A109344 a(n) consists of n 4's, n-1 8's and a single 9 (in that order).

49, 4489, 444889, 44448889, 4444488889, 444444888889, 44444448888889, 4444444488888889, 444444444888888889, 44444444448888888889, 4444444444488888888889, 444444444444888888888889, 44444444444448888888888889, 4444444444444488888888888889, 444444444444444888888888888889

Offset: 1

Author

Nicholas Protonotarios (protost(AT)hotmail.com), Aug 21 2005

Keywords

• nonn
• base
• easy

Comments

All terms are squares. The square roots are in A067275.

In the same category, we have 729, 71289, 7112889, ... with square roots in A199688. - Michel Marcus, Mar 21 2014

Examples

			a(5) = 4444488889 because the first 5 digits are 4's, the next 5 - 1 = 4 digits are 8's and the last digit is 9.
		

References

• Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 30 at p. 61.
• Italo Ghersi, Matematica dilettevole e curiosa, p. 112, Hoepli, Milano, 1967. [From Vincenzo Librandi, Dec 31 2008]
• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Example 5.5 on page 159.
• Paul Zeitz, The Art and Craft of Problem Solving, John Wiley and Sons, Inc., New York, 1999.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..500
• Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
• Michael Penn, This is always a perfect square??, YouTube video, 2021.
• StackExchange, History of 'Show that 44...88...9 is a perfect square'.
• Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Crossrefs

Cf. A067275, A199688.

Programs

Maple

a:=n->4*sum('10^i', 'i'=n..2*n-1)+8*sum('10^i', 'i'=1..n-1)+9;

Mathematica

LinearRecurrence[{111,-1110,1000},{49,4489,444889},20] (* Harvey P. Dale, Nov 28 2014 *)

PARI

a(n) = (2*10^n/3 + 1/3)^2 \\ David A. Corneth, Jan 27 2021

Formula

a(1)=49; a(n) = 4*(Sum_{i=n..2*n-1} 10^i) + 8*(Sum_{i=1..n-1} 10^i) + 9, n >= 2.

From R. J. Mathar, Jan 06 2009: (Start)

a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) = (4*100^n + 4*10^n + 1)/9.

G.f.: x*(49 - 950*x + 1000*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)

E.g.f.: (1/9)*exp(x)*(1 + 4*exp(9*x) + 4*exp(99*x)) - 1. - Stefano Spezia, Aug 22 2019

Extensions

More terms from Harvey P. Dale, Nov 28 2014

Edited by Jon E. Schoenfield, Sep 03 2018