1,1

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. [From Matthijs Coster (oeis(AT)coster.demon.nl), Aug 28 2008]

Problem 1231, Crux Mathematicorum, Vol. 13, No. 4, p. 118, 1987.

Puzzles from around the world, Richard I Hess, p. 63, H17. See: Link 3

www.primepuzzles.net/problems/prob_020.htmwww.primepuzzles.net/problems/prob_020.htm

www.ocf.berkeley.edu~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;

www.totalgadha.com/html/file.php/1/moddata/forum/17/20234/mm-hess.pdf

a(1) = 10093613546512321, because 10093613546512321 = 7^2 * 205992113194129; 10093613546512321 = 2 * 71040881^2; 10093613546512323 = 3^2 * 1121512616279147; 10093613546512324 = 2^2 * 2523403386628081 and 10093613546512325 = 5^2 * 403744541860493.

Sage program (replace leading dots by spaces):

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

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

Sequence in context: A104836 A160405 A162032 this_sequence A098143 A095432 A083216

Adjacent sequences: A141618 A141619 A141620 this_sequence A141622 A141623 A141624

hard,nonn

Matthijs Coster (oeis(AT)coster.demon.nl), Aug 23 2008

Two more terms Matthijs Coster (oeis(AT)coster.demon.nl), Aug 28 2008