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A131750
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Numbers that are both centered triangular and centered square.
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4
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1, 85, 16381, 3177721, 616461385, 119590330861, 23199907725541, 4500662508423985, 873105326726527441, 169377932722437899461, 32858445842826225967885, 6374369115575565399870121
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OFFSET
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1,2
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COMMENTS
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We solve r^2+(r+1)^2=0.5*(3*p^2+3*p+2), which is equivalent to (4*r+2)^2=3*(2*p+1)^2+1.
The Diophantine equation X^2=3*Y^2+1 gives X by A001075 and Y by A013453. The return to r gives the sequence 0,6,90,1260,17556,... which satisfies the formulas a(n+2)=14*a(n+1)-a(n)+6 and a(n+1)=7*a(n)+3+(48*a(n)^2+48*a(n)+9)^0.5 and the return to p the sequence A001921 which satisfies this new relation: a(n+1)=7*a(n)+sqrt(48*a(n)^2+48*a(n)+16). Then we obtain the present sequence.
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LINKS
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FORMULA
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a(n+2) = 195*a(n+1)-195*a(n)+a(n-1).
a(n+1) = 97*a(n) - 54 + 14*sqrt(48*a(n)^2-54*a(n)+15).
G.f.: x*(1-110*x+x^2)/((1-x)*(1-194*x+x^2)).
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MAPLE
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A131750 := proc(n) coeftayl(x*(1-110*x+x^2)/(1-x)/(1-194*x+x^2), x=0, n) ; end: seq(A131750(n), n=1..20) ; # _R. J. Mathar_, Oct 24 2007
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MATHEMATICA
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LinearRecurrence[{195, -195, 1}, {1, 85, 16381}, 20] (* _Harvey P. Dale_, Apr 26 2018 *)
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PROG
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(Magma) [n le 2 select 1 else Floor(97*Self(n-2) - 54 + 14*Sqrt(48*Self(n-2)^2-54*Self(n-2)+15)): n in [2..30]]; // _Vincenzo Librandi_, Aug 26 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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_Richard Choulet_, Sep 20 2007
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EXTENSIONS
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More terms from _R. J. Mathar_, Oct 24 2007
Recurrences corrected by _Robert Israel_, Aug 26 2015
Name corrected by _Daniel Poveda Parrilla_, Sep 19 2016
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STATUS
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approved
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