

A198680


Multiples of 3 whose sum of base3 digits are also multiples of 3.


5



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
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OFFSET

1,2


COMMENTS

It appears that Sum[k^j, 0<=k<=2^n1, k in A198680] = Sum[k^j, 0<=k<=2^n1, k in A198681] = Sum[k^j, 0<=k<=2^n1, k in A180682], for 0<=j<=n1, 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^n1, k in A198680] = {0, 0, 12636, 1108809, 94478400, 7780827681, 633724260624, 51425722195929, 4168024588857600,...}, Sum[k^3, 0<=k<=2^n1, k in A198681] = {0, 27, 14580, 1095687, 94478400, 7780827681, 633724260624, 51425722195929, 4168024588857600,..., Sum[k^3, 0<=k<=2^n1, 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. 5762.
Eric Weisstein's World of Mathematics, ProuhetTarryEscott 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



