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Search: id:A073265
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| A073265 |
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Table T(n,k) (listed antidiagonalwise in order T(1,1), T(2,1), T(1,2), T(3,1), T(2,2), ...) giving the number of ordered partitions of n into exactly k powers of 2. |
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+0
5
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| 1, 1, 0, 0, 1, 0, 1, 2, 0, 0, 0, 1, 1, 0, 0, 0, 2, 3, 0, 0, 0, 0, 2, 3, 1, 0, 0, 0, 1, 0, 4, 4, 0, 0, 0, 0, 0, 1, 6, 6, 1, 0, 0, 0, 0, 0, 2, 3, 8, 5, 0, 0, 0, 0, 0, 0, 2, 3, 13, 10, 1, 0, 0, 0, 0, 0, 0, 0, 6, 12, 15, 6, 0, 0, 0, 0, 0, 0, 0, 2, 6, 10, 25, 15, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 16, 31, 26, 7, 0, 0
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OFFSET
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1,8
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REFERENCES
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S. Lehr, J. Shallit and J. Tromp, On the vector space of the automatic reals, Theoret. Comput. Sci. 163 (1996), no. 1-2, 193-210.
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FORMULA
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T(0, k) = T(n, 0) = 0, T(n, k) = 0 if k > n, T(n, 1) = 1 if n = 2^m, 0 otherwise, and in other cases T(n, k) = Sum_{i=0..[log2(n-1)]} T(n-(2^i), k-1).
T(n, k)=coefficient of x^n in the formal power series (x+x^2+x^4+x^8+x^16+...)^k. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 04 2005
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EXAMPLE
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T(6,3) = 4 because there are four ordered partitions of 6 into 3 powers of 2, namely: 4+1+1, 1+4+1, 1+1+4 and 2+2+2, and it is recursively computed from T(5,2)+T(4,2)+T(2,2) = 2+1+1.
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CROSSREFS
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The first row is equal to the characteristic function of A000079, i.e. A036987 with offset 1 instead of 0, and the second row is A073267. The column sums give A023359. A073266 gives the upper triangular region of this array.
Adjacent sequences: A073262 A073263 A073264 this_sequence A073266 A073267 A073268
Sequence in context: A037188 A086079 A133703 this_sequence A025438 A030216 A106405
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KEYWORD
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nonn,tabl
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AUTHOR
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Antti Karttunen Jun 25 2002
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