A238746 -id:A238746 - cp's OEIS frontend

A238690 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).

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, 14, 12, 6, 15, 15, 14, 12, 13, 17, 18, 13, 14, 19, 15, 16, 16, 19, 16, 11, 15, 16, 21, 19, 17, 14, 18, 17, 24, 21, 18, 19, 19, 23, 22, 7, 22, 24, 20, 22, 27, 23, 21

Offset: 1

Matthew Vandermast, Apr 28 2014

nonn

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.)
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.
For any positive integers m and k, the following integer sequences (with n >= 0) are arithmetic progressions:
1. The sequence b(n) = a(m*(2^n)).
2. The sequence b(n) = a(m*(prime(n+k))) if prime(k) >= A006530(m).
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.
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

			The prime factorizations of integers 1 through 9, with prime factors sorted from largest to smallest:
1 - the empty factorization (no prime factors)
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 3*2
7 = 7
8 = 2*2*2
9 = 3*3
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:
1) The largest prime factor, if one exists, is not greater than 3;
2) The second-largest prime factor, if one exists, is not greater than 3;
3) The total number of prime factors (counting repeated factors) does not exceed 2.
Therefore, a(9) = 6.
From _Gus Wiseman_, Feb 24 2018: (Start)
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).
Corresponding diagrams are:
  o o o   . o o   . o o   . . o   . . o   . . o   . . .   . . .   . . .
  o o     o o     . o     o o     . o     . .     o o     . o     . .    (End)

Amiram Eldar, Table of n, a(n) for n = 1..10000

Rearrangement of A115728, A115729 and A238746. A116473(n) is the number of times n appears in the sequence.

Cf. A000041, A000085, A000720, A056239, A063834, A112798, A122111, A153452, A215366, A238689, A259478, A259480, A296150, A296188, A296561, A297388, A299925, A299926, A299966, A299967.

undptns[y_]:=Select[Tuples[Range[0,#]&/@y],OrderedQ[#,GreaterEqual]&];
primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Length[undptns[Reverse[primeMS[n]]]],{n,100}] (* Gus Wiseman, Feb 24 2018 *)

a(n) = A085082(A108951(n)) = A085082(A181821(n)).
a(n) = a(A122111(n)).
a(prime(n)) = a(2^n) = n+1.
a((prime(n))^m) = a((prime(m))^n) = binomial(n+m, n).
a(A002110(n)) = A000108(n+1).
A000005(n) <= a(n) <= n.

cp's OEIS Frontend

A238690 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).

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cp's OEIS Frontend

A238690 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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A238690 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).