common divisors of multinomial coefficients

[Richard Guy]

The reference to UPINT wd be B31, or possibly B33.

UPINT3 went to Springer on Friday.  Was just
able to stop press long enough to mention
that Ben Green & Terence Tao have (with high
probability) shown that there are arbitrarily
long arithmetic progrssions of primes.    R.

On Sun, 11 Apr 2004, David Wasserman wrote:

> Dear SeqFans,
> 
> It is know that if we take any row of Pascal's triangle, and remove 
> the two 1s at the ends, then any two of the other numbers have a 
> common divisor.  I want to find an analogue of this fact for 
> multinomial coefficients.
> 
> Let MC(n, k) = { n!/(x_1!*x_2!*...*x_k!) | x_1, x_2, ..., x_k are 
> positive integers that sum to n }.  (i.e. the n-th degree k-nomial 
> coefficients that involve all k monomials.)  Then I conjecture that 
> any k members of MC(n, k) have a common divisor.
> 
> Can anyone prove or disprove this?  I've verified k = 3 for n up to 
> 200, k = 4 for n up to 400, k = 5 for n up to 700, and k = 6 up to n 
> = 1000.  As far as I've checked, any 3 members of MC(n, 3) have a 
> common divisor, as do any 6 members of MC(n, 4), any 8 members of 
> MC(n, 5) and any 9 members of MC(n, 6).  So maybe my conjecture is 
> too weak.  (n = 119 is the first n such that there are 4 members of 
> MC(n, 3) with no common divisor.  n = 726 is the first n such that 
> there are 10 members of MC(n, 6) with no common divisor.)
> 
> I've submitted two sequences related to this question:
> 
> %I A091963
> %S A091963 
> 2,3,2,5,2,7,2,3,2,11,3,13,2,3,2,17,2,19,4,3,2,23,3,5,2,3,4,29,6,31,2,3 
> ,
> %T A091963 
> 2,5,4,37,2,3,5,41,6,43,4,3,2,47,3,7,2,3,4,53,2,5,7,3,2,59,4,61,2,7,2,5 
> ,
> %U A091963 
> 6,67,4,3,10,71,4,73,2,3,4,7,2,79,5,3,2,83,12,5,2,3,4,89,9,7,4,3,2,5,3
> %N A091963 a(n) is the smallest gcd of two interior numbers on row n 
> of Pascal's triangle ("interior" means that the 1'\
>    s at the ends of the rows are excluded).
> %C A091963 The reference contains a simple proof that there are no 1s 
> in this sequence.
> %D A091963 R. K. Guy, Unsolved Problems in Number Theory.
> %e A091963 In row 8, the interior numbers 8, 28, 56, and 70; gcd(8, 
> 28) = 4; gcd(8, 56) = 8; gcd(8, 70) = 2; gcd(28, 56\
>    ) = 28; gcd(28, 70) = 14; gcd(56, 70) = 14. The smallest of these 
> is 2, so a(8) = 2.
> %Y A091963 Cf. A014410.
> %Y A091963 Sequence in context: A020639 A079879 A071889 this_sequence 
> A067695 A079866 A080210
> %Y A091963 Adjacent sequences: A091960 A091961 A091962 this_sequence 
> A091964 A091965 A091966
> %K A091963 nonn
> %O A091963 2,1
> %A A091963 David Wasserman (dwasserm(AT)earthlink.net), Mar 13 2004
> 
> 
> 
> 
> %I A092461
> %S A092461 
> 6,12,10,30,7,28,6,30,11,66,13,91,6,12,34,102,19,38,12,22,23,46,15,65,6 
> ,12,
> %T A092461 
> 29,435,62,124,6,34,10,36,37,703,6,24,41,82,86,43,20,46,47,94,21,70,6,
> %U A092461 
> 12,53,159,10,35,21,58,59,177,61,1891,14,28,10,30,67,134,12,14,142,142
> %N A092461 Let S_n be the set {n!/(i!*j!*k!) | i, j, k > 0, i+j+k = 
> n} (i.e. trinomial coefficients that involve all th\
>    ree monomials.) Then a(n) is the smallest gcd of any three members of S_n.
> %C A092461 Are there any 1's in this sequence?
> %e A092461 S_7 = {42, 105, 140, 210}, gcd(42, 105, 140) = 7, gcd(42, 
> 105, 210) = 21, gcd(42, 140, 210) = 14, gcd(105, 1\
>    40, 210) = 35. So a(7) is the smallest of these, 7.
> %Y A092461 Cf. A091963, A046816.
> %Y A092461 Sequence in context: A028320 A074474 A048726 this_sequence 
> A060479 A028998 A040030
> %Y A092461 Adjacent sequences: A092458 A092459 A092460 this_sequence 
> A092462 A092463 A092464
> %K A092461 nonn
> %O A092461 3,1
> %A A092461 David Wasserman (dwasserm(AT)earthlink.net), Mar 25 2004
>