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A156050
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Triangle T(n,m) = binomial(n,m)+2*P(n,m) read by rows, where P(n,m) = 1+A000041(n)-A000041(m)-A000041(n-m).
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3
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1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 8, 10, 8, 1, 1, 9, 16, 16, 9, 1, 1, 14, 25, 32, 25, 14, 1, 1, 15, 35, 51, 51, 35, 15, 1, 1, 22, 48, 82, 96, 82, 48, 22, 1, 1, 25, 64, 118, 164, 164, 118, 64, 25, 1, 1, 34, 83, 170, 264, 310, 264, 170, 83, 34, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,5
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COMMENTS
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The element-by-element sum of: the Pascal triangle A007318 plus two times the elements of P(n,m).
Row sums are 2^n+2*( (n+1)*(1+A000041(n)) -2*A000070(n) ), starting 1, 2, 6, 12, 28, 52, 112, 204, 402, 744, 1414,..., always below factorial(n+1).
The remarkable thing about this sub-Eulerian numbers triangle is that it a Sierpinski gasket modulo 2.
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LINKS
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Table of n, a(n) for n=0..65.
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EXAMPLE
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P(n,m) starts in row n= 0 as
0
0, 0
0, 1, 0
0, 1, 1, 0
0, 2, 2, 2, 0
0, 2, 3, 3, 2, 0
0, 4, 5, 6, 5, 4, 0
0, 4, 7, 8, 8, 7, 4, 0
0, 7, 10, 13, 13, 13, 10, 7, 0
0, 8, 14, 17, 19, 19, 17, 14, 8, 0
0, 12, 19, 25, 27, 29, 27, 25, 19, 12, 0
to yield T(n,m) from row n=0 on:
1,
1, 1,
1, 4, 1,
1, 5, 5, 1,
1, 8, 10, 8, 1,
1, 9, 16, 16, 9, 1,
1, 14, 25, 32, 25, 14, 1,
1, 15, 35, 51, 51, 35, 15, 1,
1, 22, 48, 82, 96, 82, 48, 22, 1,
1, 25, 64, 118, 164, 164, 118, 64, 25, 1,
1, 34, 83, 170, 264, 310, 264, 170, 83, 34, 1
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MATHEMATICA
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Clear[f];
t[n_, m_] = 1 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m]; \! Table[(Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]])/2, {n, 0, 10}];
Table[Table[Binomial[n, m], {m, 0, n}] + (Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]]), {n, 0, 10}];
Flatten[%]
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CROSSREFS
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Sequence in context: A204621 A146770 A143334 * A136489 A166455 A171142
Adjacent sequences: A156047 A156048 A156049 * A156051 A156052 A156053
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KEYWORD
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nonn,tabl,easy
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AUTHOR
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Roger L. Bagula, Feb 02 2009
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EXTENSIONS
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Row sum formula and P(n,m) examples added - The Assoc. Eds. of the OEIS, Aug 29 2010
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STATUS
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approved
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