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A198680 Multiples of 3 whose sum of base-3 digits are also multiples of 3.

%I A198680 #31 Feb 16 2025 08:33:16
%S A198680 0,15,21,33,39,45,57,63,78,87,93,99,111,117,132,135,150,156,165,171,
%T A198680 186,189,204,210,222,228,234,249,255,261,273,279,294,297,312,318,327,
%U A198680 333,348,351,366,372,384,390,396,405,420,426,438,444,450,462,468,483,489,495
%N A198680 Multiples of 3 whose sum of base-3 digits are also multiples of 3.
%C A198680 It appears that Sum[k^j, 0<=k<=2^n-1, k in A198680] = Sum[k^j, 0<=k<=2^n-1, k in A198681] = Sum[k^j, 0<=k<=2^n-1, k in A180682], for 0<=j<=n-1, which has been verified numerically in a number of cases. This is a generalization of Prouhet's Theorem (see the reference).  To illustrate for j=3, we have  Sum[k^3, 0<=k<=2^n-1, k in A198680] = {0, 0, 12636, 1108809, 94478400, 7780827681, 633724260624, 51425722195929, 4168024588857600,...}, Sum[k^3, 0<=k<=2^n-1, k in A198681] =  {0, 27, 14580, 1095687, 94478400, 7780827681, 633724260624, 51425722195929, 4168024588857600,..., Sum[k^3, 0<=k<=2^n-1, k in A198682] = {0, 216, 7776, 1121931, 94478400, 7780827681, 633724260624, 51425722195929, 4168024588857600,...}, and it is seen that all three sums agree for n>=4=j+1.
%H A198680 Amiram Eldar, <a href="/A198680/b198680.txt">Table of n, a(n) for n = 1..10000</a>
%H A198680 Chris Bernhardt, <a href="http://www.jstor.org/stable/27643161">Evil twins alternate with odious twins</a>, Math. Mag. 82 (2009), pp. 57-62.
%H A198680 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Prouhet-Tarry-EscottProblem.html">Prouhet-Tarry-Escott Problem</a>
%F A198680 a(n) = 3*A079498(n). - _Charles R Greathouse IV_, Nov 02 2011
%t A198680 Select[3*Range[0,200],Divisible[Total[IntegerDigits[#,3]],3]&] (* _Harvey P. Dale_, May 31 2014 *)
%Y A198680 Cf. A000069, A001969, A157971, A157970, A198681, A198682.
%K A198680 nonn,easy,base
%O A198680 1,2
%A A198680 _John W. Layman_, Oct 28 2011
%E A198680 Offset corrected by _Amiram Eldar_, Jan 05 2020