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).

1, 1, 3, 13, 217, 57073, 3811958497, 16605534578235736513, 309708098978072051970763989442580255617, 106322990835084829467725909226560893968664147958670035553130958199430801942273

Offset: 1

Author

Gilbert Boily (sgbl(AT)escape.ca), Nov 21 2004, Sep 03 2007

Keywords

• nonn
• frac
• nice

Comments

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

This sequence gives the next number in a sequence where the sum and the product of the terms of the sequence are equal.

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.

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

Examples

			2, 2, 4/3, 16/13, 256/217, 65536/57073, 4294967296/3811958497, 18446744073709551616/16605534578235736513, ... = A001146/A100441 (essentially).
		

Links

• N. MacKinnon and N. Lord, Sums equal to products, The Mathematical Gazette, March 1986, 21-22.
• Crux Mathematicorum, Mathematical mayhem pb no. 114, Vol 30, 2004, p. 467-468. [_Robert FERREOL_, Jul 06 2015]

Crossrefs

Cf. A001146, A076628.

Programs

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

Maple

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;
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

Mathematica

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 *)

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 */

SageMath

@CachedFunction
def a(n): # a = A100441
    if (n<3): return 2*n-1
    else: return 2^(2^(n-1)) - 2^(2^(n-2))*a(n-1) + a(n-1)^2
[1]+[a(n) for n in range(1,12)] # G. C. Greubel, Apr 08 2023

Formula

Let F(n) = Product_{i=1..n} f(i) = p/q (say). Then f(n+1) = p/(p-q).

From Robert FERREOL, Jun 12 2015: (Start)

Recurrence: f(1) = f(2) = 2; f(n+1) = f(n)^2/(f(n)^2 - f(n) + 1).

Since f(n) = 2^(2^(n-2))/a(n) for n >= 2, the recurrence for a(n) is:

a(1) = a(2) = 1; a(n+1) = 2^(2^(n-1)) - 2^(2^(n-2))*a(n) + a(n)^2.

(End)

Extensions

Name edited by Michael Somos, Aug 05 2017