cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

Hugo van der Sanden, Jun 17 2005

Keywords

Comments

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)

Examples

			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.
		

Crossrefs

Cf. A107705.
Cf. A001276 (least number of prime factors for a (p_n)-rough abundant number, counted without multiplicity).

Programs

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

Formula

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

Extensions

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

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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

Formula

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

Extensions

More terms from Don Reble, Nov 10 2005

A285993 Largest odd abundant number (A005231) equal to the product of n consecutive primes.

Original entry on oeis.org

15015, 255255, 4849845, 111546435, 33426748355, 1236789689135, 50708377254535, 2180460221945005, 102481630431415235, 5431526412865007455, 320460058359035439845, 19548063559901161830545, 1309720258513377842646515, 1357656019974967471687377449, 107254825578022430263302818471
Offset: 5

Views

Author

M. F. Hasler, Apr 30 2017

Keywords

Comments

The smallest term is a(5) = 3*5*7*11*13, there is no odd abundant number (A005231) equal to the product of less than 5 consecutive primes.
The smallest odd abundant number (A005231) equal to the product of n consecutive primes is equal (when it exists, i.e., for n >= 5) to the least odd number with n (distinct) prime divisors, equal to the product of the first n odd primes = A070826(n+1) = A002110(n+1)/2.
See A188342 = (945, 3465, 15015, 692835, 22309287, ...) for the least odd primitive abundant number (A006038) with n distinct prime factors, and A275449 for the least odd primitive abundant number with n prime factors counted with multiplicity.
The terms are in general not primitive abundant numbers (A091191), in particular this cannot be the case when a(n) is a multiple of a(n-1), as is the case for most of the terms, for which a(n) = a(n-1)*A117366(a(n-1)). In the other event, spf(a(n)) = nextprime(spf(a(n-1))), and a(n) is in A007741(2,3,4...). These are exactly the primitive terms in this sequence.

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).
		

Crossrefs

A subsequence of A112643 (odd squarefree abundant numbers); see also A108227 (~ A107705) which give indices of primitive terms = those with smallest prime factor larger than that of earlier terms.

Programs

  • 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.

A337476 Position of the first occurrence of n in A337474.

Original entry on oeis.org

1, 3, 11, 23, 61, 127, 199, 331, 467, 673, 929, 1181, 1493, 1861, 2243, 2693, 3221, 3739, 4327, 4993, 5689, 6421, 7283, 8191, 9137, 10111, 11161, 12281, 13451, 14747, 16067, 17569, 19037, 20509, 22051, 23687, 25411, 27179, 29023, 31019, 32971, 34963, 37097, 39371, 41651, 44021, 46559, 49169, 51719
Offset: 0

Views

Author

Antti Karttunen, Aug 28 2020

Keywords

Crossrefs

Formula

For all n >= 0, A337474(a(n)) = n.
For all n >= 1, a(n) = A000040(A107705(n)). [Conjectured, see comment in A108227]
For all n >= 0, a(n) <= A337478(n).
Showing 1-4 of 4 results.