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 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS A178254(a(n)) = 4; union of A095990 and A096156. - Reinhard Zumkeller, May 24 2010 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 A056595(a(n)) = 4. - Reinhard Zumkeller, Aug 15 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 Also numbers n with A001222(n)=3 and A001221(n)=2. - Enrique Pérez Herrero, Jun 27 2012 A089233(a(n)) = 2. - Reinhard Zumkeller, Sep 04 2013 Subsequence of the triprimes (A014612). If a(n) is even, then a(n)/2 is semiprime (A001358). - Wesley Ivan Hurt, Sep 08 2013 From Bernard Schott, Sep 16 2017: (Start) 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 17-digit 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 11-digit 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 k-digit 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 Reinhard Zumkeller, Table of n, a(n) for n = 1..1000 Guilhem Castagnos, Antoine Joux, Fabien Laguillaumie, and Phong Q. Nguyen, Factoring pq^2 with quadratic forms: nice cryptanalyses, Advances in Cryptology - ASIACRYPT 2009. Lecture Notes in Computer Science Volume 5912 (2009), pp. 469-486. Diophante, A 350, Les Nombres d'Einstein (in French). Mathematics Stack Exchange, Sequence of numbers with prime factorization pq^2 René Peralta and Eiji Okamoto, Faster factoring of integers of a special form (1996). Index to sequences related to prime signature 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); A054753:=proc(q) local n; for n from 1 to q do if tau(n^5)=11*tau(n) then print(n); fi; od; end: A054753(10^10); # Paolo P. Lava, Mar 15 2013 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) is(n)=vecsort(factor(n)[, 2])==[1, 2]~ \\ Charles R Greathouse IV, Dec 30 2014 (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] print([k for k in range(453) if ok(k)]) # Michael S. Branicky, Dec 18 2021 CROSSREFS Cf. A001221, A001222, A001358, A014612, A056595, A089233, A095990, A096156, A178254. Numbers with 6 divisors (A030515) which are not 5th powers of primes (A050997). Subsequence of A325241. Supersequence of A096156. Table giving for each subsequence the corresponding number of groups of order p^2*q, from Bernard Schott, Jan 23 2022 ------------------------------------------------------------------------------- | Subsequence | A350638 | A143928 | A350115 | A349495 | A350245 | A350422 (*)| ------------------------------------------------------------------------------- |A000001(p^2*q)| (q+9)/2 | 5 | 5 | 4 | 3 | 2 | ------------------------------------------------------------------------------- (*) A350422 equals disjoint union of A350332 (pq). Sequence in context: A325241 A072357 A340780 * A098899 A098770 A333928 Adjacent sequences: A054750 A054751 A054752 * A054754 A054755 A054756 KEYWORD nonn AUTHOR Henry Bottomley, Apr 25 2000 EXTENSIONS Link added and incorrect Mathematica code removed by David Bevan, Sep 17 2011 STATUS approved

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Last modified September 25 15:05 EDT 2023. Contains 365648 sequences. (Running on oeis4.)