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Search: id:A134285
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| A134285 |
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Triangle of numbers obtained from the partition array A134284. |
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5
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| 1, 3, 1, 10, 3, 1, 35, 19, 3, 1, 126, 65, 19, 3, 1, 462, 331, 92, 19, 3, 1, 1716, 1190, 421, 92, 19, 3, 1, 6435, 5587, 1805, 502, 92, 19, 3, 1, 24310, 20613, 8771, 2075, 502, 92, 19, 3, 1, 92378, 92821, 35726, 10616, 2318, 502, 92, 19, 3, 1, 352716, 347930, 160205
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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This triangle is called s2(3)'.
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LINKS
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W. Lang, First 10 rows and more.
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FORMULA
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a(n,m)=sum(product(s2(3;j,1)^e(n,m,q,j),j=1..n),k=1..p(n,m)) if n>=m>=1, else 0. Here p(n,m)=A008284(n,m), the number of m parts partitions of n and e(n,m,q,j) is the exponent of j in the q-th m part partition of n. s2(3;n,1) = A035324(n,1) = A001700(n-1) = binomial(2*n-1,n).
Row sums = A001700. Triangle A134285 = A001263 * A000012 - Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 19 2007
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EXAMPLE
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[1];[3,1];[10,3,1];[35,19,3,1];[126,65,19,3,1];...
First few rows of the triangle are:
1;
2, 1;
5, 4, 1;
14, 13, 7, 1;
42, 41, 31, 11, 1;
132, 131, 116, 66, 16, 1;
429, 428, 407302, 127, 22, 1;
...
a(4,2)=19 because the m=2 parts partitions (1^1,3^1) and (2^2) of n=4 lead to 1^1*10^1 + 3^2 =19, since A001700(n-1)=[1,3,10,...], n>=1.
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CROSSREFS
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Row sums A134826. Alternating row sums A134827.
Cf. A001700.
Sequence in context: A135573 A126953 A134284 this_sequence A141811 A126954 A107870
Adjacent sequences: A134282 A134283 A134284 this_sequence A134286 A134287 A134288
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KEYWORD
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nonn,easy
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AUTHOR
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Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Nov 13 2007
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