A285993 -id:A285993 - cp's OEIS frontend

A286042 Largest prime factor of A285993(n), the largest odd abundant number (A005231) equal to the product of n consecutive primes.

13, 17, 19, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353

Offset: 5

M. F. Hasler, May 01 2017

nonn

The smallest term is a(5), there is no odd abundant number (A005231) equal to the product of less than 5 consecutive primes.

The corresponding abundant numbers are A285993(n) = prime(k-n+1)*...*prime(k), with prime(k) = a(n).

			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*...*a(n) with a(n) = prime(n+1).
For 9 <= n <= 17, the largest odd abundant number equal to the product of n consecutive primes is 5*...*a(n) with a(n) = prime(n+2).
For 18 <= n <= 30, the largest odd abundant number equal to the product of n consecutive primes is 7*...*a(n) with a(n) = prime(n+3).
For 31 <= n <= 45, the largest odd abundant number equal to the product of n consecutive primes is 11*...*a(n) with a(n) = prime(n+4).
For 46 <= n <= 66, the largest odd abundant number equal to the product of n consecutive primes is 13*...*a(n) with a(n) = prime(n+5).

Amiram Eldar, Table of n, a(n) for n = 5..10000

Cf. A285993, A005231, A006038, A007707, A007708, A007741.

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)));o[#o]} \\ Intentionally throws an error when n < 5.

a(n) = A006530(A285993(n)) >= A151800(a(n-1)) = nextprime(a(n-1)), with strict inequality for n = 9, 18, 31, 46, 67, ..., in which case a(n) = nextprime(nextprime(a(n-1))). This is the case if A285993(n) is in A007741.

a(66) corrected by Amiram Eldar, Sep 24 2019

A007741 a(n) = prime(n)...prime(m), the least product of consecutive primes which is abundant.

Original entry on oeis.org

30, 15015, 33426748355, 1357656019974967471687377449, 7105630242567996762185122555313528897845637444413640621

Offset: 1

Walter Nissen

nonn

Essentially, i.e., except for a(1), identical to A007702. All terms are primitive abundant numbers (A091191) and thus, except for the first term, odd primitive abundant (A006038). The next term is too large to be displayed here, see A007707 (and formula) for many more terms, using a more compact encoding. - M. F. Hasler, Apr 30 2017

Amiram Eldar, Table of n, a(n) for n = 1..14
G. Thimm, Emails to N. J. A. Sloane, Sep. 1994

Cf. A005101, A006038, A007702, A007707, A007708, A091191.

Cf. A108227, A285993.

a[n_] := Module[{p = Prime[n]}, r = 1; prod = 1; While[r <= 2, r *= 1 + 1/p; prod *= p; p = NextPrime[p]]; prod]; Array[a, 5] (* Amiram Eldar, Jun 29 2019 *)

a(n) = {p = prime(n); sig = p+1; prd = p; while (sig <= 2*prd, p = nextprime(p+1); sig *= p+1; prd *= p;); return (prd);} \\ Michel Marcus, Mar 10 2013

a(n) = Product_{k=n..A007707(n)} prime(k) = Product_{0 <= i < A108227(n)} prime(n+i). - M. F. Hasler, Apr 30 2017 and Jun 15 2017

More terms from Don Reble, Nov 10 2005

A108227 a(n) is the least number of prime factors for any abundant number with p_n (the n-th prime) as its least factor.

Original entry on oeis.org

3, 5, 9, 18, 31, 46, 67, 91, 122, 158, 194, 238, 284, 334, 392, 456, 522, 591, 668, 749, 835, 929, 1028, 1133, 1242, 1352, 1469, 1594, 1727, 1869, 2019, 2163, 2315, 2471, 2636, 2802, 2977, 3157, 3342, 3534, 3731, 3933, 4145, 4358, 4581, 4811

Offset: 1

Hugo van der Sanden, Jun 17 2005

nonn

If we replace "abundant" in the definition with "non-deficient", we get the same sequence with an initial 2 instead of 3, barring an astronomically unlikely coincidence with some as-yet-undiscovered odd perfect number. [This is sequence A107705. - M. F. Hasler, Jun 14 2017]
It appears that all terms >= 5 correspond to the odd primitive abundant numbers (A006038) which are products of consecutive primes (cf. A285993), i.e., of the form N = Product_{0<=iM. F. Hasler, May 08 2017
From Jianing Song, Apr 21 2021: (Start)
Let x_1 < x_2 < ... < x_k < ... be the numbers of the form p of p^2 + p, where p is a prime >= prime(n). Then a(n) is the smallest N such that Product_{i=1..N} (1 + 1/x_i) > 2. See my link below for a proof.
For example, for n = 3, we have {x_1, x_2, ..., x_k, ...} = {5, 7, 11, 13, 17, 19, 23, 29, 5^2 + 5, ...}, we have Product_{i=1..8} (1 + 1/x_i) < 2 and Product_{i=1..9} (1 + 1/x_i) > 2, so a(3) = 9. (End)

			a(2) = 5 since 945 = 3^3*5*7 is an abundant number with p_2 = 3 as its smallest prime factor, and no such number exists with fewer than 5 prime factors.

Amiram Eldar, Table of n, a(n) for n = 1..500
Jianing Song, Notes for A108227

Cf. A000040, A005101, A006038, A001222.
Cf. A107705.
Cf. A007707 (~ A007684), A007708, A007741.
Cf. A001276 (least number of prime factors for a (p_n)-rough abundant number, counted without multiplicity).

A108227(n, s=1+1/prime(n))=for(a=1, 9e9, if(2M. F. Hasler, Jun 15 2017

isform(k,q) = my(p=prime(k)); if(isprime(q) && (q>=p), 1, if(issquare(4*q+1), my(r=(sqrtint(4*q+1)-1)/2); isprime(r) && (r>=p), 0))
a(n) = my(Prod=1, Sum=0); for(i=prime(n), oo, if(isform(n,i), Prod *= (1+1/i); Sum++); if(Prod>2, return(Sum))) \\ Jianing Song, Apr 21 2021

a(n) = A007684(n)-n+1, for n>1. A007741(n) = Product_{0<=iM. F. Hasler, Jun 15 2017

Data corrected by Amiram Eldar, Aug 08 2019

A007702 a(n) = prime(n)...prime(m), the least product of consecutive primes which is non-deficient.

Original entry on oeis.org

6, 15015, 33426748355, 1357656019974967471687377449, 7105630242567996762185122555313528897845637444413640621

Offset: 1

Walter Nissen

nonn

Amiram Eldar, Table of n, a(n) for n = 1..14

Cf. A005100, A007741, A007684, A007686.

Cf. A107705, A108227, A285993.

a[n_] := Module[{p = Prime[n]}, r = 1; prod = 1; While[r < 2, r *= 1 + 1/p; prod *= p; p = NextPrime[p]]; prod]; Array[a, 5] (* Amiram Eldar, Jun 29 2019 *)

A007702(n, p=prime(n), s=1+1/p, P=p)={until(2<=s*=1+1/p,P*=p=nextprime(p+1));P} \\ M. F. Hasler, Jun 15 2017

a(n) = Product_{k = n..A007684(n)} prime(k) = Product_{0 <= i < A107705(n)} prime(n+i). - M. F. Hasler, Jun 15 2017

More terms from Don Reble, Nov 10 2005

cp's OEIS Frontend

A286042 Largest prime factor of A285993(n), the largest odd abundant number (A005231) equal to the product of n consecutive primes.

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A007741 a(n) = prime(n)*...*prime(m), the least product of consecutive primes which is abundant.

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A108227 a(n) is the least number of prime factors for any abundant number with p_n (the n-th prime) as its least factor.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A007702 a(n) = prime(n)*...*prime(m), the least product of consecutive primes which is non-deficient.

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Author

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Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A007741 a(n) = prime(n)...prime(m), the least product of consecutive primes which is abundant.

A007702 a(n) = prime(n)...prime(m), the least product of consecutive primes which is non-deficient.