Search: id:A046173
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%I A046173
%S A046173 1,99,9701,950599,93149001,9127651499,894416697901,87643708742799,
%T A046173 8588189040096401,841554882220704499,82463790268588944501,
%U A046173 8080609891439495856599,791817305570802005002201
%N A046173 Indices of square numbers which are also pentagonal.
%H A046173 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%H A046173 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 A046173 L. Euler, <a href="http://math.dartmouth.edu/~euler/pages/E029.html">
               De solutione problematum diophanteorum per numeros integros</a>, 
               par. 21
%F A046173 a(n) = 98*a(n-1) - a(n-2); g.f.: (1+x)/(1-98*x+x^2) - Warut Roonguthai 
               (warut822(AT)yahoo.com) Jan 05 2001
%F A046173 a(1-n)=-a(n). - Michael Somos Sep 05 2006
%F A046173 Define f[x,s] = s x + Sqrt[(s^2-1)x^2+1]; f[0,s]=0. a(n) = f[f[a(n-1),
               5],5]. - Marcos Carreira, Dec 27 2006
%F A046173 a(n)=((12+5*sqrt(6))/24)*(5+2*sqrt(6))^(2*n)+((12-5*sqrt(6))/24)*(5-2*sqrt(6))^(2*n) 
               for n>=0 [From Richard Choulet (richardchoulet(AT)yahoo.fr), Apr 
               29 2009]
%F A046173 a(n+1)=49*a(n)+10*sqrt(24*a(n)^2+1) for n>=0 with a(0)=1 [From Richard 
               Choulet (richardchoulet(AT)yahoo.fr), Apr 29 2009]
%o A046173 (PARI) {a(n)=subst(poltchebi(n)-poltchebi(n-1), 'x, 49)/48} /* Michael 
               Somos Sep 05 2006 */
%Y A046173 Cf. A036353, A046172.
%Y A046173 Sequence in context: A069363 A163051 A093233 this_sequence A098609 A132607 
               A093211
%Y A046173 Adjacent sequences: A046170 A046171 A046172 this_sequence A046174 A046175 
               A046176
%K A046173 nonn
%O A046173 1,2
%A A046173 Eric Weisstein (eric(AT)weisstein.com)

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