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A258169
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a(n) = a(n-1)^4/a(n-2) with a(0)=1, a(1)=2.
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1
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OFFSET
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0,2
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COMMENTS
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The next term has 235 digits.
In general, the recurrence a(n) = a(n-1)^k / a(n-2) with a(0) = 1, a(1) = m, k > 2, has a solution a(n) = m^(((k+sqrt(k^2-4))^n - (k-sqrt(k^2-4))^n) / (sqrt(k^2-4) * 2^n)).
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LINKS
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FORMULA
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a(n) = 2^(((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3))).
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MATHEMATICA
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RecurrenceTable[{a[n]==a[n-1]^4/a[n-2], a[0]==1, a[1]==2}, a, {n, 0, 6}]
nxt[{a_, b_}]:={b, b^4/a}; NestList[nxt, {1, 2}, 5][[All, 1]] (* Harvey P. Dale, Sep 04 2022 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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