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A145182
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Continued cotangent recurrence a(n+1)=a(n)^3+3*a(n) and a(1)=8
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11
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OFFSET
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1,1
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COMMENTS
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General formula for continued cotangent recurrences type:
a(n+1)=a(n)3+3*a(n) and a(1)=k is following:
a(n)=Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))]
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LINKS
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FORMULA
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a(n+1)=a(n)3+3*a(n) and a(1)=8
a(n)=Floor[((8+Sqrt[8^2+4])/2)^(3^(n-1))]
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MATHEMATICA
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a = {}; k = 7; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((8 + Sqrt[68])/2)^(3^(n - 1))], {n, 1, 5}] (*Artur Jasinski*)
RecurrenceTable[{a[1]==8, a[n]==a[n-1]^3+3a[n-1]}, a, {n, 5}] (* Harvey P. Dale, Mar 02 2018 *)
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CROSSREFS
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A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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