A153290 Designed symmetrical sequence with 2*3^n row sum and term: row(n)=3^n; f(n,m) = Floor[(m/Prime[n])*row(n)/2].

2, 3, 3, 2, 14, 2, 2, 25, 25, 2, 2, 10, 138, 10, 2, 2, 22, 219, 219, 22, 2, 2, 56, 112, 1118, 112, 56, 2, 2, 128, 256, 1801, 1801, 256, 128, 2, 2, 344, 690, 1034, 8982, 1034, 690, 344, 2, 2, 854, 1710, 2566, 14551, 14551, 2566, 1710, 854, 2, 2, 2036, 4072, 6108

Offset

0,1

Comments

This kind of 2*3^n row sum sequence is an effort to get the Sierpinski carpet scale three

level of symmetry into a triangular/ binomial like sequence.

Links

Table of n, a(n) for n=0..58.

Formula

row(n)=3^n;

f(n,m) = Floor[(m/Prime[n])*row(n)/2].

Example

{2},
{3, 3},
{2, 14, 2},
{2, 25, 25, 2},
{2, 10, 138, 10, 2},
{2, 22, 219, 219, 22, 2},
{2, 56, 112, 1118, 112, 56, 2},
{2, 128, 256, 1801, 1801, 256, 128, 2},
{2, 344, 690, 1034, 8982, 1034, 690, 344, 2},
{2, 854, 1710, 2566, 14551, 14551, 2566, 1710, 854, 2},
{2, 2036, 4072, 6108, 8144, 77374, 8144, 6108, 4072, 2036, 2}

Mathematica

Clear[v, n, row, f]; row[n_] = 3^n;
f[n_, m_] = Floor[(m/Prime[n])*row[n]/2]; v[0] = {1}; v[1] = {3/2, 3/2};
v[n_] := v[n] = If[Mod[n, 2] == 0, Join[{1}, Table[ f[n, m], {m, 1, Floor[n/2] - 1}], {row[n] - 2*Sum[ f[n, m], {m, 1, Floor[n/2] - 1}] - 2}, Table[ f[n, m], {m, Floor[n/2] - 1, 1, -1}], {1}],
Join[{1}, Table[ f[n, m], {m, 1, Floor[n/2] - 1}], {row[n]/2 - Sum[ f[n, m], {m, 1, Floor[n/2] - 1}] - 1, row[n]/2 - Sum[ f[n, m], {m, 1, Floor[n/2] - 1}] - 1}, Table[ f[n, m], {m, Floor[n/2] - 1, 1, -1}], {1}]];
Table[FullSimplify[Floor[2*v[n]]], {n, 0, 10}];
Flatten[%]

Crossrefs

Sequence in context: A175393 A184829 A338307 * A153516 A153311 A153312

Adjacent sequences: A153287 A153288 A153289 * A153291 A153292 A153293

Keyword

nonn,uned,tabl

Author

Roger L. Bagula, Dec 23 2008

Status

approved