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%I #22 Aug 04 2017 14:33:18
%S 2,420,920,1122,1218,1892,1978,2444,2914,3198,3782,4028,4136,4292,
%T 4664,4958,4960,5330,5762,5986,6020,6032,6710,6834,6864,6882,6954,
%U 6956,6968,7106,7130,7140,7238,7254,7448,7616,8178,8190,8400,8692,9462,9506,10712,11060,11288
%N Numbers k such that A000984(k) mod k = 0 and A080383(k) != 7.
%C Numbers arising in A067348 and not present in A080385.
%C Even numbers n such that n divides binomial(n, [n/2]) and A010551(n) does not divide j!*(n-j)! exactly 7 times for j = 0..n. - _Peter Luschny_, Aug 04 2017
%H David A. Corneth, <a href="/A080392/b080392.txt">Table of n, a(n) for n = 1..274</a> (Terms <= 60000)
%e A080383(2) = 3;
%e A080383(420) = 11;
%e A080383(920) = 11;
%e A080383(1122) = 9;
%e A080383(1218) = 9.
%p isa := proc(n) local bn, bm;
%p if n mod 2 = 0 then bn := binomial(n, iquo(n,2)):
%p if modp(bn, n) = 0 then
%p bm := (n, j) -> `if`(modp(bn, binomial(n, j)) = 0, 1, 0):
%p return 1 <> add(bm(n, j), j=2..iquo(n,2)-1)
%p fi fi; false end:
%p select(isa, [$1..5000]); # _Peter Luschny_, Aug 04 2017
%t Do[s=Count[Table[IntegerQ[Binomial[n, Floor[n/2]]/ Binomial[n, j]], {j, 0, n}], True]; s1=IntegerQ[Binomial[n, n/2]/n]; If[ !Equal[s, 7] && Equal[s1, True], Print[n]], {n, 1, 10000}]
%t (* Second program: *)
%t Select[Range@ 5000, Function[n, And[Divisible[Binomial[n, n/2], n], Count[Table[Divisible[Binomial[n, Floor[n/2]], Binomial[n, j]], {j, 0, n}], True] != 7]]] (* _Michael De Vlieger_, Jul 30 2017 *)
%Y Cf. A000984, A001405, A067348, A080383, A080385.
%K nonn
%O 1,1
%A _Labos Elemer_, Mar 17 2003
%E More terms from _Michael De Vlieger_, Jul 30 2017
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