OFFSET
1,2
LINKS
Colin Barker, Table of n, a(n) for n = 1..558
S. C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).
FORMULA
Define x(n) + y(n)*sqrt(15) = (5+sqrt(15))*(4+sqrt(15))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+5*(s(n)^2-s(n))).
From Richard Choulet, Sep 19 2007: (Start)
We must solve 3*p^2-p=5*r^2+5*r+2, which gives X^2=15*Y^2+10 where X=6*p-1 and Y=2*r+1.
Four other sequences are obtained at the same time:
X is given by 5,35,275,2165,... with the recurrence a(n+2)=8*a(n+1)-a(n) and also a(n+1)=4*a(n)+(15*a(n)^2-150)^(1/2) (numbers such that 15*X^2-150 is a square).
Y is given by 1,9,71,559,... with the recurrence a(n+2)=8*a(n+1)-a(n) and also a(n+1)=4*a(n)+(15*a(n)^2+10)^(1/2) (numbers such that 15*Y^2+10 is a square).
p is given by 1,6,46,361,... with the recurrence a(n+2)=8*a(n+1)-a(n)-1 and also a(n+1)=4*a(n)-0.5+0.5*(60*a(n)^2-20*a(n)-15)^(1/2) (numbers such that 15*(6*p-1)^2-150 is a square).
r is given by 0,4,35,279,... with the recurrence a(n+2)=8*a(n+1)-a(n)+3 and also a(n+1)=4*a(n)+1.5+0.5*(60*a(n)^2+60*a(n)+25)^(1/2) (numbers such that 15*(2*r+1)^2+10 is a square).
a(n+2) = 62*a(n+1)-a(n)-10, a(n+1)=31*a(n)-5+(960*a(n)^2-320*a(n)-45)^(1/2).
G.f.: z*(1-12*z+z^2)/((1-z)*(1-62*z+z^2)). (End)
a(n) = 63*a(n-1)-63*a(n-2)+a(n-3). - Colin Barker, Jan 07 2015
EXAMPLE
a(1)=51 because 51 is the fifth centered pentagonal number and the sixth pentagonal number.
MAPLE
CP := n -> 1+1/2*5*(n^2-n): N:=10: u:=4: v:=1: x:=5: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+15*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp), CP(s)]: end do: k_pcp;
MATHEMATICA
LinearRecurrence[{63, -63, 1}, {1, 51, 3151}, 20] (* Harvey P. Dale, Nov 26 2022 *)
PROG
(PARI) Vec(-x*(x^2-12*x+1)/((x-1)*(x^2-62*x+1)) + O(x^100)) \\ Colin Barker, Jan 07 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Steven Schlicker, Apr 24 2007
EXTENSIONS
Edited by N. J. A. Sloane, Sep 25 2007
More terms from R. J. Mathar, Oct 31 2007
STATUS
approved