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

1,2

COMMENT

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]

PROGRAM

(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 this_sequence A085695 A049293 A116430

Adjacent sequences: A129554 A129555 A129556 this_sequence 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 November 30 13:13 EST 2009. Contains 167758 sequences.


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