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A136190
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The 4rd order Zeckendorf array, T(n,k), read by antidiagonals.
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1
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1, 2, 6, 3, 9, 8, 4, 13, 12, 11, 5, 18, 17, 16, 15, 7, 24, 23, 22, 21, 20, 10, 33, 31, 30, 29, 28, 25, 14, 46, 43, 41, 40, 39, 35, 27, 19, 64, 60, 57, 55, 54, 49, 38, 32, 26, 88, 83, 79, 76, 74, 68, 53, 45, 34, 36, 121, 114, 109, 105, 102, 93, 73, 63, 48, 37, 50, 167, 157, 150
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Rows satisfy this recurrence: T(n,k)=T(n,k-1)+T(n,k-4) for all k>=5. Except for initial terms, (row 1) = A003269 (row 2) = A014101. 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
| C. Kimberling, "The Zeckendorf array equals the Wythoff array," Fibonacci Quarterly 33 (1995) 3-8.
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FORMULA
| Row 1 is the 4rd order Zeckendorf basis, given by initial terms b(1)=1,b(2)=2,b(3)=3,b(4)=4 and recurrence b(k)=b(k-1)+b(k-4) for k>=5 Every positive integer has a unique 4-Zeckendorf representation: n=b(i(1))+b(i(2))+...+b(i(p)), where |i(h)-i(j))>=4. 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
| Northwest corner:
1 2 3 4 5 7 10 14 ...
6 9 13 18 24 33 46 64 ...
8 12 17 23 31 43 60 83 ...
11 16 22 30 41 57 79 109 ...
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CROSSREFS
| Cf. A003269, A134564.
Sequence in context: A016637 A133917 A078340 * A079297 A143219 A109465
Adjacent sequences: A136187 A136188 A136189 * A136191 A136192 A136193
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KEYWORD
| nonn,tabl
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu), Dec 20 2007
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