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A176228
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A symmetrical triangle sequence:t(n,m)=Binomial[n, m] + Fibonacci[n] + 1
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0
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2, 3, 3, 3, 4, 3, 4, 6, 6, 4, 5, 8, 10, 8, 5, 7, 11, 16, 16, 11, 7, 10, 15, 24, 29, 24, 15, 10, 15, 21, 35, 49, 49, 35, 21, 15, 23, 30, 50, 78, 92, 78, 50, 30, 23, 36, 44, 71, 119, 161, 161, 119, 71, 44, 36, 57, 66, 101, 176, 266, 308, 266, 176, 101, 66, 57
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table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,1
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COMMENTS
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Row sums are:
{2, 6, 10, 20, 36, 68, 127, 240, 454, 862, 1640,...}.
The sequence is designed as an leading ones "adjustable" sequence
that will give Pascal's triangle.
Replacing Fibonacci[n] with any a(n) will still adjust back to the original
symmetrical triangle sequence.
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LINKS
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Table of n, a(n) for n=0..65.
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FORMULA
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t(n,m)=Binomial[n, m] + Fibonacci[n] + 1
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EXAMPLE
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{2},
{3, 3},
{3, 4, 3},
{4, 6, 6, 4},
{5, 8, 10, 8, 5},
{7, 11, 16, 16, 11, 7},
{10, 15, 24, 29, 24, 15, 10},
{15, 21, 35, 49, 49, 35, 21, 15},
{23, 30, 50, 78, 92, 78, 50, 30, 23},
{36, 44, 71, 119, 161, 161, 119, 71, 44, 36},
{57, 66, 101, 176, 266, 308, 266, 176, 101, 66, 57}
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MATHEMATICA
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t[n_, m_] = Binomial[n, m] + Fibonacci[n] + 1;
Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[%]
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CROSSREFS
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Sequence in context: A049837 A098201 A175239 * A129574 A130193 A084516
Adjacent sequences: A176225 A176226 A176227 * A176229 A176230 A176231
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KEYWORD
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nonn,tabl,uned
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AUTHOR
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Roger L. Bagula, Apr 12 2010
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STATUS
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approved
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