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A128917 Pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891). 2
1, 51, 3151, 195301, 12105501, 750345751, 46509331051, 2882828179401, 178688837791801, 11075825114912251, 686522468286767751, 42553317208664688301, 2637619144468923906901, 163489833639864617539551, 10133732066527137363545251, 628127898291042651922266001 (list; graph; refs; listen; history; text; internal format)



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).


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)).


a(n) = 1/6+(5/48)*sqrt(15)*[31+8*sqrt(15)]^(n-1)-(5/48)*sqrt(15)*[31-8*sqrt(15)]^(n-1)+(5/12)*[31+8 *sqrt(15)]^(n-1)+(5/12)*[31-8*sqrt(15)]^(n-1), with n>=1. - Paolo P. Lava, Jul 15 2008

a(n) = 63*a(n-1)-63*a(n-2)+a(n-3). - Colin Barker, Jan 07 2015


a(1)=51 because 51 is the fifth centered pentagonal number and the sixth pentagonal number.


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;


(PARI) Vec(-x*(x^2-12*x+1)/((x-1)*(x^2-62*x+1)) + O(x^100)) \\ Colin Barker, Jan 07 2015


Cf. A000326, A005891, A128917, A253654.

Sequence in context: A267786 A267733 A238603 * A172742 A172821 A172868

Adjacent sequences:  A128914 A128915 A128916 * A128918 A128919 A128920




Steven Schlicker, Apr 24 2007


Edited by N. J. A. Sloane, Sep 25 2007

More terms from R. J. Mathar, Oct 31 2007



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