A182856 a(0) = 1; for n > 0, a(n) = smallest positive integer whose prime signature contains, for k = 1 to n, exactly one positive number appearing exactly k times.

1, 2, 60, 1801800, 11657093261814000, 7167827541370578634694420017740000, 291943326350524088652207164949980988754136887856059678357800000

Offset: 0

Author

Matthew Vandermast, Jan 05 2011

Keywords

• nonn
• easy

Comments

a(n) = smallest number m such that A181819(m) = A006939(n). a(n) belongs to A182855 iff n > 1.

Next term has 105 digits.

Smallest number k with A323022(k) = n, where A323022(m) is the number of distinct multiplicities in the prime signature of m. - Gus Wiseman, Jan 03 2019

Examples

			The canonical prime factorization of a(3) = 1801800 is 2^3*3^2*5^2*7*11*13. The prime signature of 1801800 is therefore (3,2,2,1,1,1). Note that (3,2,2,1,1,1) contains exactly one number that appears once (3), one number that appears twice (2), and one number that appears three times (1).
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..12
• Dario Alpern, Factorization using the Elliptic Curve Method

Crossrefs

Cf. A006939, A181819, A182855.

Cf. A056239, A059404, A062770, A071625, A118914, A181821, A182857, A323022.

Programs

Mathematica

Table[Product[Times@@Prime[i*(i-1)/2+Ceiling[Range[i*(n-i)]/(n-i)]],{i,n-1}],{n,6}] (* Gus Wiseman, Jan 03 2019 *)

PARI

a(n) = if(n == 0, return(1)); my(f = matrix(binomial(n+1,2), 2)); f[, 1] = primes(#f~ )~; f[, 2] = Vecrev(concat(vector(n, i, vector(n+1-i, j, i))))~; factorback(f) \\ David A. Corneth, Jan 03 2019

Formula

Partial products of A113511.

log a(n) ~ (1/3) n^3 log n. [Charles R Greathouse IV, Jan 13 2012]

A001222(a(n)) = A000292(n). - Gus Wiseman, Jan 03 2019

a(0) = 1; a(n + 1) = A002110(binomial(n + 2, 2)) * a(n). - David A. Corneth, Jan 03 2019