A157651 a(n) = 100*n^2 - 49*n + 6.

57, 308, 759, 1410, 2261, 3312, 4563, 6014, 7665, 9516, 11567, 13818, 16269, 18920, 21771, 24822, 28073, 31524, 35175, 39026, 43077, 47328, 51779, 56430, 61281, 66332, 71583, 77034, 82685, 88536, 94587, 100838, 107289, 113940, 120791, 127842

Offset

1,1

Comments

The identity (80000*n^2 -39200*n +4801)^2 - (100*n^2 -49*n +6)*(8000*n -1960)^2 = 1 can be written as A157653(n)^2 - a(n)*A157652(n)^2 = 1.

The continued fraction expansion of sqrt(a(n)) is [10n-3; {1, 1, 4, 1, 1, 20n-6}]. - Magus K. Chu, Sep 09 2022

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Vincenzo Librandi, X^2-AY^2=1

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).

G.f.: x*(57 + 137*x + 6*x^2)/(1-x)^3.

E.g.f.: (6 + 51*x + 100*x^2)*exp(x) - 6. - G. C. Greubel, Nov 17 2018

Mathematica

LinearRecurrence[{3, -3, 1}, {57, 308, 759}, 40]

Prog

(Magma) I:=[57, 308, 759]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
(PARI) a(n) = 100*n^2 - 49*n + 6.
(Sage) [100*n^2-49*n+6 for n in (1..40)] # G. C. Greubel, Nov 17 2018
(GAP) List([1..40], n -> 100*n^2-49*n+6); # G. C. Greubel, Nov 17 2018

Crossrefs

Cf. A157652, A157653.

Sequence in context: A145296 A176635 A048422 * A251263 A367277 A371515

Adjacent sequences: A157648 A157649 A157650 * A157652 A157653 A157654

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 03 2009

Status

approved