A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!*2!*...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

1, 2, 0, 2, 0, 0, 0, 3, 4, 0, 0, 0, 6, 0, 8, 9, 0, 8, 9, 0, 7, 0, 10, 0, 0, 0, 12, 0, 0, 0, 14, 0, 0, 0, 15, 16, 0, 18, 0, 18, 0, 0, 0, 20, 0, 0, 0, 22, 0, 0, 0, 24, 25, 0, 0, 0, 26, 0, 0, 0, 28, 0, 0, 0, 30, 0, 32, 0, 32, 0, 0, 0, 34, 0, 0, 0, 35, 36, 0, 0, 0, 38, 0, 0, 0, 40

Offset: 1

Author

Bernard Schott, Oct 30 2021

Keywords

• nonn
• tabf

Comments

This sequence is the generalization of a problem proposed during the 17th Tournament of Towns (Spring 1996) and also during the first stage of the Moscow Mathematical Olympiad (1995-1996); the problem asked the question for n = 100 (see Andreescu-Gelca reference, Norman Do link, and Examples section).

Exhaustive results coming from Mabry-McCormick's link and adapted for OEIS:

-> n$ (A000178) is never a square if n > 1.

-> There is no solution if n is odd > 1, hence row(2q+1) = 0 when q > 0.

-> When n is even and there is a solution, then m belongs to {n/2 - 2, n/2 - 1, n/2, n/2 + 1, n/2 + 2}.

-> If 4 divides n (A008536), m = n/2 is always a solution because

(n$) / (n/2)! = ( 2^(n/4) * Product_{j=1..n/2} ((2j-1)!) )^2.

-> For other cases, see Formula section.

-> When n is even, there are 0, 1 or 2 solutions, so, the maximal length of a row is 2.

-> It is not possible to get more than three consecutive 0 terms, and three consecutive 0 terms correspond to three consecutive rows such that (n, n+1, n+2) = (4u+1, 4u+2, 4u+3) for some u >= 1.

Examples

			For n = 4, 4$ / 3! = 48, 4$ / 4! = 12 but 4$ / 2! = 12^2, hence, m = 2.
For n = 8, 8$ / 2! is not a square, but m_1 = 3 because 8$ / 3! = 29030400^2 and m_2 = 4 because 8$ / 4! = 14515200^2.
For n = 14, m_1 = 8 because 14$ / 8! = 1309248519599593818685440000000^2 and m_2 = 9 because 14$ / 9! = 436416173199864606228480000000^2.
For n = 16, m_1 = 8 because 16$ / 8! = 6848282921689337839624757371207680000000000^2 and m_2 = 9 because 16$ / 9! = 2282760973896445946541585790402560000000000^2.
For n = 18, m = 7 because 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2 and there is no other solution.
For n = 100, m = 50, unique solution to the Olympiad problems.
Triangle begins:
    1;
    2;
    0;
    2;
    0;
    0;
    0;
    8,  9;
    0;
    ...
		

References

• Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 725, pp. 253 and 686.
• Peter J. Taylor and A. M. Storozhev, Tournament of Towns 1993-1997, Book 4, Tournament 17, Spring 1996, O Level, Senior questions, Australian Mathematics Trust, 1998, problem 3, p. 96.

Links

• Diophante, A1963 - Le vilain petit canard (in French).
• Norman Do, Factorial fun, Puzzle Corner 13, Gaz. Aust. Math. Soc. 36, 2009, 176-179, page 178.
• Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.
• Tournament of Towns, Tournament 17, 1995-1996, Spring 1996, O Level, Senior questions, question 3 (in Russian).
• Index to sequences related to Olympiads and other Mathematical competitions.

Crossrefs

Cf. A000178, A001541, A008536, A035008, A060626, A139098.

Programs

PARI

sf(n)=prod(k=2, n, k!); \\ A000178
row(n) = my(s=sf(n)); Vec(select(issquare, vector(n, k, s/k!), 1));
lista(nn) = {my(list = List()); for (n=1, nn, my(r=row(n)); if (#r, for (k=1, #r, listput(list, r[k])), listput(list, 0));); Vec(list);} \\ Michel Marcus, Oct 30 2021

Formula

When there are two such integers m, then m_1 < m_2.

If n = 8*q^2 (A139098), then m_1 = n/2 - 1 = 4q^2-1 (see example for n=8).

If n = 8q*(q+1) (A035008), then m_2 = n/2 + 1 = (2q+1)^2 (see example for n=16).

if n = 4q^2 - 2 (A060626), then m_1 = n/2 + 1 = 2q^2 (see example for n=14).

If n = 2q^2, q>1 in A001541, then m = n/2 - 2 = q^2-2 (see example for n=18).

If n = 2q^2-4, q>1 in A001541, then m_2 = n/2 + 2 = q^2 (see example for n=14).