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Sum of the digits of n*(n+1).
0

%I #6 Nov 21 2013 12:49:03

%S 0,2,6,3,2,3,6,11,9,9,2,6,12,11,3,6,11,9,9,11,6,12,11,12,6,11,9,18,11,

%T 15,12,20,12,6,11,9,9,11,15,12,11,12,15,20,18,9,11,15,12,11,12,15,20,

%U 18,18,11,15,12,11,12,15,20,18,9,11,15,12,20,21,15,20,9,18,11,15,12,20,12

%N Sum of the digits of n*(n+1).

%F a(n)=Sum_digits[n*(n+1)], with n>=0

%p P:=proc(n) local i,k,w; for i from 0 by 1 to n do w:=0;k:=i*(i+1); while k>0 do w:=w+k-(trunc(k/10)*10); k:=trunc(k/10); od; print(w); od; end: P(100);

%t Table[Total[IntegerDigits[n(n+1)]],{n,0,80}] (* _Harvey P. Dale_, May 14 2011 *)

%K easy,nonn,base

%O 0,2

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, May 03 2007