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%I #34 Apr 17 2023 23:53:36
%S 2,6,210,9260790,794226015149981778210,
%T 500993769952171221242360491304282624082148055373236400176882790
%N a(n) = a(n-1)*(a(n-1)^2 - 1) with a(0) = 2.
%C a(6) has 189 digits so it is not displayed here.
%C It appears that the sequence might be squarefree, but this is unproved so far.
%C If n is the least index such that p^2 | a(n+1) for some prime p, then one must have p^2 | a(n)+1 or p^2 | a(n)-1, and p >= 5.
%C If p is prime and p^2 | a(n) for some n, then p > 10^6. - _Robert Israel_, Apr 16 2023
%C The constant c in the formula can be obtained as exp(log(a(n))/3^n) with more than 3^n/2 significant digits. - _M. F. Hasler_, Apr 13 2023
%H Winston de Greef, <a href="/A361086/b361086.txt">Table of n, a(n) for n = 0..7</a>
%H Dan Asimov, <a href="https://mailman.xmission.com/hyperkitty/list/math-fun@mailman.xmission.com/thread/5TNAYAUATD2D5IHWQQIIBPF3QE2O3ILR/">Interesting sequence on MathOverflow</a>, math-fun mailing list (access restricted to subscribers), Mar 28 2023.
%H Fredrick M. Nelson, <a href="https://mathoverflow.net/questions/443446/">Does a(0)=6, a(n+1)=a(n)^3-a(n), define a square-free sequence?</a>, MathOverflow, Mar 24 2023.
%F a(n) ~ c^(3^n), where c = 1.8114401993215336517307679103877676944735007990106818615618948774060863424... - _Vaclav Kotesovec_, Apr 13 2023
%e After a(0) = 2, we have:
%e a(1) = a(0) * 3 = 6 = A002110(2),
%e a(2) = a(1) * 5 * 7 = 210 = A002110(4),
%e a(3) = a(2) * 11 * 19 * 211 = A002110(5) * 19 * 211,
%e a(4) = a(3) * 23 * 137 * 2939 * 101 * 91691, (*)
%e a(5) = a(4) * 2381 * 4547 * 73360073674487 * 4111 * 9463 * 20415865295227, (*)
%e a(6) = a(5) * (a(5)-1) * (a(5)+1) where a(5)-1 = 8297 * 2809343 * 1697219017859557 * 12663932268383565339458540396093810087 and
%e a(5)+1 = 6529 * 11220511276757421196620288973 * 6838691459094922770231096737923,
%e etc.
%e (*) We list first the prime factors of a(n-1)-1, then those of a(n-1)+1.
%p a:= proc(n) option remember; procname(n-1)*(procname(n-1)-1)*(procname(n-1)+1) end proc:
%p a(0):= 2:
%p seq(a(i),i=0..7); # _Robert Israel_, Apr 16 2023
%t RecurrenceTable[{a[n] == a[n-1]*(a[n-1]^2 - 1), a[0] == 2}, a, {n, 0, 6}] (* _Vaclav Kotesovec_, Apr 13 2023 *)
%o (PARI) A361086_first(N)=vector(N,i,N=if(i>1,N*(N^2-1),2))
%Y Cf. A002110, A007018, A228649, A361085.
%K nonn,less
%O 0,1
%A _M. F. Hasler_, Mar 28 2023
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