OFFSET
1,1
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
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]
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).
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
EXAMPLE
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.
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 *)
PROG
(PARI) a(n) = (2*10^n/3 + 1/3)^2 \\ David A. Corneth, Jan 27 2021
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Nicholas Protonotarios (protost(AT)hotmail.com), Aug 21 2005
EXTENSIONS
More terms from Harvey P. Dale, Nov 28 2014
Edited by Jon E. Schoenfield, Sep 03 2018
STATUS
approved