%I #22 Jul 25 2022 18:32:39
%S 1,6,22,72,114,148,164,260,261,780,1078,1184,1266,2952,4674,21868
%N Numbers m where m^2 divides A159553(m), where A159553(m) = Sum_{k=0..m} binomial(m,k) * gcd(m,k).
%C For the purpose of this sequence, gcd(m,0) = m.
%C No other term up to 15000. - _Michel Marcus_, Sep 06 2019
%p A159068 := proc(n) option remember; add(binomial(n, k)*gcd(k, n), k=1..n) ; end: A159553 := proc(n) option remember ; A159068(n)+n; end: isA159555 := proc(n) if A159553(n) mod ( n^2) = 0 then true; else false; fi; end: for n from 1 do if isA159555(n) then printf("%d,\n",n) ; fi; od: # _R. J. Mathar_, Apr 29 2009
%o (PARI) f(n) = sum(k=0, n, binomial(n,k) * gcd(n,k)); \\ A159553
%o isok(n) = !(f(n) % n^2); \\ _Michel Marcus_, Sep 05 2019
%Y Cf. A159458, A159553, A159554.
%K nonn,more
%O 1,2
%A _Leroy Quet_, Apr 15 2009
%E Extended by _R. J. Mathar_, Apr 29 2009
%E a(14)-a(15) from _Ray Chandler_, Jun 18 2009
%E a(16) from _Jinyuan Wang_, Jul 25 2022