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Smallest of three consecutive integers divisible respectively by three consecutive squares greater than 1.
2

%I #12 Sep 09 2019 11:56:46

%S 2223,5823,9423,13023,16623,20223,23823,27423,31023,32975,34623,38223,

%T 41823,45423,49023,52623,56223,59823,63423,67023,70623,74223,77075,

%U 77823,81423,85023,88623,92223,95823,99423,103023,106623,110223

%N Smallest of three consecutive integers divisible respectively by three consecutive squares greater than 1.

%H Amiram Eldar, <a href="/A178919/b178919.txt">Table of n, a(n) for n = 1..10000</a>

%e 5823 is a term as 5823, 5824 and 5825 are divisible by 9, 16 and 25 respectively.

%p with(numtheory):for n from 1 to 200000 do: k:=0:q:=floor(sqrt(n)):for m from

%p 2 to q do: p1:=m^2:p2:=(m+1)^2:p3:=(m+2)^2:if irem(n,p1)=0 and irem(n+1,p2)=0

%p and irem(n+2,p3)=0 and k=0 then k:=1:printf(`%d, `,n):else fi:od:od:

%t f[p_, e_] := p^Floor[e/2]; maxsq[n_] := Times@@ (f @@@ FactorInteger[n]); aQ[n_] := (s = maxsq[n]) > 1 && AnyTrue[Rest @ Divisors[s], Divisible[n+1, (#+1)^2] && Divisible[n+2, (#+2)^2] &]; Select[Range[50000], aQ] (* _Amiram Eldar_, Sep 09 2019 *)

%o (Sage) is_A178919 = lambda n: any(all(((k+x)**2).divides(n+x) for x in range(3)) for k in divisors(n) if k > 1) # _D. S. McNeil_, Dec 29 2010

%Y Cf. A178918.

%K nonn

%O 1,1

%A _Michel Lagneau_, Dec 29 2010