1,2

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]

Contribution from Weisenhorn Paul (paulweisenhorn(AT)online.de), May 15 2009: (Start)

a(n)*(3*a(n)-1)/2=m*m is eqivalent to tht pell-equation

(6*a(n)-1)^2-6*(2*m)^2=1 or x(n)^2-6*y(n)^2=1

(End)

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

L. Euler, De solutione problematum diophanteorum per numeros integros, par. 21

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

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]

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]

Contribution from Weisenhorn Paul (paulweisenhorn(AT)online.de), May 15 2009: (Start)

x(n+2)=98*x(n+1)-x(n) with x(1)=5,x(2)=485

y(n+2)=98*y(n+1)-y(n) with y(n)=A046173(n)*2

m(n+2)=98*m(n+1)-m(n) with m(n)=A046173(n)

a(n)=A072256(n)^2

(End)

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]

Cf. A036353, A046173.

Contribution from Weisenhorn Paul (paulweisenhorn(AT)online.de), May 15 2009: (Start)

A046172(n)=A072256(n)^2

(End)

Adjacent sequences: A046169 A046170 A046171 this_sequence A046173 A046174 A046175

Sequence in context: A089683 A099372 A036515 this_sequence A123847 A115443 A034993

nonn

Eric Weisstein (eric(AT)weisstein.com)