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

2,3

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

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

Sequence in context: A167580 A080213 A200091 * A334504 A161151 A146383

Adjacent sequences: A180730 A180731 A180732 * A180734 A180735 A180736

Keyword

nonn,easy

Author

Alonso del Arte, Jan 21 2011

Status

approved