A109344 - OEIS

%I #49 Aug 01 2024 01:17:27

%S 49,4489,444889,44448889,4444488889,444444888889,44444448888889,

%T 4444444488888889,444444444888888889,44444444448888888889,

%U 4444444444488888888889,444444444444888888888889,44444444444448888888888889,4444444444444488888888888889,444444444444444888888888888889

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

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

%C In the same category, we have 729, 71289, 7112889, ...  with square roots in A199688. - _Michel Marcus_, Mar 21 2014

%D Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 30 at p. 61.

%D Italo Ghersi, Matematica dilettevole e curiosa, p. 112, Hoepli, Milano, 1967. [From _Vincenzo Librandi_, Dec 31 2008]

%D Paul Zeitz, The Art and Craft of Problem Solving, John Wiley and Sons, Inc., New York, 1999.

%H Seiichi Manyama, <a href="/A109344/b109344.txt">Table of n, a(n) for n = 1..500</a>

%H Emile Fourrey, <a href="https://gallica.bnf.fr/ark:/12148/bpt6k875411n/f82.image">Récréations arithmétiques</a>, Vuibert, 1899 and after, Paris, pages 72-73.

%H Michael Penn, <a href="https://www.youtube.com/watch?v=5Z8OXvbJftQ">This is always a perfect square??</a>, YouTube video, 2021.

%H StackExchange, <a href="http://math.stackexchange.com/questions/16946">History of 'Show that 44...88...9 is a perfect square'</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).

%F 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.

%F From _R. J. Mathar_, Jan 06 2009: (Start)

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

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

%F E.g.f.: (1/9)*exp(x)*(1 + 4*exp(9*x) + 4*exp(99*x)) - 1. - _Stefano Spezia_, Aug 22 2019

%e 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.

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

%t LinearRecurrence[{111,-1110,1000},{49,4489,444889},20] (* _Harvey P. Dale_, Nov 28 2014 *)

%o (PARI) a(n) = (2*10^n/3 + 1/3)^2 \\ _David A. Corneth_, Jan 27 2021

%Y Cf. A067275, A199688.

%K nonn,base,easy

%O 1,1

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

%E More terms from _Harvey P. Dale_, Nov 28 2014

%E Edited by _Jon E. Schoenfield_, Sep 03 2018