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A129557
Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).
4
1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851, 24792361368566254
OFFSET
1,2
COMMENTS
Corresponding numbers m such that centered pentagonal number A005891(m) = (5*m^2 + 5*m + 2)/2 is a perfect square are listed in A129556 = {0, 2, 21, 95, 816, 3626, 31005, ...}.
Also positive integers x in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of y being A254332. - Colin Barker, Jan 28 2015
LINKS
Eric Weisstein's World of Mathematics, Centered Pentagonal Number.
FORMULA
a(n) = sqrt( (5*A129556(n)^2 + 5*A129556(n) + 2)/2 ).
For n >= 5, a(n) = 38*a(n-2) - a(n-4). - Max Alekseyev, May 08 2009
G.f.: x*(1-x)*(1 + 5*x + x^2)/((1 + 6*x - x^2)*(1 - 6*x - x^2)). - Colin Barker, Apr 11 2012
From Andrea Pinos, Oct 07 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7 + 2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13 + 4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).
a(n) = (A005667(n) - (-1)^n*A005667(n-1))/4. (End)
MATHEMATICA
Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n, 1, 40000} ]
CoefficientList[Series[(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 11 2012 *)
PROG
(PARI) A129557()={ for(n=1, 1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print1(sqrtint(f), ", ") ; ); ) ; } \\ R. J. Mathar, Oct 11 2007
(PARI) Vec(x*(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)) + O(x^100)) \\ Colin Barker, Jan 28 2015
CROSSREFS
Cf. A005891 (centered pentagonal numbers).
Cf. A129556 (k such that A005891(k) is a perfect square).
Sequence in context: A002101 A241231 A297715 * A231518 A196908 A197075
KEYWORD
nonn,easy
AUTHOR
Alexander Adamchuk, Apr 20 2007
EXTENSIONS
More terms from R. J. Mathar, Oct 11 2007
More terms from Max Alekseyev, May 08 2009
STATUS
approved