A285993 Largest odd abundant number (A005231) equal to the product of n consecutive primes.
15015, 255255, 4849845, 111546435, 33426748355, 1236789689135, 50708377254535, 2180460221945005, 102481630431415235, 5431526412865007455, 320460058359035439845, 19548063559901161830545, 1309720258513377842646515, 1357656019974967471687377449, 107254825578022430263302818471
Offset: 5
Keywords
Examples
For n < 5, there is no odd abundant number equal to the product of n distinct primes. For 5 <= n <= 8, the largest odd abundant number equal to the product of n consecutive primes is 3*...*prime(n+1). For 9 <= n <= 17, the largest odd abundant number equal to the product of n consecutive primes is 5*...*prime(n+2). For 18 <= n <= 30, the largest odd abundant number equal to the product of n consecutive primes is 7*...*prime(n+3). For 31 <= n <= 45, the largest odd abundant number equal to the product of n consecutive primes is 11*...*prime(n+4). For 46 <= n <= 66, the largest odd abundant number equal to the product of n consecutive primes is 13*...*prime(n+5).
Links
- Amiram Eldar, Table of n, a(n) for n = 5..341
Crossrefs
Programs
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PARI
a(r,f=vector(r,i,prime(i+1)),o)={ while(sigma(factorback(f),-1)>2, o=f; f=concat(f[^1],nextprime(f[r]+1)));factorback(o)} \\ Intentionally throws an error when n < 5.
Formula
a(n) >= a(n-1)*p where p = A117366(a(n-1)) = A151800(A006530(a(n-1))) = nextprime(gpf(a(n-1))), an odd abundant number equal to the product of n consecutive primes. We have strict inequality for n = 9, 18, 31, 46, 67, ..., in which case a(n) = a(n-1)*p*p'/q, where p' = nextprime(p), q = least prime factor of a(n-1). This is the case if a(n) is in A007741.
Comments