 1}, {3, 47, 659}, 20]] (* Harvey P. Dale, Feb 02 2015 *)
CoefficientList[Series[x (1 + x) (3 - x) / ((1 - x) (1 - 14 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 03 2015 *)
PROG
(PARI) isok(n) = issquare((n*(n+1)/2)^2 + n^2*(n^2+1)/2); \\ Michel Marcus, Mar 19 2014
(Magma) I:=[0, 3, 47, 659]; [n le 4 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..60]]; // Vincenzo Librandi, Feb 03 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alex Ratushnyak, Mar 19 2014
EXTENSIONS
a(11)-a(20) from Giovanni Resta, Mar 19 2014
STATUS
approved