A141621 Numbers that begin a run of 5 consecutive integers of the form p^2*q where p and q are distinct primes.

10093613546512321, 14414905793929921, 266667848769941521, 562672865058083521, 1579571757660876721, 1841337567664174321, 2737837351207392721, 4456162869973433521, 4683238426747860721, 4993613853242910721, 5037980611623036721, 5174116847290255921

Offset: 1

Author

Matthijs Coster, Aug 23 2008

Keywords

• hard
• nonn

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

Examples

			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]
		

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)
• Richard I. Hess, Problem 1231, Crux Mathematicorum, Vol. 13, No. 4, p. 118, 1987. (takes a long time to download)
• Richard I. Hess, Puzzles from Around the World, p. 63, H17.
• Carlos Rivera, Problem 20.- Divisors (II) K consecutive numbers with the same number of divisors, The Prime Puzzles & Problems Connection.
• StackExchange, Sequence of numbers with prime factorization pq^2
• wu :: forums, Same Number of Divisors, Oct 05 2007.
• Index to sequences related to prime signature

Crossrefs

Cf. A119479, A006558, A005237, A005238, A006601.

Cf. A054753, A074172, A178032, A308683.

Programs

Sage

## 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)

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