A180733 Largest element of n-th row of Pascal's triangle that is not a multiple of n.

1, 1, 6, 1, 20, 1, 70, 84, 252, 1, 495, 1, 3432, 5005, 12870, 1, 48620, 1, 184756, 293930, 705432, 1, 2704156, 3268760, 10400600, 17383860, 40116600, 1, 145422675, 1, 601080390, 193536720, 2333606220, 2319959400, 9075135300, 1

Offset: 2

Author

Alonso del Arte, Jan 21 2011

Keywords

• nonn
• easy

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.

Examples

			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.
		

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

Crossrefs

Cf. A007318, A080211 Binomial(n, smallest prime factor of n).

Programs

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}]