OFFSET
0,2
COMMENTS
For n > 1, a(n) = 98*a(n-1) - a(n-2) + 15. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1) = A014105(m) and for n > 1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m). Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).
LINKS
Colin Barker, Table of n, a(n) for n = 0..500
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).
FORMULA
From Colin Barker, Mar 02 2016: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n>2.
G.f.: 15*x / ((1-x)*(1-98*x+x^2)). (End)
a(n) = (-10+(5-2*sqrt(6))*(49+20*sqrt(6))^(-n)+(5+2*sqrt(6))*(49+20*sqrt(6))^n)/64. - Colin Barker, Mar 03 2016
EXAMPLE
2*0 = 3*0.
2*15 = 3*10.
2*1485 = 3*990.
2*145530 = 3*97020.
MATHEMATICA
LinearRecurrence[{99, -99, 1}, {0, 15, 1485}, 20] (* T. D. Noe, Feb 15 2012 *)
PROG
(PARI) concat(0, Vec(15*x/((1-x)*(1-98*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016
(Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(15*x/((1-x)*(1-98*x+x^2)))); // G. C. Greubel, Jul 15 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Charlie Marion, Feb 15 2012
STATUS
approved