A331547 Numbers k such that k and k! - 1 have the same number of divisors.

3, 7, 8, 10, 26, 27, 34, 85, 93, 104, 143, 152

Offset: 1

Author

Matthew Niemiro, Jan 20 2020

Keywords

• nonn
• hard
• more

Comments

The sequence also includes: 143, 152, 186, 230, 379, 381, 543, 573, 602. - Daniel Suteu, Jan 21 2020

The sequence also includes 2881. Even though the complete factorization of 136!-1 is not known, 136 is not a term, since 136!-1 is known to be the product of 2 distinct primes and a composite number, so it has at least 4 prime factors and 3 distinct prime factors, thus the number of divisors >= 12, whereas 136 has 8 divisors. - Chai Wah Wu, Feb 26 2020

Similar reasoning (considering only small prime factors of k! - 1) shows that the next terms (> a(12) = 152) can only be within the set {154, 160, 162, 164, 176, 180, 182, 186, 187, 188, 192, 195, 196, 198, 204, ...}. - M. F. Hasler, Feb 26 2020

Links

• factordb.com, Status of 154!-1.

Crossrefs

Supersequence of A103317.

Cf. A000005, A002982, A064145.

Programs

Mathematica

Select[Range[50], DivisorSigma[0, #] - DivisorSigma[0, Factorial[#] - 1] == 0 &]

PARI

isok(k) = k>1 && numdiv(k)==numdiv(k!-1); \\ Jinyuan Wang, Jan 20 2020

PARI

{is(n)=my(f); n>2&& numdiv(n)>=numdiv(f=factor(n!-1,0))&& if( ispseudoprime(vecmax(f[,1])), numdiv(n)==numdiv(f), numdiv(n)<2*numdiv(f), 0, numdiv(n)==numdiv(n!-1))} \\ Avoids complete factorization if possible. - M. F. Hasler, Feb 26 2020

Formula

A331547 = { n > 1 | A000005(n) = A064145(n) }. - M. F. Hasler, Feb 26 2020

Extensions

a(8)-a(9) from Jinyuan Wang, Jan 20 2020

a(10) from Amiram Eldar, Jan 20 2020

a(11)-a(12) from Chai Wah Wu, Feb 26 2020