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Search: id:A136189
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| A136189 |
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The 3rd order Zeckendorf array, T(n,k), read by antidiagonals. |
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+0
5
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| 1, 2, 5, 3, 8, 7, 4, 12, 11, 10, 6, 17, 16, 15, 14, 9, 25, 23, 22, 21, 18, 13, 37, 34, 32, 31, 27, 20, 19, 54, 50, 47, 45, 40, 30, 24, 28, 79, 73, 69, 66, 58, 44, 36, 26, 41, 116, 107, 101, 97, 85, 64, 53, 39, 29, 60, 170, 157, 148, 142, 125, 94, 77, 57, 43, 33, 88, 249, 230
(list; table; graph; listen)
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OFFSET
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1,2
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COMMENT
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Rows satisfy this recurrence: T(n,k)=T(n,k-1)+T(n,k-3) for all k>=4. Except for initial terms, (row 1) = A000930 (column 1) = A020942 (column 2) = A064105 (column 3) = A064106. As a sequence, the array is a permutation of the natural numbers. As an array, T is an interspersion (hence also a dispersion).
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REFERENCES
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C. Kimberling, "The Zeckendorf array equals the Wythoff array," Fibonacci Quarterly 33 (1995) 3-8.
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FORMULA
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Row 1 is the 3rd order Zeckendorf basis, given by initial terms b(1)=1,b(2)=2,b(3)=3 and recurrence b(k)=b(k-1)+b(k-3) for k>=4. Every positive integer has a unique 3-Zeckendorf representation: n=b(i(1))+b(i(2))+...+b(i(p)), where |i(h)-i(j))>=3. Rows of T are defined inductively: T(n,1) is the least positive integer not in an earlier row. T(n,2) is obtained from T(n,1) as follows: if T(n,1)=b(i(1))+b(i(2))+...+b(i(p)), then T(n,k+1)=b(i(1+k))+b(i(2+k))+...+b(i(p+k)) for k=1,2,3... .
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EXAMPLE
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Northwest corner:
1 2 3 4 6 9 13 19 ...
5 8 12 17 25 37 54 79 ...
7 11 16 23 34 50 73 107 ...
10 15 22 32 47 69 101 148 ...
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CROSSREFS
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Cf. A035513, A134563, A136175.
Sequence in context: A063955 A097594 A050171 this_sequence A081146 A082652 A065222
Adjacent sequences: A136186 A136187 A136188 this_sequence A136190 A136191 A136192
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KEYWORD
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nonn,tabl
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu), Dec 20 2007
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