|
| |
|
|
A046172
|
|
Indices of pentagonal numbers which are also square.
|
|
6
|
|
|
|
1, 81, 7921, 776161, 76055841, 7452696241, 730288175761, 71560788528321, 7012226987599681, 687126683996240401, 67331402804643959601, 6597790348171111800481, 646516122717964312487521
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
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, Apr 28 2009]
Contribution from Paul Weisenhorn, May 15 2009: a(n)*(3*a(n)-1)/2=m*m is equivalent to the Pell equation (6*a(n)-1)^2-6*(2*m)^2=1 or x(n)^2-6*y(n)^2=1
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = (sqrt(2) + sqrt(3))^4 = 49 + 20 * sqrt(6). - Ant King, Nov 07 2011
|
|
|
LINKS
|
Table of n, a(n) for n=1..13.
L. Euler, De solutione problematum diophanteorum per numeros integros, par. 21
W. Sierpinski, Sur les nombres pentagonaux, Bull. Soc. Roy. Sci. Liege 33 (1964) 513-517.
Eric Weisstein's World of Mathematics, Pentagonal Square Number.
Index to sequences with linear recurrences with constant coefficients, signature (99,-99,1).
|
|
|
FORMULA
|
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 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, 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, Apr 29 2009]
Contribution from Paul Weisenhorn, 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]
From Ant King, Nov 07 2011: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
a(n) = ceiling(1/12*(sqrt(3) + sqrt(2))^(4*n-2)).
(End)
|
|
|
MATHEMATICA
|
LinearRecurrence[{99, -99, 1}, {1, 81, 7921}, 13] (* Ant King, Nov 07 2011 *)
|
|
|
CROSSREFS
|
Cf. A036353, A046173.
Sequence in context: A205056 A186132 A206504 * A123847 A115443 A186527
Adjacent sequences: A046169 A046170 A046171 * A046173 A046174 A046175
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
Eric W. Weisstein
|
|
|
STATUS
|
approved
|
| |
|
|
