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a(n) = a(n-1)^4 + a(n-2)^4 for n >= 2 with a(0) = 0, a(1) = 1.
10

%I #18 Sep 15 2023 12:00:27

%S 0,1,1,2,17,83537,48698490414981559682,

%T 5624216052381164150697569400035392464306474190030694298257552124199835791859537

%N a(n) = a(n-1)^4 + a(n-2)^4 for n >= 2 with a(0) = 0, a(1) = 1.

%C A quartic Fibonacci sequence.

%C This is the quartic (or biquadratic) analog of the Fibonacci sequence similarly to A000283 being the quadratic analog of the Fibonacci sequence. The primes in this sequence begin a(3), a(4), a(5).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/QuarticEquation.html">Quartic Equation</a>.

%F a(n) ~ c^(4^n), where c = 1.0111288972169538887655499395580320278253918666919181401824606983217263409... . - _Vaclav Kotesovec_, Dec 18 2014

%e a(3) = 1^4 + 1^4 = 2.

%e a(4) = 1^4 + 2^4 = 17.

%e a(5) = 2^4 + 17^4 = 83537.

%e a(6) = 17^4 + 83537^4 = 48698490414981559682.

%t RecurrenceTable[{a[1] ==1, a[2] == 1, a[n] == a[n-1]^4 + a[n-2]^4}, a, {n, 1, 8}] (* _Vaclav Kotesovec_, Dec 18 2014 *)

%Y Cf. A000045, A000283.

%K easy,nonn

%O 0,4

%A _Jonathan Vos Post_, Jan 02 2006

%E Name edited by _Petros Hadjicostas_, Nov 03 2019

%E a(0)=0 prepended by _Alois P. Heinz_, Sep 15 2023