Search: id:A046172
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%I A046172
%S A046172 1,81,7921,776161,76055841,7452696241,730288175761,71560788528321,
%T A046172 7012226987599681,687126683996240401,67331402804643959601,
%U A046172 6597790348171111800481,646516122717964312487521
%N A046172 Indices of pentagonal numbers which are also square.
%C A046172 if P_x=y^2 is a pentagonal number which is also a square, the least both 
               pentagonal and square number which is greater as P_x, is P_(49*x+40*y-8)=(60*x+49*y-10)^2 
               (in fact P_(49*x+40*y-8)-(60*x+49*y-10)^2=1.5*x^2-0.5*x-y^2). [From 
               Richard Choulet (richardchoulet(AT)yahoo.fr), Apr 28 2009]
%C A046172 Contribution from Weisenhorn Paul (paulweisenhorn(AT)online.de), May 
               15 2009: (Start)
%C A046172 a(n)*(3*a(n)-1)/2=m*m is eqivalent to tht pell-equation
%C A046172 (6*a(n)-1)^2-6*(2*m)^2=1 or x(n)^2-6*y(n)^2=1
%C A046172 (End)
%H A046172 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               PentagonalSquareNumber.html">Link to a section of The World of Mathematics.</
               a>
%H A046172 L. Euler, <a href="http://math.dartmouth.edu/~euler/pages/E029.html">
               De solutione problematum diophanteorum per numeros integros</a>, 
               par. 21
%F A046172 a(n) = 98*a(n-1) - a(n-2) - 16; g.f.: (1-18*x+x^2)/((1-x)*(1-98*x+x^2)) 
               - Warut Roonguthai (warut822(AT)yahoo.com) Jan 05 2001
%F A046172 a(n+1)=49*a(n)-8+10*sqrt(8*(3a(n)^2-a(n)) with a(1)=1 [From Richard Choulet 
               (richardchoulet(AT)yahoo.fr), Apr 28 2009]
%F A046172 a(n)=1/6+((5+2*sqrt(6))^(2*n+1)/12)+((5-2*sqrt(6))^(2*n+1)/12) for n>
               =0 [From Richard Choulet (richardchoulet(AT)yahoo.fr), Apr 29 2009]
%F A046172 Contribution from Weisenhorn Paul (paulweisenhorn(AT)online.de), May 
               15 2009: (Start)
%F A046172 x(n+2)=98*x(n+1)-x(n) with x(1)=5,x(2)=485
%F A046172 y(n+2)=98*y(n+1)-y(n) with y(n)=A046173(n)*2
%F A046172 m(n+2)=98*m(n+1)-m(n) with m(n)=A046173(n)
%F A046172 a(n)=A072256(n)^2
%F A046172 (End)
%F A046172 a(n)=b(n)*b(n) b(n)=10*b(n-1)- b(n-2) b(1)=1 b(2)=9 b(n)=((5+sqrt(24))^n-(5-sqrt(24))^n)/
               (2*sqrt(24)) [From Sture Sjostedt (sture.sjostedt(AT)spray.se), Sep 
               21 2009]
%Y A046172 Cf. A036353, A046173.
%Y A046172 Contribution from Weisenhorn Paul (paulweisenhorn(AT)online.de), May 
               15 2009: (Start)
%Y A046172 A046172(n)=A072256(n)^2
%Y A046172 (End)
%Y A046172 Sequence in context: A089683 A099372 A036515 this_sequence A123847 A115443 
               A034993
%Y A046172 Adjacent sequences: A046169 A046170 A046171 this_sequence A046173 A046174 
               A046175
%K A046172 nonn
%O A046172 1,2
%A A046172 Eric Weisstein (eric(AT)weisstein.com)

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