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A160695
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a(n) such that 3*a(n)+1 and 7*a(n)+1 are both perfect squares.
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1
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0, 5, 120, 2760, 63365, 1454640, 33393360, 766592645, 17598237480, 403992869400, 9274237758725, 212903475581280, 4887505700610720, 112199727638465285, 2575706229984090840, 59129043561995624040, 1357392295695915262085, 31160893757444055403920
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| The ansatz 3*a(n)+1=A^2, 7*a(n)+1=B^2 is equivalent to the Pell equation x^2-21*y^2=1
(see A077232 for d=21), with x=(21*a(n)+5)/2 and y=A*B/2.
The associated A are in A004253, the B in A030221.
Bisection of A089927 [R. J. Mathar mathar(AT)strw.leidenuniv.nl), Jul 10 2009]
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FORMULA
| a(n)=24*a(n-1)-24*a(n-2)+a(n-3).
a(n)=( A004253(n)^2-1)/3 = (A030221(n)^2-1)/7.
a(n)=((5+w)/2*((23+5*w)/2)^(j-1)+((5+w)/2*((23+5*w)/2)^(j-1)-5)/21; where w=sqrt(21).
G.f.: 5*x^2/((1-x)*(x^2-23*x+1)) [R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 10 2009]
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MAPLE
| j:=0: for n from 0 to 1000000 do a:=sqrt(3*n+1): b:=sqrt(7*n+1):
if (trunc(a)=a) and (trunc(b)=b) then j:=j+1: print(j, n, a, b): end if:
end do:
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CROSSREFS
| Cf. A160682
Sequence in context: A168599 A193328 A002008 * A158564 A192639 A099123
Adjacent sequences: A160692 A160693 A160694 * A160696 A160697 A160698
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KEYWORD
| nonn
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AUTHOR
| Weisenhorn Paul (paulweisenhorn(AT)online.de), May 24 2009
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EXTENSIONS
| Edited and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 10 2009
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