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A294306
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Irregular triangle read by rows: T(n, m) = total number of times the different values appear in row n of A280269, where 0 <= m <= A280274(n).
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1
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1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1, 1, 1, 5, 2, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 4, 1, 1, 1, 5, 2, 1, 1, 1, 1, 1, 5, 2, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 7, 3, 1, 2, 1, 3, 1, 1, 1, 1, 3, 1, 5, 2, 1, 1, 1, 7, 6, 3, 1, 1, 1, 1, 5
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OFFSET
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1,7
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COMMENTS
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The numbers i in A162306(n) divide n^k with k >= 0; these k are listed in row n of A280269.
Row 1 = 1 and T(n, 0) = 1 for all n, since 1 is the empty product and divides n^0.
Row p = 1, 1, (row length = 2) since the only divisors of p are 1 and p; 1 | p^0, and p | p^1.
Row p^e = 1, e, since the only numbers in A162306(p^e) are 1 and p^k for 1 <= k <= e.
Row length of a(n) > 2 for n with omega(n) > 1.
Sum of terms of T(n, m) with m <= 1 in row n = A000005(n).
Sum of terms of T(n, m) with m > 1 = A243822(n).
Terms in row n of A294306 start at 1, generally quickly rise to a maximum, then gradually decline at m = A280274(n).
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LINKS
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EXAMPLE
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Row n of A280269(10) = 0, 1, 2, 1, 3, 1, corresponding to A162306(10) = 1, 2, 4, 5, 8, 10, since 1 | 10^0, 2 | 10^1, 4 | 10^2, 5 | 10^1, 8 | 10^3, and 10 | 10^1. There is 1 zero, 3 ones, 1 two, and 1 three, thus a(10) = 1, 3, 1, 1. sum(a(10)) = A010846(10) = 6. Length of a(10) = A280274(10) + 1 = 4.
Triangle begins:
1: 1
2: 1 1
3: 1 1
4: 1 2
5: 1 1
6: 1 3 1
7: 1 1
8: 1 3
9: 1 2
10: 1 3 1 1
11: 1 1
12: 1 5 2
13: 1 1
14: 1 3 1 1
15: 1 3 1
16: 1 4
17: 1 1
18: 1 5 2 1 1
19: 1 1
20: 1 5 2
...
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MATHEMATICA
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Table[Tally[#][[All, -1]] &@ Map[SelectFirst[Range[0, Floor@ Log2@ n], Function[k, Divisible[n^k, #]]] &, Select[Range@ n, PowerMod[n, Floor@ Log2@ n, #] == 0 &]], {n, 32}] // Flatten (* Michael De Vlieger, Oct 30 2017 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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