%I
%S 3,10,21,36,55,78,105,136,171,210,253,300,351,406,465,18,595,666,741,
%T 820,903,990,1081,1176,1275,1378,1485,1596,1711,1830,1953,2080,2211,4,
%U 2485,2628,2775,12,3081,3240,3403,3570,3741,3916,4095,4278,4465,4656,4851,60
%N Least number k such that Sum_{j=k..k+n1}{j^2} = Sum_{j=k+n..t}{j^2}, for some t >= k+n.
%C With n = 17 consecutive numbers we can start from k = 18 but also from k = 528. The sequence considers only the least number: a(17) = 18.
%C In general t = k + 2*(n1) but sometimes it differs, e.g., for n = 17, 35, 39, 51, 93, 127, 382, etc.
%H Paolo P. Lava, <a href="/A281153/a281153.txt">First 250 terms with values for n, k and t</a>
%e a(2) = 3 because 3^2 + 4^2 = 5^2 and 3 is the least number to have this property;
%e a(3) = 10 because 10^2 + 11^2 + 12^2 = 13^2 + 14^2 and 10 is the least number to have this property.
%e a(4) = 21 because 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2 and 21 is the least number to have this property.
%e a(5) = 36 because 36^2 + 37^2 + 38^2 + 39^2 + 40^2 = 41^2 + 42^2 + 43^2 + 44^2 and 36 is the least number to have this property.
%p P:=proc(q,h) local a,b,c,j,k,n; for n from 2 to q do for k from 1 to q do a:=add(j^h,j=k..k+n1); b:=0;
%p c:=k+n1; while b<a do c:=c+1; b:=b+c^h; od; if a=b then print(k); break; fi; od; od; end: P(10^6,2);
%Y Cf. A014105, A059255, A281152.
%K nonn,easy
%O 2,1
%A _Paolo P. Lava_, Jan 16 2017
