A214728 - OEIS

%I #16 Mar 14 2016 19:00:41

%S 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 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 15,107,109,111,8,10,117,2,121,24,125,1,129,131,19,135,25,139,6

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

%C a(n) is the number of consecutive integers starting from n needed to sum up to a perfect square.

%C Indices of 1's: A000290.

%C Indices of 2's: A046092(k), k!=A001108(m).

%C 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 Terms of A108269 never appear in a(n).

%H Harvey P. Dale, <a href="/A214728/b214728.txt">Table of n, a(n) for n = 0..1000</a>

%e a(2): 2+3+4 = 9, three summands, so a(2)=3.

%e a(3): 3+4+5+6+7 = 25, five summands, so a(3)=5.

%e a(12): 12+13 = 25, so a(12)=2.

%t 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 (C)

%o int main() {  // OK with GCC

%o   unsigned long long i, n, sum, sr;

%o   for (n=0; n<333; ++n) {

%o     for (sum=0, i=n; i==n || sr*sr!=sum; ++i)  sr=sqrt(sum+=i);

%o     printf("%llu, ", i-n);

%o   }

%o }

%Y Cf. A000290, A108269.

%K nonn,easy

%O 0,3

%A _Alex Ratushnyak_, Jul 27 2012