Ok, dear BillIt is correct, do you understand me? a [n+1]=2^(2*n*m+5*m)*(1-(1-(a [n])^(1/m)/2^(2*n+4))^(1/2))^m a [0]=2^(3*m) ; n----> infinity a [n]----> Pi^m Envoyé depuis Yahoo Mail pour Android Le ven., janv. 24, 2020 à 4:49 AM, Bill Gosper<billgosper@gmail.com> a écrit : Hi François, collecting snippets from disorganized Mathematica windows, do you agreeOut[145]= a[1 + n] == 2^(3 + 2 n) (1 - Sqrt[1 - 2^(-2 - 2 n) a[n]]) Changing variables In[182]:= a[n_] :> 2^(2 n + 1) b[n] In[147]:= %% /. % Out[147]= 2^(1 + 2 (1 + n)) b[1 + n] == 2^(3 + 2 n) (1 - Sqrt[1 - b[n]/2]) In[148]:= FullSimplify@% Out[148]= 4^n (-2 + Sqrt[4 - 2 b[n]] + 2 b[1 + n]) == 0 In[149]:= %/4^n Out[149]= -2 + Sqrt[4 - 2 b[n]] + 2 b[1 + n] == 0 Out[152]= 2 - 2 b[1 + n] == Sqrt[4 - 2 b[n]] In[153]:= %^2 Out[153]= (2 - 2 b[1 + n])^2 == 4 - 2 b[n] Changing variables again, In[64]:= -2 + Sqrt[4 - 2 b[n]] + 2 b[1 + n] == 0 /. b[x_] :> (4 - c[x])/2 Out[64]= 2 + Sqrt[c[n]] - c[1 + n] == 0
From another window, In[169]:= %[[1, 1 ;; 2]] - %
Out[169]= c[1 + n] == 2 + Sqrt[c[n]] Please tell me if this makes no sense.—Bill On Wed, Jan 22, 2020 at 4:57 PM françois mendzina essomba2 <m_essob@yahoo.fr> wrote: Hi, Dear Bill, I didn't quite understand your reccurence.It may be obvious, but not for me. here is another one I propose for 1 / Pi ^ 2. It is simpler to obtain and the method is disconcertingly simple. a[n+1]=2^(-2*n-5)/(1-sqrt(1-2^(-2*n-4)/a[n])); a[0]=1/8 n----- infinity a[n] ------- 1/Pi^2 Best regards