cp's OEIS Frontend

A182433 Smallest number such that the next n integers each have the square of one of the first n primes as a factor in order.

%I A182433 #11 Aug 09 2015 15:19:29
%S A182433 7,547,29347,1308247,652312447,180110691547,65335225716547,
%T A182433 38733853511213647,4368761145612023947,1804216772228848838647,
%U A182433 14884872991210984993091647,9816873967836575781598117447,143397994078495393809327283088347
%N A182433 Smallest number such that the next n integers each have the square of one of the first n primes as a factor in order.
%C A182433 These are found by an application of the Chinese remainder theorem. The remainders are the numbers prime(n)^2 - n (A182174), and the moduli are the squares of primes (A001248).
%C A182433 This guarantees a run of at least n nonsquarefree numbers. But just as n! + 1 guarantees a run of at least n - 1 composite numbers, this might not be the smallest run of n nonsquarefree numbers (for that, see A045882).
%C A182433 Marmet credits Erick Bryce Wong with the idea of applying the Chinese remainder theorem and a sieving process to obtain upper limits for squarefree gaps. From this it occurred to me to just apply the Chinese remainder theorem to find these squarefree gaps exhibiting the squares of primes in order.
%C A182433 Also, beyond a(4), that is n > 4, we will observe that some of the numbers in the run of nonsquarefree numbers are divisible by more than one prime power, e.g., a(n) + 5 is divisible both by 49 (the square of the fourth prime) and 4.
%H A182433 A. Bogomolny, <a href="http://www.cut-the-knot.org/blue/chinese.shtml">Chinese Remainder Theorem</a> from Interactive Mathematics Miscellany and Puzzles
%H A182433 Louis Marmet, <a href="http://www.marmet.org/louis/sqfgap/">First occurrences of squarefree gaps...</a>
%e A182433 a(3) = 547 as that is the solution to the simultaneous congruences x = 3 mod 4 = 7 mod 9 = 22 mod 25. We verify that the next 3 integers meet the requirement: 548 = 4 * 137, 549 = 9 * 61, 550 = 25 * 2 * 11.
%e A182433 a(4) = 29347 as that is the solution to the simultaneous congruences x = 3 mod 4 = 7 mod 9 = 22 mod 25 = 45 mod 49. We verify that the next 4 integers meet the requirement: 29348 = 4 * 11 * 23 * 29, 29349 = 9 * 3 * 1087, 29350 = 25 * 2 * 587, 29351 = 49 * 599.
%t A182433 Table[ChineseRemainder[Prime[Range[n]]^2 - Range[n], Prime[Range[n]]^2], {n, 2, 14}]
%Y A182433 Cf. A069021, A051681.
%K A182433 nonn
%O A182433 2,1
%A A182433 _Alonso del Arte_, Apr 28 2012