OFFSET
2,3
COMMENTS
If n is prime, then a(n) = 1, because all other elements of the n-th row of Pascal's triangle are multiples of that prime.
If n is composite, then the inequality 1 < gcd(n, a(n)) < n holds; in other words, n and a(n) are not coprime, but n does not divide a(n) evenly.
a(n) does not always equal binomial(n, gpf(n)), where gpf(n) is the greatest prime factor function. For example, in the twelfth row of Pascal's triangle, binomial(12, 3) = 220, but binomial(12, 4) = 495.
REFERENCES
Vladimir Andreevich Uspenskii, Pascal's Triangle. Translated and adapted from the Russian by David J. Sookne and Timothy McLarnan. University of Chicago Press, 1974, p. 11.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 2..1000
EXAMPLE
a(4) = 6 because in the fourth row of Pascal's triangle, 1 and 6 are not multiples of 4, and 6 is the largest of those.
a(5) = 1 because in the fifth row all the other terms are multiples of 5.
MAPLE
a:= proc(n) local mx, t, i, r;
mx:=1;
t:=n;
for i from 2 to floor(n/2) do
t:= t* (n-i+1)/i;
if irem(t, n)>0 and t>mx then mx:=t fi
od; mx
end;
seq(a(n), n=2..100); # Alois P. Heinz, Jan 22 2011
MATHEMATICA
Table[Max[Select[Table[Binomial[n, m], {m, 0, n}], GCD[#, n] < n &]], {n, 2, 30}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alonso del Arte, Jan 21 2011
STATUS
approved