OFFSET
0,1
COMMENTS
This sequence was the subject of the 1st problem of the 27th British Mathematical Olympiad in 1991 (see the link BMO).
Proposition: a(n) is never a perfect square.
Proof (by induction): the unit digits of a(n) follow the pattern 3773, 3773, ...
Generalization: Steve Dinh proves that for nonnegative integers k, m, u and v, the numbers (10^k*u + 3)^n + 2*(10^m*v + 7)^n are never a perfect square for n >= 0 (see reference). - Bernard Schott, Dec 27 2021
REFERENCES
S. Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 of British Mathematical Olympiad 1991, page 186.
A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 1 pp. 57 and 115 (1991).
LINKS
Colin Barker, Table of n, a(n) for n = 0..800
British Mathematical Olympiad, Problem 1, 1991.
Index entries for linear recurrences with constant coefficients, signature (20,-51).
FORMULA
From Colin Barker, Mar 18 2020: (Start)
G.f.: (3 - 23*x) / ((1 - 3*x)*(1 - 17*x)).
a(n) = 20*a(n-1) - 51*a(n-2) for n>1.
(End)
EXAMPLE
a(4) = 3^4 + 2 * 17^4 = 167123 = 7 * 19 * 1031 is not a perfect square.
MAPLE
S:=seq(3^n+2*17^n, n=0..40);
MATHEMATICA
a[n_] := 3^n + 2 * 17^n ; Array[a, 18, 0] (* Amiram Eldar, Mar 18 2020 *)
PROG
(PARI) Vec((3 - 23*x) / ((1 - 3*x)*(1 - 17*x)) + O(x^20)) \\ Colin Barker, Mar 18 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bernard Schott, Mar 18 2020
STATUS
approved