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a(0) = 4; a(n) = a(n-1)^2 - 1.
3

%I #34 Jan 09 2025 09:51:11

%S 4,15,224,50175,2517530624,6337960442777829375,

%T 40169742574216538983356186036612890624,

%U 1613608218478824775913354216413699241125577233045500390244103887844987109375

%N a(0) = 4; a(n) = a(n-1)^2 - 1.

%C This is the next analog of A003096 with different initial value a(0), as starting with a(0) = 2 is A003096 and a(0) = 3 is A003096 with first term omitted. It alternates between even and odd values, specifically between 4 mod 10 and 5 mod 10 and is always composite (by difference of squares factorization).

%C a(n+2) is divisible by a(n)^2. A007814(a(2 n)) = A153893(n). - _Robert Israel_, Jul 20 2015

%H Robert Israel, <a href="/A139244/b139244.txt">Table of n, a(n) for n = 0..10</a>

%H A. V. Aho and N. J. A. Sloane, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.

%H A. V. Aho and N. J. A. Sloane, <a href="/A000058/a000058.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)

%H <a href="/index/Aa#AHSL">Index entries for sequences of form a(n+1)=a(n)^2 + ...</a>

%F a(n-1) = ceiling(c^(2^n)) where c is a constant between 1 and 2.

%F More specifically, c=1.9668917617901763653335057202... (sequence A260315). - _Chayim Lowen_, Jul 17 2015

%p A[0]:= 4:

%p for n from 1 to 10 do A[n]:= A[n-1]^2-1 od:

%p seq(A[i],i=0..10); # _Robert Israel_, Jul 20 2015

%t a=4; lst={a}; Do[b=a^2-1; AppendTo[lst,b]; a=b, {n,10}]; lst (* _Vladimir Joseph Stephan Orlovsky_, Dec 28 2010 *)

%o (PARI) a(n)=if(n,a(n-1)^2-1,4) \\ _Charles R Greathouse IV_, Jul 23 2015

%Y Cf. A003096, A007814, A153893, A260315.

%K easy,nonn,changed

%O 0,1

%A _Jonathan Vos Post_, Jun 06 2008