A198682 -id:A198682 - cp's OEIS frontend

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

John W. Layman, Oct 28 2011

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.

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

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

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

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

Offset corrected by Amiram Eldar, Jan 05 2020

A198681 Nonnegative multiples of 3 whose sum of base-3 digits are of the form 3k+1.

Original entry on oeis.org

3, 9, 24, 27, 42, 48, 60, 66, 72, 81, 96, 102, 114, 120, 126, 138, 144, 159, 168, 174, 180, 192, 198, 213, 216, 231, 237, 243, 258, 264, 276, 282, 288, 300, 306, 321, 330, 336, 342, 354, 360, 375, 378, 393, 399, 408, 414, 429, 432, 447, 453, 465, 471, 477, 492, 498

Offset: 1

John W. Layman, Oct 28 2011

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.

Michel Marcus, Table of n, a(n) for n = 1..1001 (after Harvey P. Dale).
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

Cf. A000069, A001969, A157971, A158704, A198680, A198682.

Select[3Range[200],IntegerQ[(Total[IntegerDigits[#,3]]-1)/3]&] (* Harvey P. Dale, Feb 05 2012 *)

Offset corrected by Michel Marcus, Mar 02 2016

cp's OEIS Frontend

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

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