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A129557
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Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).
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4
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1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851, 24792361368566254
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
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).
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A005891 (centered pentagonal numbers).
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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