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A157659
a(n) = 100*n^2 - n.
4
99, 398, 897, 1596, 2495, 3594, 4893, 6392, 8091, 9990, 12089, 14388, 16887, 19586, 22485, 25584, 28883, 32382, 36081, 39980, 44079, 48378, 52877, 57576, 62475, 67574, 72873, 78372, 84071, 89970, 96069, 102368, 108867, 115566, 122465, 129564
OFFSET
1,1
COMMENTS
The identity (200*n - 1)^2 - (100*n^2 - n)*(20)^2 = 1 can be written as A157955(n)^2 - a(n)*(20)^2 = 1 (see Barbeau's paper).
Also, the identity (80000*n^2 - 800*n + 1)^2 - (100*n^2 - n)*(8000*n - 40)^2 = 1 can be written as A157661(n)^2 - a(n)*A157660(n)^2 = 1 (see also the second part of the comment at A157661). - Vincenzo Librandi, Jan 28 2012
LINKS
E. J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 14 in the first table at p. 85, case d(t) = t*(10^2*t-1)).
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(99 + 101*x)/(1-x)^3.
E.g.f.: x*(99 + 100*x)*exp(x). - G. C. Greubel, Nov 17 2018
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {99, 398, 897}, 50]
PROG
(Magma) I:=[99, 398, 897]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
(PARI) a(n)=100*n^2-n \\ Charles R Greathouse IV, Dec 28 2011
(Sage) [100*n^2-n for n in (1..40)] # G. C. Greubel, Nov 17 2018
(GAP) List([1..40], n -> 100*n^2-n); # G. C. Greubel, Nov 17 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 04 2009
STATUS
approved