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%I #31 Jul 08 2023 20:57:55
%S 1,2,60,1801800,11657093261814000,7167827541370578634694420017740000,
%T 291943326350524088652207164949980988754136887856059678357800000
%N 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.
%C a(n) = smallest number m such that A181819(m) = A006939(n). a(n) belongs to A182855 iff n > 1.
%C Next term has 105 digits.
%C 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
%H David A. Corneth, <a href="/A182856/b182856.txt">Table of n, a(n) for n = 0..12</a>
%H Dario Alpern, <a href="https://www.alpertron.com.ar/ECM.HTM">Factorization using the Elliptic Curve Method</a>
%F Partial products of A113511.
%F log a(n) ~ (1/3) n^3 log n. [_Charles R Greathouse IV_, Jan 13 2012]
%F A001222(a(n)) = A000292(n). - _Gus Wiseman_, Jan 03 2019
%F a(0) = 1; a(n + 1) = A002110(binomial(n + 2, 2)) * a(n). - _David A. Corneth_, Jan 03 2019
%e 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).
%t 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 *)
%o (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
%Y Cf. A006939, A181819, A182855.
%Y Cf. A056239, A059404, A062770, A071625, A118914, A181821, A182857, A323022.
%K nonn,easy
%O 0,2
%A _Matthew Vandermast_, Jan 05 2011
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