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%I #19 Oct 07 2019 11:02:49
%S 1,2,3,3,4,5,5,4,6,7,6,7,7,9,9,5,8,9,9,10,12,11,10,9,10,13,10,13,11,
%T 14,12,6,15,15,14,12,13,17,18,13,14,19,15,16,16,19,16,11,15,16,21,19,
%U 17,14,18,17,24,21,18,19,19,23,22,7,22,24,20,22,27,23,21
%N Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)*prime_m(2)*...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).
%C Equivalently, a(n) equals the number of values of m such that each value of A238689 T(m,k) <= A238689 T(n,k). (Since the prime factorization of 1 is the empty factorization, we consider each prime_1(i) not to be greater than prime_n(i) for all positive integers n.)
%C Suppose we say that n "covers" m iff both m and n are factorized as described in the sequence definition and each prime_m(i) <= prime_n(i). At least three sequences (A037019, A108951 and A181821) have the property that a(m) divides a(n) iff n "covers" m. These sequences are also divisibility sequences (i.e., sequences with the property that a(m) divides a(n) if m divides n), since any positive integer "covers" each of its divisors.
%C For any positive integers m and k, the following integer sequences (with n >= 0) are arithmetic progressions:
%C 1. The sequence b(n) = a(m*(2^n)).
%C 2. The sequence b(n) = a(m*(prime(n+k))) if prime(k) >= A006530(m).
%C Also, a(n) = the number of distinct prime signatures that occur among the divisors of any integer m such that A181819(m) = n and/or A238745(m) = n.
%C Number of skew partitions whose numerator has Heinz number n, where a skew partition is a pair y/v of integer partitions such that the diagram of v fits inside the diagram of y. The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). - _Gus Wiseman_, Feb 24 2018
%H Amiram Eldar, <a href="/A238690/b238690.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A085082(A108951(n)) = A085082(A181821(n)).
%F a(n) = a(A122111(n)).
%F a((prime(n)) = a(2^n) = n+1; a((prime(n))^m) = a((prime(m))^n) = binomial (n+m, n); a(002110(n)) = A000108(n+1).
%F A000005(n) <= a(n) <= n.
%e The prime factorizations of integers 1 through 9, with prime factors sorted from largest to smallest:
%e 1 - the empty factorization (no prime factors)
%e 2 = 2
%e 3 = 3
%e 4 = 2*2
%e 5 = 5
%e 6 = 3*2
%e 7 = 7
%e 8 = 2*2*2
%e 9 = 3*3
%e To find a(9), we consider 9 = 3*3. There are 6 positive integers (1, 2, 3, 4, 6 and 9) which satisfy the following criteria:
%e 1) The largest prime factor, if one exists, is not greater than 3;
%e 2) The second-largest prime factor, if one exists, is not greater than 3;
%e 3) The total number of prime factors (counting repeated factors) does not exceed 2.
%e Therefore, a(9) = 6.
%e From _Gus Wiseman_, Feb 24 2018: (Start)
%e Heinz numbers of the a(15) = 9 partitions contained within the partition (32) are 1, 2, 3, 4, 5, 6, 9, 10, 15. The a(15) = 9 skew partitions are (32)/(), (32)/(1), (32)/(11), (32)/(2), (32)/(21), (32)/(22), (32)/(3), (32)/(31), (32)/(32).
%e Corresponding diagrams are:
%e o o o . o o . o o . . o . . o . . o . . . . . . . . .
%e o o o o . o o o . o . . o o . o . . (End)
%t undptns[y_]:=Select[Tuples[Range[0,#]&/@y],OrderedQ[#,GreaterEqual]&];
%t primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
%t Table[Length[undptns[Reverse[primeMS[n]]]],{n,100}] (* _Gus Wiseman_, Feb 24 2018 *)
%Y Rearrangement of A115728, A115729 and A238746. A116473(n) is the number of times n appears in the sequence.
%Y Cf. A000041, A000085, A000720, A056239, A063834, A112798, A122111, A153452, A215366, A238689, A259478, A259480, A296150, A296188, A296561, A297388, A299925, A299926, A299966, A299967.
%K nonn
%O 1,2
%A _Matthew Vandermast_, Apr 28 2014
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