A134803 - OEIS

%I #3 Aug 24 2012 10:50:00

%S 3,8,119,288,4059,9800

%N Numbers n such that the sum of all numbers of the same parity <= n is equal to the sum of numbers of the opposite parity from n+1 to n+m, where m is odd and > 1.

%e 3 -> 1+3 = 4 = 4

%e 8 -> 2+4+6+8 = 20 = 9+11

%e 119 -> 1+3+5+...+119 = 3600 = 120+122+...+168

%p P:=proc(n) local a,k,i,s1,s2; for i from 1 by 1 to n do if 2*trunc(i/2)=i then s1:=sum('2*k','k'=1..(i/2)); else s1:=sum('2*k-1','k'=1..(i+1)/2); fi; a:=1; s2:=i+1; while s1>s2 do a:=a+2; s2:=s2+i+a; od; if s1=s2 then lprint(i,s1); fi; od; end: P(10000);

%Y Cf. A001109, A084068, A089895.

%K easy,nonn

%O 1,1

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jan 28 2008