%I #10 Oct 06 2019 18:56:08
%S 10,36,78,120,136,165,210,300,406,462,528,666,820,924,969,990,1140,
%T 1176,1378,1596,1716,1830,2080,2346,2380,2628,2926,3060,3240,3276,
%U 3570,3654,3876,3916,4278,4656,5050,5460,5886,6328,6786,7260,7750,7770,8256,8436
%N Numbers k such that binomial(d(i),d(j)) = k for some proper divisors d(i) > d(j) of k.
%C The corresponding pairs of divisors (d(i), d(j)) are (5, 2), (9, 2), (13, 2), (10, 3), (17, 2), (11, 3), {(21, 2) and (10, 6)}, (25, 2), (29, 2), (11, 6), ...
%C We observe that the term 210 of the sequence generates two pairs of divisors having the property that binomial(d(i),d(j)) = 210 (see the example).
%H Robert Israel, <a href="/A328146/b328146.txt">Table of n, a(n) for n = 1..10000</a>
%e 10 is in the sequence because the proper divisors of 10 are {2, 5} and binomial(5, 2) = 10.
%e 210 is in the sequence because the proper divisors of 210 are { 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105} with binomial(21, 2) = 210 and binomial(10, 6) = 210 where 2, 6, 10, 21 are divisors of 210.
%p with(numtheory):
%p for n from 3 to 10000 do:
%p ii:=0:d:=divisors(n):n0:=nops(d):
%p for i from n0-1 by -1 to 2 while (ii=0) do:
%p for j from i-1 by -1 to 2 while(ii=0) do:
%p f:=binomial(d[i],d[j]):
%p if f=n
%p then
%p ii:=1:printf(`%d, `,n):
%p else
%p fi:
%p od:
%p od:
%p od:
%p # Alternative:
%p N:= 10^6: # for all terms <= N
%p S:= {}:
%p for n from 2 while (n+1)*(n+2)/2 <= N do
%p for m from n+2 do
%p t:= binomial(m,n);
%p if t > N then break fi;
%p if t mod n = 0 and t mod m = 0 and t > m then S:= S union {t} fi
%p od;
%p od:
%p sort(convert(S,list)); # _Robert Israel_, Oct 06 2019
%o (PARI) isok(m) = {my(d=divisors(m)); for (i=1, #d-1, for (j=1, i-1, if (binomial(d[i], d[j]) == m, return(1));););} \\ _Michel Marcus_, Oct 05 2019
%Y Cf. A007318, A027751, A032741
%K nonn
%O 1,1
%A _Michel Lagneau_, Oct 05 2019