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A129556
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Numbers n such that centered pentagonal number A005891(n) = (5n^2+5n+2)/2 is a perfect square.
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5
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0, 2, 21, 95, 816, 3626, 31005, 137711, 1177392, 5229410, 44709909, 198579887, 1697799168, 7540806314, 64471658493, 286352060063, 2448225223584, 10873837476098, 92968086837717, 412919472031679, 3530339074609680, 15680066099727722, 134059916748330141
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OFFSET
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1,2
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COMMENTS
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Corresponding numbers k>0 such that k^2 is a centered pentagonal number are listed in A129557(n) = {1, 4, 34, 151, 1291, 5734, 49024, ...}.
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LINKS
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Table of n, a(n) for n=1..23.
Eric Weisstein, Link to a section of The World of Mathematics, Centered Pentagonal Number.
Index to sequences with linear recurrences with constant coefficients, signature (1,38,-38,-1,1).
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FORMULA
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For n>=5, a(n) = 38*a(n-2) - a(n-4) + 18 [Max Alekseyev, May 08 2009]
G.f.: x^2*(x^3+2*x^2-19*x-2) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)). [Colin Barker, Feb 21 2013]
2*a(n)+1 = A221874(n). [Bruno Berselli, Feb 21 2013]
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MAPLE
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A005891 := proc(n) (5*n^2+5*n+2)/2 ; end: n := 0 : while true do if issqr(A005891(n)) then print(n) ; fi ; n := n+1 ; od : - R. J. Mathar, Jun 06 2007
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MATHEMATICA
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Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[n] ], {n, 1, 40000} ]
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CROSSREFS
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Cf. A005891 (centered pentagonal numbers), A129557 (numbers k>0 such that k^2 is a centered pentagonal number), A221874.
Sequence in context: A034520 A111128 A213827 * A077209 A068045 A188530
Adjacent sequences: A129553 A129554 A129555 * A129557 A129558 A129559
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KEYWORD
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nonn,easy
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AUTHOR
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Alexander Adamchuk, Apr 20 2007
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EXTENSIONS
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More terms from R. J. Mathar, Jun 06 2007
Formula and further terms from Max Alekseyev, May 08 2009
a(22)-a(23) from Colin Barker, Feb 21 2013
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STATUS
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approved
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