cp's OEIS Frontend

A214728 Least k such that n + (n+1) + ... + (n+k-1) is a square.

%I A214728 #16 Mar 14 2016 19:00:41
%S A214728 1,1,3,5,1,9,11,13,15,1,19,3,2,25,27,29,1,33,5,37,39,8,43,45,2,1,3,53,
%T A214728 55,57,59,61,9,65,67,6,1,8,75,11,2,81,83,5,87,9,13,3,95,1,99,101,18,
%U A214728 15,107,109,111,8,10,117,2,121,24,125,1,129,131,19,135,25,139,6
%N A214728 Least k such that n + (n+1) + ... + (n+k-1) is a square.
%C A214728 a(n) is the number of consecutive integers starting from n needed to sum up to a perfect square.
%C A214728 Indices of 1's: A000290.
%C A214728 Indices of 2's: A046092(k), k!=A001108(m).
%C A214728 If a(n) is bigger than previous terms then a(n)=n*2-1, for example a(5)=9 is bigger than previous maximum, and a(5)=5*2-1.
%C A214728 Terms of A108269 never appear in a(n).
%H A214728 Harvey P. Dale, <a href="/A214728/b214728.txt">Table of n, a(n) for n = 0..1000</a>
%e A214728 a(2): 2+3+4 = 9, three summands, so a(2)=3.
%e A214728 a(3): 3+4+5+6+7 = 25, five summands, so a(3)=5.
%e A214728 a(12): 12+13 = 25, so a(12)=2.
%t A214728 lks[n_]:=Module[{k=1},While[!IntegerQ[Sqrt[Total[Range[n,n+k-1]]]], k++];k]; lks/@Range[0,80] (* _Harvey P. Dale_, Mar 14 2016 *)
%o A214728 (C)
%o A214728 int main() {  // OK with GCC
%o A214728   unsigned long long i, n, sum, sr;
%o A214728   for (n=0; n<333; ++n) {
%o A214728     for (sum=0, i=n; i==n || sr*sr!=sum; ++i)  sr=sqrt(sum+=i);
%o A214728     printf("%llu, ", i-n);
%o A214728   }
%o A214728 }
%Y A214728 Cf. A000290, A108269.
%K A214728 nonn,easy
%O A214728 0,3
%A A214728 _Alex Ratushnyak_, Jul 27 2012