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A125651
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Numbers k such that A125650(k) is a perfect square.
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4
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1, 3, 24, 147, 864, 5043, 29400, 171363, 998784, 5821347, 33929304, 197754483, 1152597600, 6717831123, 39154389144, 228208503747, 1330096633344, 7752371296323, 45184131144600, 263352415571283, 1534930362283104
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Corresponding numbers m such that m^2 = A125650(a(n)) are listed in A125652.
3 divides a(n) for n>1. For n>1 a(n) = 3*A001108(n-1), where A001108(k) = {0, 1, 8, 49, 288, 1681, ...}, A001108(k)-th triangular number is a square. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jan 19 2007
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FORMULA
| For n>1, a(n+2) = 6*a(n+1) - a(n) + 6.
For n>1, a(n) = ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))*3/4 - 3/2.
a(2k) = 3*A002315(n)^2; a(2k+1) = 6*A001542(n)^2.
a(n) = 3*A001108(n-1) for n>1. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jan 19 2007
For n>1, a(2)=3, a(3)=24, a(4)=147, a(n)=7*a(n-1)-7*a(n-2)+a(n-3) [From Harvey P. Dale, May 15 2011]
G.f.: (-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x)x)) [From Harvey P. Dale, May 15 2011]
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EXAMPLE
| a(2)=3 because A125650(3)=9=3^2; a(3)=24 because A125650(24)=81=9^2.
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MATHEMATICA
| Join[{1}, LinearRecurrence[{7, -7, 1}, {3, 24, 147}, 35]] (* or *) CoefficientList[Series[(-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x) x)), {x, 0, 35}], x] (* From Harvey P. Dale, May 15 2011 *)
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CROSSREFS
| Cf. A125650, A125652.
Cf. A001108.
Sequence in context: A056350 A056344 A201231 * A043017 A003443 A119581
Adjacent sequences: A125648 A125649 A125650 * A125652 A125653 A125654
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KEYWORD
| nonn
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AUTHOR
| Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 29 2006
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EXTENSIONS
| Edited by Max Alekseyev (maxale(AT)gmail.com), Jan 11 2007
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