A039685 Numbers m such that m^2 ends in 444.

38, 462, 538, 962, 1038, 1462, 1538, 1962, 2038, 2462, 2538, 2962, 3038, 3462, 3538, 3962, 4038, 4462, 4538, 4962, 5038, 5462, 5538, 5962, 6038, 6462, 6538, 6962, 7038, 7462, 7538, 7962, 8038, 8462, 8538, 8962, 9038, 9462, 9538, 9962, 10038, 10462

Offset: 1

Author

Felice Russo

Keywords

• nonn
• base
• easy

Comments

No square can end in more than three 4's.

When a square ends in exactly three identical digits, these digits are necessarily 444. - Bernard Schott, Oct 31 2019

References

• Albert H. Beiler, "Recreations in the Theory of Numbers", Dover Publ., 2nd Ed. 1966, Chapter XV, "On The Square", p. 139. ISBN 0-486-21096-0.
• A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 1 pp. 55 and 95-96 (1995)
• David Wells, "Curious and Interesting Numbers", Revised Ed. Penguin Books, p. 152. ISBN 0-14-026149-4.

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• British Mathematical Olympiad, 1995 - Problem 1.
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A328886 (squares that end in 444).

Programs

Mathematica

Drop[ Flatten[ Table[{500n-38, 500n+38}, {n, 0, 21}]], 1] (* Robert G. Wilson v, Nov 27 2004 *)
Sqrt[#]&/@Select[Range[15000]^2,Mod[#,1000]==444&] (* or *) LinearRecurrence[{1,1,-1},{38,462,538},50] (* Harvey P. Dale, Dec 26 2023 *)

Formula

a(2n+1) = 500n + 38 and a(2n+2) = 500n - 38.

From Bruno Berselli, Oct 27 2010: (Start)

a(n) = 250*n + 87*(-1)^n - 125.

G.f.: 2*x*(19 + 212*x + 19*x^2)/((1+x)*(1-x)^2).

a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3. (End)

E.g.f.: 38 + (250*x - 38)*cosh(x) + (250*x - 212)*sinh(x). - Stefano Spezia, Sep 15 2024

Extensions

More terms from Patrick De Geest, Jun 15 1999