|
|
A072880
|
|
A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6).
|
|
9
|
|
|
1, 1, 1, 1, 1, 1, 5, 29, 869, 756029, 571580604869, 326704387862983487112029, 21347151409785350408171299054974277225256721769, 15713823217665540462976624783900822313284439536736221766688609460305249837839107387688348185
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,7
|
|
COMMENTS
|
Any six successive terms satisfy the Markoff-Hurwitz equation a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6*a*b*c*d*e*f. - Bruno Langlois, Aug 09 2016
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 6*a(n-1)*a(n-2)*a(n-3)*a(n-4)*a(n-5) - a(n-6). - Bruno Langlois, Aug 09 2016
|
|
MATHEMATICA
|
RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 3]^2 + a[n - 4]^2 + a[n - 5]^2)/a[n - 6], a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 1, 14}] (* Michael De Vlieger, Aug 11 2016 *)
nxt[{a_, b_, c_, d_, e_, f_}] := {b, c, d, e, f, (b^2+c^2+d^2+e^2+f^2)/a}; NestList[ nxt, Table[1, 6], 20][[All, 1]] (* Harvey P. Dale, Mar 18 2018 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|