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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).
1

%I #16 Jan 06 2018 18:44:33

%S 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,

%T 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,

%U 1,1,1,3,1,5,2,1,1,1,7,6,3,1,1,1,1,5

%N 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).

%C The numbers i in A162306(n) divide n^k with k >= 0; these k are listed in row n of A280269.

%C Row 1 = 1 and T(n, 0) = 1 for all n, since 1 is the empty product and divides n^0.

%C Row p = 1, 1, (row length = 2) since the only divisors of p are 1 and p; 1 | p^0, and p | p^1.

%C Row p^e = 1, e, since the only numbers in A162306(p^e) are 1 and p^k for 1 <= k <= e.

%C Row length of a(n) > 2 for n with omega(n) > 1.

%C Total of row n = A010846(n).

%C Sum of terms of T(n, m) with m <= 1 in row n = A000005(n).

%C Sum of terms of T(n, m) with m > 1 = A243822(n).

%C Terms in row n of A294306 start at 1, generally quickly rise to a maximum, then gradually decline at m = A280274(n).

%H Michael De Vlieger, <a href="/A294306/b294306.txt">Table of n, a(n) for n = 1..11355</a> (rows 1 <= n <= 2000).

%H Michael De Vlieger, <a href="/A294306/a294306.txt">A294306 and indices of records in A294306</a>.

%e 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.

%e Triangle begins:

%e 1: 1

%e 2: 1 1

%e 3: 1 1

%e 4: 1 2

%e 5: 1 1

%e 6: 1 3 1

%e 7: 1 1

%e 8: 1 3

%e 9: 1 2

%e 10: 1 3 1 1

%e 11: 1 1

%e 12: 1 5 2

%e 13: 1 1

%e 14: 1 3 1 1

%e 15: 1 3 1

%e 16: 1 4

%e 17: 1 1

%e 18: 1 5 2 1 1

%e 19: 1 1

%e 20: 1 5 2

%e ...

%t 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 *)

%Y Cf. A000005, A010846, A162306, A243822, A280269, A280274.

%K nonn,tabf

%O 1,7

%A _Michael De Vlieger_, Oct 30 2017