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A297440
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Rectangular array R by antidiagonals: row n shows the positive integers whose base-3 digits have total variation n, for n>=0. See Comments.
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3
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1, 2, 3, 4, 5, 6, 8, 7, 10, 11, 13, 9, 16, 15, 20, 26, 12, 18, 19, 34, 33, 40, 14, 21, 29, 46, 47, 60, 80, 17, 23, 30, 56, 61, 100, 101, 121, 22, 24, 32, 57, 87, 142, 141, 182, 242, 25, 28, 35, 59, 92, 168, 181, 304, 303, 364, 27, 31, 38, 62, 96, 173, 263
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OFFSET
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1,2
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COMMENTS
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Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0). The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1). The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b). See A297330 for a guide to related sequences and partitions of the natural numbers.
Every positive integer occurs exactly once in the array, so that as a sequence this is a permutation of the positive integers.
Conjecture: each column, after some number of initial terms, satisfies the linear recurrence relation c(n) = c(n-1) + 2*c(n-2) - 2*c(n-3) + 3*c(n-4) - 3*c(n-5).
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LINKS
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EXAMPLE
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Northwest corner:
1 2 4 8 13 26 40 80
3 5 7 9 12 14 17 22
6 10 16 18 21 23 24 28
11 15 19 29 30 32 35 38
20 34 46 56 57 59 62 65
33 47 61 87 92 96 99 102
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MATHEMATICA
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a[n_, b_] := Differences[IntegerDigits[n, b]];
b = 3; z = 250000; t = Table[a[n, b], {n, 1, z}];
u = Map[Total, Map[Abs, t]]; p[n_] := Position[u, n];
TableForm[Table[Take[Flatten[p[n]], 15], {n, 0, 9}]]
v[n_, k_] := p[k - 1][[n]]
Table[v[k, n - k + 1], {n, 12}, {k, n, 1, -1}] // Flatten
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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