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

0, 15, 21, 33, 39, 45, 57, 63, 78, 87, 93, 99, 111, 117, 132, 135, 150, 156, 165, 171, 186, 189, 204, 210, 222, 228, 234, 249, 255, 261, 273, 279, 294, 297, 312, 318, 327, 333, 348, 351, 366, 372, 384, 390, 396, 405, 420, 426, 438, 444, 450, 462, 468, 483, 489, 495

Offset

1,2

Comments

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.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Chris Bernhardt, Evil twins alternate with odious twins, Math. Mag. 82 (2009), pp. 57-62.

Eric Weisstein's World of Mathematics, Prouhet-Tarry-Escott Problem

Formula

a(n) = 3*A079498(n). - Charles R Greathouse IV, Nov 02 2011

Mathematica

Select[3*Range[0, 200], Divisible[Total[IntegerDigits[#, 3]], 3]&] (* Harvey P. Dale, May 31 2014 *)

Crossrefs

Cf. A000069, A001969, A157971, A157970, A198681, A198682.

Sequence in context: A225709 A020162 A046404 * A300117 A214044 A177516

Adjacent sequences: A198677 A198678 A198679 * A198681 A198682 A198683

Keyword

nonn,easy,base

Author

John W. Layman, Oct 28 2011

Extensions

Offset corrected by Amiram Eldar, Jan 05 2020

Status

approved