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Triangular numbers with arithmetic mean of digits = 1 (sum of digits = number of digits).
2

%I #18 Jul 28 2019 19:32:20

%S 1,120,210,300,112101,100600020,101111310,110120220,200130021,

%T 200310120,1000051003,1010004040,1130002030,1411000003,2002021003,

%U 3200200003,5000050000,100110002070,111111101310,111202101003,180000300000,211104100200,231201020001,500001500001,501001000500,100021000424010

%N Triangular numbers with arithmetic mean of digits = 1 (sum of digits = number of digits).

%C The sum of the digits of a triangular number is 0, 1, 3 or 6 (mod 9).

%C From _Robert Israel_, Aug 24 2018: (Start)

%C Suppose A007953(x) + A007953(2*x^2) - A055642(2*x^2) is even and

%C A007953(x) + A007953(2*x^2) >= 2*A055642(x) + A055642(2*x^2).

%C Then 10^k*x*(1+2*10^k*x) is in the sequence, where k = (A007953(x) + A007953(2*x^2) - A055642(2*x^2))/2.

%C In particular, x = 10^j-2 satisfies this criterion for all j>=1, with k = j. Thus the sequence is infinite. - _Robert Israel_, Aug 24 2018

%H Jon E. Schoenfield, <a href="/A069790/b069790.txt">Table of n, a(n) for n = 1..341</a> (all terms < 10^23)

%p T:= proc(n,k) option remember;

%p if n*9 < k then return {} fi;

%p if n = 1 then return {k} fi;

%p `union`(seq(map(t -> 10*t+j, procname(n-1,k-j)),j=0..min(9,k)))

%p end proc:

%p T(1,0):= {}:

%p sort(convert(select(t -> issqr(8*t+1), `union`(seq(seq(T(9*i+j,9*i+j),j=[0,1,3,6]),i=0..1))),list)); # _Robert Israel_, Aug 24 2018

%t s=Select[Range[500000], Length[z=IntegerDigits[ #(#+1)/2]]==Plus@@z&]; s(s+1)/2

%t Select[Accumulate[Range[500000]],Mean[IntegerDigits[#]]==1&] (* _Harvey P. Dale_, May 05 2011 *)

%Y Cf. A007953, A055642.

%K base,nonn

%O 1,2

%A _Amarnath Murthy_, Apr 08 2002

%E Edited by _Dean Hickerson_ and _Robert G. Wilson v_, Apr 10 2002

%E More terms from _Robert Israel_, Aug 24 2018