OFFSET
1,2
COMMENTS
a(1) = 1, a(2) = 35, for n >= 3; a(n) = the smallest number h > a(n-1) such that [[a(n-2) + a(n-1)] * [a(n-2) + h] * [a(n-1) + h]] / [a(n-2) * a(n-1) * h] is integer (= 130). (conjectured)
10^(floor((n - 1)/2)) | a(n), for n>=1. - G. C. Greubel, Jan 11 2018
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,90).
FORMULA
a(2n) = 35* a(2n-1), a(2n+1) = 18/7 * a(2n).
a(2n) = 35*90^(n-1), a(2n+1) = 90^n.
EXAMPLE
For n = 4; a(2) = 35, a(3) = 90, a(4) = 3150 before [(35+90)*(35+3150)*(90+3150)] / (35*90*3150) = 130.
MATHEMATICA
LinearRecurrence[{0, 90}, {1, 35}, 50] (* or *) CoefficientList[Series[(1 + 35*x)/(1-90*x^2), {x, 0, 50}], x] (* G. C. Greubel, Jan 11 2018 *)
PROG
(PARI) Vec((1+35*x)/(1-90*x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012
(Magma)
I:=[1, 35]; [n le 2 select I[n] else 90*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 11 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jaroslav Krizek, Nov 27 2010
EXTENSIONS
Terms a(12) onward added by G. C. Greubel, Jan 11 2018
STATUS
approved