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A109344
a(n) consists of n 4's, n-1 8's and a single 9 (in that order).
7
49, 4489, 444889, 44448889, 4444488889, 444444888889, 44444448888889, 4444444488888889, 444444444888888889, 44444444448888888889, 4444444444488888888889, 444444444444888888888889, 44444444444448888888888889, 4444444444444488888888888889, 444444444444444888888888888889
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
Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
Michael Penn, This is always a perfect square??, YouTube video, 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
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
Sequence in context: A219061 A218317 A283229 * A129207 A243944 A144928
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