cp's OEIS Frontend

A100441 a(n) is the denominator of f(n) where f(1) = 2 and f(n+1) is the solution of x + Sum_{i=1..n} f(i) = x * Product_{i=1..n} f(i).

%I A100441 #57 Apr 28 2024 11:28:21
%S A100441 1,1,3,13,217,57073,3811958497,16605534578235736513,
%T A100441 309708098978072051970763989442580255617,
%U A100441 106322990835084829467725909226560893968664147958670035553130958199430801942273
%N A100441 a(n) is the denominator of f(n) where f(1) = 2 and f(n+1) is the solution of x + Sum_{i=1..n} f(i) = x * Product_{i=1..n} f(i).
%C A100441 Let E(0) = x + 1, let E(n+1) = 1 - E(n) + E(n)^2. Let e(n) = discrim(E(n),x) and let f(n) = e(n+1)/e(n)^2. Then f(1,2,3,...) = -3,13,217,57073,381195849,... which looks like this sequence (I do not have a proof yet). - Daniel R. L. Brown (dbrown(AT)certicom.com), Nov 18 2005
%C A100441 This sequence gives the next number in a sequence where the sum and the product of the terms of the sequence are equal.
%C A100441 It happens that the sum or product of the terms of this sequence match A001146 for the numerator of the sum or product and A076628 for the denominator of the sum or product of the sequence.
%C A100441 Let g(x) = x^2 - x + 1 be the map producing Sylvester's sequence A000058. Then for n >= 0, g^n(1/2) = 1/f(n+2), where g^n is the n-th iterate of g, so a(n+2) is the numerator of g^n(1/2). - _Curtis Bechtel_, Apr 05 2024
%H A100441 N. MacKinnon and N. Lord, <a href="http://www.jstor.org/stable/3615819">Sums equal to products</a>, The Mathematical Gazette, March 1986, 21-22.
%H A100441 Crux Mathematicorum, <a href="https://cms.math.ca/crux/v30/n8/public_page463-470.pdf">Mathematical mayhem pb no. 114</a>, Vol 30, 2004, p. 467-468. [_Robert FERREOL_, Jul 06 2015]
%F A100441 Let F(n) = Product_{i=1..n} f(i) = p/q (say). Then f(n+1) = p/(p-q).
%F A100441 From _Robert FERREOL_, Jun 12 2015: (Start)
%F A100441 Recurrence: f(1) = f(2) = 2; f(n+1) = f(n)^2/(f(n)^2 - f(n) + 1).
%F A100441 Since f(n) = 2^(2^(n-2))/a(n) for n >= 2, the recurrence for a(n) is:
%F A100441 a(1) = a(2) = 1; a(n+1) = 2^(2^(n-1)) - 2^(2^(n-2))*a(n) + a(n)^2.
%F A100441 (End)
%e A100441 2, 2, 4/3, 16/13, 256/217, 65536/57073, 4294967296/3811958497, 18446744073709551616/16605534578235736513, ... = A001146/A100441 (essentially).
%p A100441 f:=proc(n) option remember; local i,k,k1,k2; if n = 1 then return(2); fi; k:=mul(f(i),i=1..n-1); k1:=numer(k); k2:=denom(k); k1/(k1-k2); end;
%p A100441 f:=n-> if n=1 or n=2 then 2 else f(n-1)^2/(f(n-1)^2-f(n-1)+1) fi; # _Robert FERREOL_, Jun 12 2015
%t A100441 f[n_] := f[n] = (frac = Product[f[i], {i, 1, n-1}]; p = Numerator[frac]; q = Denominator[frac]; p/(p-q)); f[1] = 2; (* or, after _Robert FERREOL_ *) f[n_] := f[n] = If[n == 1 || n == 2, 2, f[n-1]^2/(f[n-1]^2-f[n-1]+1)]; Table[f[n], {n, 1, 10}] // Denominator (* _Jean-François Alcover_, Sep 19 2012, updated Jun 15 2015 *)
%o A100441 (Magma) I:=[1,3]; [1] cat  [n le 2 select I[n] else 2^(2^(n-1))-2^(2^(n-2))*Self(n-1)+Self(n-1)^2: n in [1..10]]; // _Vincenzo Librandi_, Jun 13 2015
%o A100441 (PARI) {a(n) = my(s, t); if( n<3, n>0, t = a(n-1); s = 2^(2^(n-3)); s*s -s*t +t*t)}; /* _Michael Somos_, Aug 05 2017 */
%o A100441 (SageMath)
%o A100441 @CachedFunction
%o A100441 def a(n): # a = A100441
%o A100441     if (n<3): return 2*n-1
%o A100441     else: return 2^(2^(n-1)) - 2^(2^(n-2))*a(n-1) + a(n-1)^2
%o A100441 [1]+[a(n) for n in range(1,12)] # _G. C. Greubel_, Apr 08 2023
%Y A100441 Cf. A001146, A076628.
%K A100441 nonn,frac,nice
%O A100441 1,3
%A A100441 Gilbert Boily (sgbl(AT)escape.ca), Nov 21 2004, Sep 03 2007
%E A100441 Name edited by _Michael Somos_, Aug 05 2017