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 A141621 Numbers that begin a run of 5 consecutive integers of the form p^2*q where p and q are distinct primes. 8
 10093613546512321, 14414905793929921, 266667848769941521, 562672865058083521, 1579571757660876721, 1841337567664174321, 2737837351207392721, 4456162869973433521, 4683238426747860721, 4993613853242910721, 5037980611623036721, 5174116847290255921 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Old name was "The first number of a series of 5 consecutive numbers with the same signature, i.e., all numbers have the format p^2*q, where p and q are primes. Therefore the number of divisors is the same (6)." [That name could have been confusing in that not every sequence of 5 consecutive integers having the same prime signature has the prime signature p^2*q; e.g., 204323 is the first of 5 consecutive numbers of the form p^2*q*r. - Jon E. Schoenfield, Jun 05 2018] Each of the five numbers in each such sequence has 6 divisors. It is easy to prove that any number in this sequence must be congruent to 1 modulo 240. The program below calculates only an element of the sequence. Since the reference A119479 it is the smallest one. If we assume that the first element has the format 7^2*n49, the second number has the format 2*p^2, the third element has the format 3^2*n9 and the fifth element has the format 5^2*n25, then p must be modulo 22050 one out of 1181, 3719, 4219, 9119, 12931, 17831, 18331 or 20869. It is unclear if these numbers are the smallest ones. - Matthijs Coster, Aug 28 2008 [The terms listed in the Data section are, in fact, the smallest numbers matching the definition. - Jon E. Schoenfield, Jun 05 2018] The first quintuple not of the aforementioned form starts with 5344962129269790721 = 23^2*prime. - Ivan Neretin, Feb 08 2016 Among the first 200 terms, the frequency with which the squared prime factor p is {7, 17, 23, 31, 41, 47, 73, 127, 193, 1039, 1399} is {171, 10, 6, 4, 3, 1, 1, 1, 1, 1, 1}, respectively. - Jon E. Schoenfield, Jun 09 2018 REFERENCES Puzzles from around the world, Richard I Hess, p. 63, H17. See: Link 3. LINKS Ray Chandler, Table of n, a(n) for n = 1..2000 (first 25 terms from Ivan Neretin, through 200 terms from Jon E. Schoenfield) Carlos Rivera, Problem 20.- Divisors (II) K consecutive numbers with the same number of divisors, PrimePuzzles. StackExchange, Sequence of numbers with prime factorization pq^2 wu :: forums, Same Number of Divisors, Oct 05 2007. EXAMPLE a(1) = 10093613546512321, because 10093613546512321 = 7^2 * 205992113194129, 10093613546512322 = 2 * 71040881^2, 10093613546512323 = 3^2 * 1121512616279147, 10093613546512324 = 2^2 * 2523403386628081, and 10093613546512325 = 5^2 * 403744541860493, so each of the five consecutive integers is of the form p^2*q, and no smaller run of five consecutive integers has this property. [corrected by Jon E. Schoenfield, Jun 05 2018] PROG (Sage) # (replace leading dots by spaces): ## Warning: this program appears to be incorrect [Joerg Arndt, Feb 29 2016] for m in range(5000): p = 22050*m+17831 if is_prime(p): n = 2*p^2-2 n4 = n/4+1 if is_prime(n4): n49 = floor((n+1)/49) if (49*n49 == n+1) and is_prime(n49): n9 = floor((n+3)/9) if (9*n9 == n+3) and is_prime(n9): n25 = floor((n+5)/25) if (25*n25 == n+5) and is_prime(n25): print(n+1, n49, p, n9, n4, n25) CROSSREFS Cf. A119479, A006558, A005237, A005238, A006601. Cf. A054753, A074172, A178032, A308683. Sequence in context: A329948 A160405 A162032 * A095432 A238358 A288280 Adjacent sequences: A141618 A141619 A141620 * A141622 A141623 A141624 KEYWORD hard,nonn AUTHOR Matthijs Coster, Aug 23 2008 EXTENSIONS Two more terms Matthijs Coster, Aug 28 2008 Missing terms added and extended by Ivan Neretin, Feb 08 2016 New name from Jon E. Schoenfield, Jun 05 2018 STATUS approved

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Last modified February 7 12:58 EST 2023. Contains 360123 sequences. (Running on oeis4.)