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A129557 Numbers k>0 such that k^2 is a centered pentagonal number. 2
1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851 (list; graph; refs; listen; history; internal format)
OFFSET

1,2

COMMENTS

Corresponding numbers n such that centered pentagonal number A005891(n) = (5n^2+5n+2)/2 is a perfect square are listed in A129556(n) = {0, 2, 21, 95, 816, 3626, 31005, ...}.

LINKS

Eric Weisstein, Link to a section of The 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). [From Max Alekseyev (maxale(AT)gmail.com), May 08 2009]

MATHEMATICA

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n, 1, 40000} ]

q=5; s=0; lst={}; Do[s+=n; If[Sqrt[q*s+1]==Floor[Sqrt[q*s+1]], AppendTo[lst, Sqrt[q*s+1]]], {n, 0, 8!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Apr 02 2009]

PROG

(PARI) A129557()={ for(n=1, 1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print(round(sqrt(f))) ; ); ) ; } A129557() ; - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 11 2007

CROSSREFS

Cf. A005891 = Centered pentagonal numbers: (5n^2+5n+2)/2. Cf. A129556 = numbers n such that centered pentagonal number A005891(n) = (5n^2+5n+2)/2 is a perfect square.

Sequence in context: A053902 A054464 A002101 * A196908 A197075 A085695

Adjacent sequences:  A129554 A129555 A129556 * A129558 A129559 A129560

KEYWORD

nonn

AUTHOR

Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 20 2007

EXTENSIONS

More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 11 2007

Formula and further terms from Max Alekseyev (maxale(AT)gmail.com), May 08 2009

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Last modified February 14 23:16 EST 2012. Contains 205687 sequences.