

A054753


Numbers which are the product of a prime and the square of a different prime (p^2 * q).


76



12, 18, 20, 28, 44, 45, 50, 52, 63, 68, 75, 76, 92, 98, 99, 116, 117, 124, 147, 148, 153, 164, 171, 172, 175, 188, 207, 212, 236, 242, 244, 245, 261, 268, 275, 279, 284, 292, 316, 325, 332, 333, 338, 356, 363, 369, 387, 388, 404, 412, 423, 425, 428, 436, 452
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OFFSET

1,1


COMMENTS

Numbers with prime signature (2,1) = union of numbers with ordered prime signature (1,2) and numbers with ordered prime signature (2,1) (restating second part of above comment).  Daniel Forgues, Feb 05 2011
Sum_{n>=1} 1/a(n)^k = P(k) * P(2*k)  P(3*k), where P is the Prime Zeta function.  Enrique Pérez Herrero, Jun 27 2012
These numbers are called "Nombres d'Einstein" on the French site "Diophante" (see link) because a(n) = m * c^2 where m and c are two different primes.
The numbers 44 = 2^2 * 11 and 45 = 3^2 * 5 are the two smallest consecutive "Einstein numbers"; 603, 604, 605 are the three smallest consecutive integers in this sequence. It's not possible to get more than five such consecutive numbers (proof in the link); the first set of five such consecutive numbers begins at the 17digit number 10093613546512321. Where does the first sequence of four consecutive "Einstein numbers" begin? (End) [corrected by Jon E. Schoenfield, Sep 20 2017]
The first set of four consecutive integers in this sequence begins at the 11digit number 17042641441. (Each such set must include two even numbers, one of which is of the form 2^2*q, the other of the form p^2*2; a quick search, taking the factorizations of consecutive integers before and after numbers of the latter form, shows that the number of sets of four consecutive kdigit integers in this sequence is 1, 7, 12, 18 for k = 11, 12, 13, 14, respectively.)  Jon E. Schoenfield, Sep 16 2017
The first 13 sets of 5 consecutive integers in this sequence have as their first terms 10093613546512321, 14414905793929921, 266667848769941521, 562672865058083521, 1579571757660876721, 1841337567664174321, 2737837351207392721, 4456162869973433521, 4683238426747860721, 4993613853242910721, 5037980611623036721, 5174116847290255921, 5344962129269790721. Each of these numbers except for the last is 7^2 times a prime; the last is 23^2 times a prime.  Jon E. Schoenfield, Sep 17 2017


LINKS



FORMULA

Solutions of the equation tau(n^5) = 11*tau(n).  Paolo P. Lava, Mar 15 2013


EXAMPLE

a(1) = 12 because 12 = 2^2*3 is the smallest number of the form p^2*q.


MAPLE

with(numtheory);
for n from 1 to q do if tau(n^5)=11*tau(n) then print(n); fi; od; end:


MATHEMATICA

Select[Range[12, 452], {1, 2}==Sort[Last/@FactorInteger[ # ]]&] (* Zak Seidov, Jul 19 2009 *)
With[{nn=60}, Take[Union[Flatten[{#[[1]]#[[2]]^2, #[[1]]^2 #[[2]]}&/@ Subsets[ Prime[Range[nn]], {2}]]], nn]] (* Harvey P. Dale, Dec 15 2014 *)


PROG

(PARI) for(n=1, 1e3, if(numdiv(n)  bigomega(n) == 3, print1(n, ", "))) \\ Altug Alkan, Nov 24 2015
(Python)
from sympy import factorint
def ok(n): return sorted(factorint(n).values()) == [1, 2]


CROSSREFS

Numbers with 6 divisors (A030515) which are not 5th powers of primes (A050997).
Table giving for each subsequence the corresponding number of groups of order p^2*q, from Bernard Schott, Jan 23 2022


A000001(p^2*q) (q+9)/2  5  5  4  3  2 



KEYWORD

nonn


AUTHOR



EXTENSIONS

Link added and incorrect Mathematica code removed by David Bevan, Sep 17 2011


STATUS

approved



