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A070234 Let u(k), v(k), w(k) satisfy the recursions u(1) = v(1) = w(1) = 1, u(k+1) = u(k) + v(k) + w(k), v(k+1) = u(k)*v(k) + v(k)*w(k) + w(k)*u(k), and w(k+1) = u(k)*v(k)*w(k); then a(n) = v(n). 8

%I #34 May 11 2020 19:15:17

%S 1,3,15,303,325023,2896797882687,10689080432835089614170716799,

%T 1051462916692114532403603811392745230616355871287492722818364671

%N Let u(k), v(k), w(k) satisfy the recursions u(1) = v(1) = w(1) = 1, u(k+1) = u(k) + v(k) + w(k), v(k+1) = u(k)*v(k) + v(k)*w(k) + w(k)*u(k), and w(k+1) = u(k)*v(k)*w(k); then a(n) = v(n).

%H Petros Hadjicostas, <a href="/A070234/b070234.txt">Table of n, a(n) for n = 1..11</a>

%F Let C be the positive root of x^3 + x^2 - 2*x - 1 = 0; that is, C = 1.246979603717... = A255249. Then Lim_{n -> infinity} u(n)^(C+1)/w(n) = Lim_{n -> infinity} v(n)^C/w(n) = Lim_{n -> infinity} u(n)^B/v(n) = 1 with B = C + 1 - 1/(1 + C) = 1.8019377... = A160389. [corrected by _Vaclav Kotesovec_, May 11 2020]

%F a(n) ~ gv^((C + 1)^n), where C is defined above and gv = 1.250231610564761084... The relation between constants gu (see A070231), gv and gw (see A070233) is gu^(1 + C) = gv^C = gw. - _Vaclav Kotesovec_, May 11 2020

%t u[1] = 1; a[1] = 1; w[1] = 1; u[k_] := u[k] = u[k - 1] + a[k - 1] + w[k - 1]; a[k_] := a[k] = u[k - 1]*a[k - 1] + a[k - 1]*w[k - 1] + w[k - 1]*u[k - 1]; w[k_] := w[k] = u[k - 1]*a[k - 1]*w[k - 1]; Table[a[n], {n, 1, 9}] (* _Vaclav Kotesovec_, May 11 2020 *)

%o (PARI)_lista(nn) = {my(u = vector(nn)); my(v = vector(nn)); my(w = vector(nn)); u[1] = 1; v[1] = 1; w[1] = 1; for (n=2, nn, u[n] = u[n-1] + v[n-1] + w[n-1]; v[n] = u[n-1]*v[n-1] + v[n-1]*w[n-1] + w[n-1]*u[n-1]; w[n] = u[n-1]*v[n-1]*w[n-1]; ); v; } \\ _Petros Hadjicostas_, May 11 2020

%Y Cf. A003686, A064183, A064526, A064847, A070231 (= u), A070233 (= w), A094303, A160389, A255249.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, May 08 2002

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