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).
1, 3, 15, 303, 325023, 2896797882687, 10689080432835089614170716799, 1051462916692114532403603811392745230616355871287492722818364671
Offset: 1
Links
- Petros Hadjicostas, Table of n, a(n) for n = 1..11
Crossrefs
Programs
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Mathematica
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 *) -
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
Formula
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]
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