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%I #24 Nov 14 2014 10:54:50
%S 1,1,6,1,20,1,70,84,252,1,495,1,3432,5005,12870,1,48620,1,184756,
%T 293930,705432,1,2704156,3268760,10400600,17383860,40116600,1,
%U 145422675,1,601080390,193536720,2333606220,2319959400,9075135300,1
%N Largest element of n-th row of Pascal's triangle that is not a multiple of n.
%C 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.
%C 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.
%C 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.
%D 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.
%H Alois P. Heinz, <a href="/A180733/b180733.txt">Table of n, a(n) for n = 2..1000</a>
%e 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.
%e a(5) = 1 because in the fifth row all the other terms are multiples of 5.
%p a:= proc(n) local mx, t, i, r;
%p mx:=1;
%p t:=n;
%p for i from 2 to floor(n/2) do
%p t:= t* (n-i+1)/i;
%p if irem(t,n)>0 and t>mx then mx:=t fi
%p od; mx
%p end;
%p seq(a(n), n=2..100); # _Alois P. Heinz_, Jan 22 2011
%t Table[Max[Select[Table[Binomial[n, m], {m, 0, n}], GCD[#, n] < n &]], {n, 2, 30}]
%Y Cf. A007318, A080211 Binomial(n, smallest prime factor of n).
%K nonn,easy
%O 2,3
%A _Alonso del Arte_, Jan 21 2011
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