A333385 a(n) = 3^n + 2 * 17^n for n >= 0.

3, 37, 587, 9853, 167123, 2839957, 48275867, 820679533, 13951521443, 237175772677, 4031987859947, 68543792792413, 1165244474990963, 19809156067406197, 336755653123584827, 5724846103033980493, 97322383751376783683, 1654480523772802668517

Offset: 0

Author

Bernard Schott, Mar 18 2020

Keywords

• nonn
• easy

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

Examples

			a(4) = 3^4 + 2 * 17^4 = 167123 = 7 * 19 * 1031 is not a perfect square.
		

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 to sequences related to Olympiads.
• Index entries for linear recurrences with constant coefficients, signature (20,-51).

Crossrefs

Cf. A000244 (3^n), A001026 (17^n), A330770.

Subsequence of A000037.

Programs

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 *)

PARI

Vec((3 - 23*x) / ((1 - 3*x)*(1 - 17*x)) + O(x^20)) \\ Colin Barker, Mar 18 2020

Formula

a(n) = A000244(n) + 2 * A001026(n).

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)