A358413 -id:A358413 - cp's OEIS frontend

A358412 Least number k coprime to 2 and 3 such that sigma(k)/k >= n.

1, 5391411025, 5164037398437051798923642083026622326955987448536772329145127064375

Offset: 1

Jianing Song, Nov 14 2022

Data copied from the Hi.gher. Space link where Mercurial, the Spectre calculated the terms. We have a(2) = 5^2*7*...*29 and a(3) = 5^4*7^3*11^2*13^2*17*...*157 ~ 5.16404*10^66. a(4) = 5^5*7^4*11^3*13^3*17^2*19^2*23^2*29^2*31^2*37^2*41*...*853 ~ 1.83947*10^370 is too large to display.

			a(2) = A047802(2) = 5391411025 is the smallest abundant number coprime to 2 and 3.
Even if there is a number k coprime to 2 and 3 with sigma(k)/k = 3, we have that k is a square since sigma(k) is odd. If omega(k) = m, then 3 = sigma(k)/k < Product_{i=3..m+2} (prime(i)/(prime(i)-1)) => m >= 33, and we have k >= prime(3)^2*...*prime(35)^2 ~ 6.18502*10^112 > A358413(2) ~ 5.16403*10^66. So a(3) = A358413(2).
Even if there is a number k coprime to 2 and 3 with sigma(k)/k = 4, there can be at most 2 odd exponents in the prime factorization of k (see Theorem 2.1 of the Broughan and Zhou link). If omega(k) = m, then 4 = sigma(k)/k < Product_{i=3..m+2} (prime(i)/(prime(i)-1)) => m >= 140, and we have k >= prime(3)^2*...*prime(140)^2*prime(141)*prime(142) ~ 2.65585*10^669 > A358414(2) ~ 1.83947*10^370. So a(4) = A358414(2).

Jianing Song, Table of n, a(n) for n = 1..4
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, author’s version, Research Commons.
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, Journal of Number Theory 128 (2008) 1566-1575.
Mercurial, the Spectre, Abundant numbers coprime to n, Hi.gher. Space.

Smallest k-abundant number which is not divisible by any of the first n primes: A047802 (k=2), A358413 (k=3), A358414 (k=4).

Least p-rough number k such that sigma(k)/k >= n: A023199 (p=2), A119240 (p=3), this sequence (p=5), A358418 (p=7), A358419 (p=11).

A358414 Smallest 4-abundant number (sigma(x) > 4x) which is not divisible by any of the first n primes.

Original entry on oeis.org

27720, 1853070540093840001956842537745897243375

Offset: 0

Jianing Song, Nov 14 2022

Data copied from the Hi.gher. Space link where Mercurial, the Spectre calculated the terms. We have a(0) = 2^3*3^2*5*7*11 and a(1) = 3^5*5^3*7^2*11^2*13*...*89 ~ 1.85307*10^39. a(2) = 5^5*7^4*11^3*13^3*17^2*19^2*23^2*29^2*31^2*37^2*41*...*853 ~ 1.83947*10^370, a(3) = 7^5*11^3*13^3*17^3*19^3*23^2*...*97^2*101*...*4561 ~ 1.11116*10^1986, and a(4) = 11^4*13^4*17^3*19^3*23^3*29^3*31^3*37^2*...*181^2*191*...*18493 ~ 2.99931*10^8063 are too large to display.

			a(1) = A119240(4) = 1853070540093840001956842537745897243375 is the smallest 4-abundant odd number.
a(2) = A358412(4) ~ 1.83947*10^370 is the smallest 4-abundant number that is coprime to 2 and 3.

Jianing Song, Table of n, a(n) for n = 0..2
Mercurial, the Spectre, Abundant numbers coprime to n, Hi.gher. Space.

Cf. A068404 (4-abundant numbers).
Smallest k-abundant number which is not divisible by any of the first n primes: A047802 (k=2), A358413 (k=3), this sequence (k=4).
Least p-rough number k such that sigma(k)/k >= n: A023199 (p=2), A119240 (p=3), A358412 (p=5), A358418 (p=7), A358419 (p=11).

A358418 Least number k coprime to 2, 3, and 5 such that sigma(k)/k >= n.

Original entry on oeis.org

1, 20169691981106018776756331

Offset: 1

Jianing Song, Nov 14 2022

Data copied from the Hi.gher. Space link where Mercurial, the Spectre calculated the terms. We have a(2) = 7^2*11^2*13*...*67 ~ 2.01697*10^25. a(3) = 7^3*11^3*13^2*17^2*19^2*23^2*29^2*31*...*569 ~ 2.54562*10^239 and a(4) = 7^5*11^3*13^3*17^3*19^3*23^2*...*97^2*101*...*4561 ~ 1.11116*10^1986 are too large to display.

			a(2) = A047802(3) = 20169691981106018776756331 is the smallest abundant number coprime to 2, 3, and 5.
Even if there is a number k coprime to 2, 3, and 5 with sigma(k)/k = 3, we have that k is a square since sigma(k) is odd. If omega(k) = m, then 3 = sigma(k)/k < Product_{i=4..m+3} (prime(i)/(prime(i)-1)) => m >= 97, and we have k >= prime(4)^2*...*prime(100)^2 ~ 2.46692*10^436 > A358413(3) ~ 2.54562*10^239. So a(3) = A358413(3).
Even if there is a number k coprime to 2, 3, and 5 with sigma(k)/k = 4, there can be at most 2 odd exponents in the prime factorization of k (see Theorem 2.1 of the Broughan and Zhou link). If omega(k) = m, then 4 = sigma(k)/k < Product_{i=4..m+3} (prime(i)/(prime(i)-1)) => m >= 606, and we have k >= prime(4)^2*...*prime(607)^2*prime(608)*prime(609) ~ 6.54355*10^3814 > A358414(3) ~ 1.11116*10^1986. So a(4) = A358414(3).

Jianing Song, Table of n, a(n) for n = 1..3
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, author’s version, Research Commons.
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, Journal of Number Theory 128 (2008) 1566-1575.
Mercurial, the Spectre, Abundant numbers coprime to n, Hi.gher. Space.

Smallest k-abundant number which is not divisible by any of the first n primes: A047802 (k=2), A358413 (k=3), A358414 (k=4).

Least p-rough number k such that sigma(k)/k >= n: A023199 (p=2), A119240 (p=3), A358412 (p=5), this sequence (p=7), A358419 (p=11).

A358419 Least number k coprime to 2, 3, 5, and 7 such that sigma(k)/k >= n.

Original entry on oeis.org

1, 49061132957714428902152118459264865645885092682687973

Offset: 1

Jianing Song, Nov 14 2022

Data copied from the Hi.gher. Space link where Mercurial, the Spectre calculated the terms. We have a(2) = 11^2*13^2*17*...137 ~ 4.90611*10^52. a(3) = 11^3*13^3*17^2*...*47^2*53*...*1597 ~ 3.99515*10^688 and a(4) = 11^4*13^4*17^3*19^3*23^3*29^3*31^3*37^2*...*181^2*191*...*18493 ~ 2.99931*10^8063 are too large to display.

			a(2) = A047802(4) = 49061132957714428902152118459264865645885092682687973 is the smallest abundant number coprime to 2, 3, 5, and 7.
Even if there is a number k coprime to 2, 3, 5, and 7 with sigma(k)/k = 3, we have that k is a square since sigma(k) is odd. If omega(k) = m, then 3 = sigma(k)/k < Product_{i=5..m+4} (prime(i)/(prime(i)-1)) => m >= 240, and we have k >= prime(5)^2*...*prime(244)^2 ~ 1.60834*10^1297 > A358413(4) ~ 3.99515*10^688. So a(3) = A358413(4).
Even if there is a number k coprime to 2, 3, 5, and 7 with sigma(k)/k = 4, there can be at most 2 odd exponents in the prime factorization of k (see Theorem 2.1 of the Broughan and Zhou link). If omega(k) = m, then 4 = sigma(k)/k < Product_{i=5..m+4} (prime(i)/(prime(i)-1)) => m >= 2096, and we have k >= prime(5)^2*...*prime(2098)^2*prime(2099)*prime(2100) ~ 6.21439*10^15801 > A358414(4) ~ 2.99931*10^8063. So a(4) = A358414(4).

Jianing Song, Table of n, a(n) for n = 1..3
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, author’s version, Research Commons.
Kevin A. Broughan and Qizhi Zhou, Odd multiperfect numbers of abundancy 4, Journal of Number Theory 128 (2008) 1566-1575.
Mercurial, the Spectre, Abundant numbers coprime to n, Hi.gher. Space.

Smallest k-abundant number which is not divisible by any of the first n primes: A047802 (k=2), A358413 (k=3), A358414 (k=4).

Least p-rough number k such that sigma(k)/k >= n: A023199 (p=2), A119240 (p=3), A358412 (p=5), A358418 (p=7), this sequence (p=11).

A358395 Odd numbers k such that sigma(k) + sigma(k+2) > 2*sigma(k+1); odd terms in A053228.

Original entry on oeis.org

1125, 1573, 1953, 2205, 2385, 3465, 5185, 5353, 5773, 6433, 6613, 6825, 7245, 7425, 7665, 7693, 8505, 8925, 9133, 9205, 9405, 9945, 10393, 10773, 11473, 11653, 12285, 12493, 12705, 13473, 13585, 13725, 14025, 15013, 15145, 15433, 16065, 16245, 16905, 17253, 17325, 17953

Offset: 1

Jianing Song, Nov 13 2022

nonn

Odd numbers k such that A053223(k) > 0.
Terms congruent to 5 modulo 6 exist but must be very large: for example A053223(670173643268502741420822977335461337017377351999597045900203591953125) = 1311786588705365455963902347308393766941056366825184647502989937872.
A number m coprime to 2 and 3 such that sigma(m)/m >= 3 (m = A358412(3) = A358413(2) = 5^4*7^3*11^2*13^2*17*...*157 ~ 5.16404*10^66 is the smallest such number; see the link from Mercurial, the Spectre) produces a family of infinitely many terms congruent to 5 modulo 6 in this sequence, by Dirichlet's theorem on arithmetic progressions. Concretely, let k == 5 (mod 6), N(t) = t*k*(k+2) + (k+1)/6 for t >= 0, then:
(i) If sigma(k)/k >= 3. If N(t) is prime and 6*N(t)+1 is composite, then sigma(6*N(t)-1) >= 3*(6*N(t)-1), sigma(6*N(t)) = 12*(N(t)+1) and sigma(6*N(t)+1) >= 1+sqrt(6*N(t)+1)+(6*N(t)+1), so A053223(6*N(t)-1) >= sqrt(6*N(t)+1) - 25 >= sqrt(k+2) - 25 > 0.
(ii) If sigma(k+2)/(k+2) >= 3. If N(t) is prime and 6*N(t)-1 is composite, then sigma(6*N(t)+1) >= 3*(6*N(t)+1), sigma(6*N(t)) = 12*(N(t)+1) and sigma(6*N(t)-1) >= 1+sqrt(6*N(t)-1)+(6*N(t)-1), so A053223(6*N(t)-1) >= sqrt(6*N(t)-1) - 21 >= sqrt(k) - 21 > 0.

			1125 is a term since sigma(1126) = 1692 is smaller than the average of sigma(1125) = 2028 and sigma(1127) = 1368.

Jianing Song, Table of n, a(n) for n = 1..20000
Mercurial, the Spectre, Abundant numbers coprime to n, Hi.gher. Space.
Wikipedia, Dirichlet's theorem on arithmetic progressions

Cf. A053228, A053223, A000203 (sigma), A358396.

Cf. also A358412, A358413.

isA358395(n) = (n%2) && (sigma(n) + sigma(n+2) > 2*sigma(n+1))

cp's OEIS Frontend

A358412 Least number k coprime to 2 and 3 such that sigma(k)/k >= n.

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A358414 Smallest 4-abundant number (sigma(x) > 4x) which is not divisible by any of the first n primes.

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A358418 Least number k coprime to 2, 3, and 5 such that sigma(k)/k >= n.

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A358419 Least number k coprime to 2, 3, 5, and 7 such that sigma(k)/k >= n.

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A358395 Odd numbers k such that sigma(k) + sigma(k+2) > 2*sigma(k+1); odd terms in A053228.

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