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A287692 Triangle read by rows: T(n,k) is the greatest difference between prime factors among squarefree numbers A002110(n) <= m <= (A002110(n+1)-1) such that A001221(m) = n and m is divisible by A002110(k). 0

%I #9 Jul 14 2017 22:23:54

%S 3,2,5,2,3,9,2,3,5,18,2,2,4,7,30,2,2,3,5,10,42,2,2,3,4,6,13,60,2,2,3,

%T 4,5,8,17,77,2,2,3,3,4,6,10,22,113,2,2,2,3,4,5,8,12,25,145,2,2,2,3,4,

%U 5,6,9,15,32,179,2,2,2,3,4,4,6,7,11,19,36,229

%N Triangle read by rows: T(n,k) is the greatest difference between prime factors among squarefree numbers A002110(n) <= m <= (A002110(n+1)-1) such that A001221(m) = n and m is divisible by A002110(k).

%C Let p_n# = A002110(n).

%C T(n,1) is the greatest index of the smallest prime divisor p of terms m in row n.

%C T(n,n) = A120941(n).

%C Consider the use of A287352 as a method for formulating squarefree numbers with n distinct prime factors. The values in row n serve as a limit beyond which we need not search further for terms p_n# <= m <= (p_(n+1)# - 1). A287352 defines a squarefree number using a sequence of nonzero positive terms, beginning with the index of the smallest prime factor, then listing differences between indexes of subsequent prime factors in order of their magnitude. We can direct increment to the largest prime index as long as the number m < p_(n+1), then increment the index before it, etc. to produce the entire tree of factors that code numbers m.

%e Triangle begins:

%e n\k| 1 2 3 4 5 6 7 8 9 10 11 12

%e ---------------------------------------------------------

%e 1 | 3

%e 2 | 2 5

%e 3 | 2 3 9

%e 4 | 2 3 5 18

%e 5 | 2 2 4 7 30

%e 6 | 2 2 3 5 10 42

%e 7 | 2 2 3 4 6 13 60

%e 8 | 2 2 3 4 5 8 17 77

%e 9 | 2 2 3 3 4 6 10 22 113

%e 10 | 2 2 2 3 4 5 8 12 25 145

%e 11 | 2 2 2 3 4 5 6 9 15 32 179

%e 12 | 2 2 2 3 4 4 6 7 11 19 36 229

%e ...

%e Let p_n# = A002110(n). For n = 2, there are A287484(2) = 7 squarefree numbers p_2# <= m <= (p_3# - 1) such that omega(m) = n. These are {6, 10, 14, 22, 26, 15, 21}. These numbers m have A287352(m) = {{1,1}, {1,2}, {1,3}, {1,4}, {1,5}, {2,1}, {2,2}} respectively; the largest values in both positions are {2,5}, thus row n = 2 of a(n) is {2,5}.

%t f[n_] := If[n == 0, {{1}}, Block[{P = Product[Prime@ i, {i, n}], lim, k = 1, c, w = ConstantArray[1, n]}, lim = Prime[n + 1] P; {w}~Join~Reap[Do[w = If[k == 1, MapAt[# + 1 &, w, -k], Join[Drop[MapAt[# + 1 &, w, -k], -k + 1], ConstantArray[1, k - 1]]]; c = Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ w]; If[c < lim, Sow[w]; k = 1, If[k == n, Break[], k++]], {i, Infinity}] ][[-1, 1]] ] ]; Table[Max /@ Transpose@ f@ n, {n, 14}] // Flatten (* _Michael De Vlieger_, Jun 15 2017 *)

%Y Cf. A001221, A002110, A120941, A287483, A287484, A287691.

%K nonn,tabl

%O 1,1

%A _Michael De Vlieger_, Jun 15 2017

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Last modified March 28 04:13 EDT 2024. Contains 371235 sequences. (Running on oeis4.)