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A156853
a(n) = 2025*n^2 - 649*n + 52.
4
1428, 6854, 16330, 29856, 47432, 69058, 94734, 124460, 158236, 196062, 237938, 283864, 333840, 387866, 445942, 508068, 574244, 644470, 718746, 797072, 879448, 965874, 1056350, 1150876, 1249452, 1352078, 1458754, 1569480, 1684256
OFFSET
1,1
COMMENTS
The identity (32805000*n^2 - 55096200*n + 23133601)^2 - (2025*n^2 - 649*n + 52)*(729000*n - 612180)^2 = 1 can be written as A157078(n)^2 - a(n)*A156865(n)^2 = 1.
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(1428 + 2570*x + 52*x^2)/(1-x)^3.
E.g.f.: -52 + (52 + 1376*x + 2025*x^2)*exp(x). - G. C. Greubel, Jan 27 2022
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1428, 6854, 16330}, 40]
PROG
(Magma) I:=[1428, 6854, 16330]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..30]];
(PARI) a(n)=2025*n^2-649*n+52 \\ Charles R Greathouse IV, Dec 23 2011
(Sage) [(25*n -4)*(81*n -13) for n in (1..30)] # G. C. Greubel, Jan 27 2022
CROSSREFS
Sequence in context: A163589 A250582 A260283 * A292704 A321036 A094230
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 17 2009
STATUS
approved