

A348692


Triangle whose nth 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 nth row = 0.


9



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
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OFFSET

1,2


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 (19951996); the problem asked the question for n = 100 (see AndreescuGelca reference, Norman Do link, and Examples section).
Exhaustive results coming from MabryMcCormick'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} ((2j1)!) )^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.


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 19931997, Book 4, Tournament 17, Spring 1996, O Level, Senior questions, Australian Mathematics Trust, 1998, problem 3, p. 96.


LINKS

Norman Do, Factorial fun, Puzzle Corner 13, Gaz. Aust. Math. Soc. 36, 2009, 176179, page 178.


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^21 (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^22 (see example for n=18).
If n = 2q^24, q>1 in A001541, then m_2 = n/2 + 2 = q^2 (see example for n=14).


EXAMPLE

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;
...


PROG

(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


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



