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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)
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) = 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

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;

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

Cf. A000326, A005891, A128917, A253654.

Sequence in context: A267786 A267733 A238603 * A172742 A172821 A172868

Adjacent sequences:  A128914 A128915 A128916 * A128918 A128919 A128920

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

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Last modified October 6 05:47 EDT 2022. Contains 357261 sequences. (Running on oeis4.)