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A075442 Slowest-growing sequence of primes whose reciprocals sum to 1. 7

%I #23 Sep 06 2023 21:08:19

%S 2,3,7,43,1811,654149,27082315109,153694141992520880899,

%T 337110658273917297268061074384231117039,

%U 8424197597064114319193772925959967322398440121059128471513803869133407474043

%N Slowest-growing sequence of primes whose reciprocals sum to 1.

%C This sequence was mentioned by K. S. Brown. The sequence is generated by a greedy algorithm given by the Mathematica program. The sum converges quadratically.

%C It is easily shown that this sequence is infinite. For suppose there was a finite representation of unity as a sum of unit fractions with distinct prime denominators. Multiply the equation by the product of all denominators to obtain this product of prime numbers on one side of the equation and a sum of products consisting of this product with always exactly one of the prime numbers removed on the other side. Then each of the prime numbers divides one side of the equation but not the other, since it divides all the products added except exactly one. Contradiction. - Peter C. Heinig (algorithms(AT)gmx.de), Sep 22 2006

%C {a(n)} = 2, 3, 7, ..., so A225671(1) = 3. - _Jonathan Sondow_, May 13 2013

%D R. K. Guy, Unsolved Problems in Number Theory, D11.

%H Robert G. Wilson v, <a href="/A075442/b075442.txt"> Table of n, a(n) for n = 1..14 </a>

%H K. S. Brown, <a href="http://www.mathpages.com/home/kmath454.htm">Odd, Greedy and Stubborn (Unit Fractions)</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>

%t x=1; lst={}; Do[n=Ceiling[1/x]; If[PrimeQ[n], n++ ]; While[ !PrimeQ[n], n++ ]; x=x-1/n; AppendTo[lst, n], {10}]; lst

%t a[n_] := a[n] = Block[{sm = Sum[1/(a[i]), {i, n - 1}]}, NextPrime[ Max[ a[n - 1], 1/(1 - sm)]]]; a[0] = 1; Array[a, 10] (* _Robert G. Wilson v_, Oct 28 2010 *)

%o (PARI) a(n)=if(n<3, return(prime(n))); my(x=1.); for(i=1,n-1,x-=1/a(i)); nextprime(1/x) \\ _Charles R Greathouse IV_, Apr 29 2015

%o (PARI) a_vector(N=10)= my(r=1, v=vector(N)); for(i=1, N, v[i]= nextprime(1+1/r); r-= 1/v[i]); v; \\ Ruud H.G. van Tol, Jul 29 2023

%Y Cf. A000058, A225669, A225671.

%K nice,nonn

%O 1,1

%A _T. D. Noe_, Sep 16 2002

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Last modified April 23 11:35 EDT 2024. Contains 371912 sequences. (Running on oeis4.)