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A128917 Pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891).

%I A128917 #38 Sep 20 2024 05:43:25
%S A128917 1,51,3151,195301,12105501,750345751,46509331051,2882828179401,
%T A128917 178688837791801,11075825114912251,686522468286767751,
%U A128917 42553317208664688301,2637619144468923906901,163489833639864617539551,10133732066527137363545251,628127898291042651922266001
%N A128917 Pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891).
%H A128917 Colin Barker, <a href="/A128917/b128917.txt">Table of n, a(n) for n = 1..558</a>
%H A128917 S. C. Schlicker, <a href="http://www.jstor.org/stable/10.4169/math.mag.84.5.339">Numbers Simultaneously Polygonal and Centered Polygonal</a>, Mathematics Magazine,  Vol. 84, No. 5, December 2011
%H A128917 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (63,-63,1).
%F A128917 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))).
%F A128917 From _Richard Choulet_, Sep 19 2007: (Start)
%F A128917 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.
%F A128917 Four other sequences are obtained at the same time:
%F A128917 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).
%F A128917 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).
%F A128917 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).
%F A128917 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).
%F A128917 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).
%F A128917 G.f.: z*(1-12*z+z^2)/((1-z)*(1-62*z+z^2)). (End)
%F A128917 a(n) = 63*a(n-1)-63*a(n-2)+a(n-3). - _Colin Barker_, Jan 07 2015
%e A128917 a(1)=51 because 51 is the fifth centered pentagonal number and the sixth pentagonal number.
%p A128917 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;
%t A128917 LinearRecurrence[{63,-63,1},{1,51,3151},20] (* _Harvey P. Dale_, Nov 26 2022 *)
%o A128917 (PARI) Vec(-x*(x^2-12*x+1)/((x-1)*(x^2-62*x+1)) + O(x^100)) \\ _Colin Barker_, Jan 07 2015
%Y A128917 Cf. A000326, A005891, A128917, A253654.
%K A128917 easy,nonn
%O A128917 1,2
%A A128917 _Steven Schlicker_, Apr 24 2007
%E A128917 Edited by _N. J. A. Sloane_, Sep 25 2007
%E A128917 More terms from _R. J. Mathar_, Oct 31 2007