OFFSET
1,2
COMMENTS
We solve r^2+(r+1)^2=0.5*(3*p^2+3*p+2), which is equivalent to (4*r+2)^2=3*(2*p+1)^2+1.
The Diophantine equation X^2=3*Y^2+1 gives X by A001075 and Y by A013453. The return to r gives the sequence 0,6,90,1260,17556,... which satisfies the formulas a(n+2)=14*a(n+1)-a(n)+6 and a(n+1)=7*a(n)+3+(48*a(n)^2+48*a(n)+9)^0.5 and the return to p the sequence A001921 which satisfies this new relation: a(n+1)=7*a(n)+sqrt(48*a(n)^2+48*a(n)+16). Then we obtain the present sequence.
LINKS
Robert Israel, Table of n, a(n) for n = 1..394
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).
FORMULA
a(n+2) = 195*a(n+1)-195*a(n)+a(n-1).
a(n+1) = 97*a(n) - 54 + 14*sqrt(48*a(n)^2-54*a(n)+15).
G.f.: x*(1-110*x+x^2)/((1-x)*(1-194*x+x^2)).
MAPLE
A131750 := proc(n) coeftayl(x*(1-110*x+x^2)/(1-x)/(1-194*x+x^2), x=0, n) ; end: seq(A131750(n), n=1..20) ; # R. J. Mathar, Oct 24 2007
MATHEMATICA
LinearRecurrence[{195, -195, 1}, {1, 85, 16381}, 20] (* Harvey P. Dale, Apr 26 2018 *)
PROG
(Magma) [n le 2 select 1 else Floor(97*Self(n-2) - 54 + 14*Sqrt(48*Self(n-2)^2-54*Self(n-2)+15)): n in [2..30]]; // Vincenzo Librandi, Aug 26 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Richard Choulet, Sep 20 2007
EXTENSIONS
More terms from R. J. Mathar, Oct 24 2007
Recurrences corrected by Robert Israel, Aug 26 2015
Name corrected by Daniel Poveda Parrilla, Sep 19 2016
STATUS
approved