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.

7, 547, 29347, 1308247, 652312447, 180110691547, 65335225716547, 38733853511213647, 4368761145612023947, 1804216772228848838647, 14884872991210984993091647, 9816873967836575781598117447, 143397994078495393809327283088347

Offset: 2

Author

Alonso del Arte, Apr 28 2012

Keywords

• nonn

Comments

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).

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).

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.

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.

Examples

			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.
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.
		

Links

• A. Bogomolny, Chinese Remainder Theorem from Interactive Mathematics Miscellany and Puzzles
• Louis Marmet, First occurrences of squarefree gaps...

Crossrefs

Cf. A069021, A051681.

Programs

Mathematica

Table[ChineseRemainder[Prime[Range[n]]^2 - Range[n], Prime[Range[n]]^2], {n, 2, 14}]