

A307540


Irregular triangle T(n,k) such that squarefree m with gpf(m) = prime(n) in each row are arranged according to increasing values of phi(m)/m.


5



1, 2, 6, 3, 30, 10, 15, 5, 210, 42, 70, 14, 105, 21, 35, 7, 2310, 330, 462, 66, 770, 110, 154, 1155, 22, 165, 231, 33, 385, 55, 77, 11, 30030, 2730, 4290, 6006, 390, 546, 858, 10010, 78, 910, 1430, 2002, 130, 15015, 182, 286, 1365, 2145, 26, 3003, 195, 273, 429
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OFFSET

0,2


COMMENTS

Row n contains m in A005117 such that A000720(A006530(m)) = n, sorted such that phi(m)/m increases as k increases.
Let m be the squarefree kernel A007947(m') of m'. We only consider squarefree m since phi(m)/m = phi(m')/m'. Let prime p  n and prime q be a nondivisor of n.
Since m is squarefree, we might encode the multiplicities of its prime divisors in a positional notation M that is finite at n significant digits. For example, m = 42 can be encoded reverse(A067255(42)) = 1,0,1,1 = 7^1 * 5^0 * 3^1 * 2^1. It is necessary to reverse row m of A067255 (hereinafter simply A067255(m)) so as to preserve zeros in M = A067255(m) pertaining to small nondivisor primes q < p. The code M is a series of 0's and 1's since m is squarefree. Then it is clear that row n contains all m such that A067255(m) has n terms, and there are 2^(n  1) possible terms for n >= 1.
We may use an approach that generates the binary expansion of the range 2^(n  1) < M < 2^n  1, or we may append 1 to the reversed (n  1)tuples of {1, 0} to achieve codes M > m for each row n, which is tantamount to ordering according to A059894.
Originally it was thought that the codes M were in order of the latter algorithm, and we could avoid sorting. Observation shows that the m still require sorting by the function phi(m)/m indeed to be in increasing order in row n. Still, the latter approach is slightly more efficient than the former in generating the sequence.


LINKS



FORMULA

For n > 0, row lengths = A000079(n  1).
T(n, 2) = A306237(n) = p_n#/prime(n  1).
T(n, 2^(n  1)) = A000040(n) = prime(n) for n >= 1.
Last even term in row n = A077017(n).
First odd term in row n = A070826(n).


EXAMPLE

Triangle begins:
1;
2;
6, 3;
30, 10, 15, 5;
210, 42, 70, 14, 105, 21, 35, 7;
...
First terms of this sequence appear bottom to top in the chart below. The
values of n appear in the header, values m = T(n,k) followed
parenthetically by phi(m)/m appear in column n. The x axis plots
according to primepi(gpf(m)), while the y axis plots k according to
phi(m)/m:
0 1 2 3 4
. . . . .
 1 
(1/1) . . . .
. . . . .
. . . . .
. . . . 7
. . . 5 (6/7)
. . . (4/5) .
. . . . .
. . . . 35
. . 3 . (24/35)
. . (2/3) . .
. . . . .
. . . . .
. . . . 21
. . . . (4/7)
. . . 15 .
. . . (8/15) .
. 2 . . .
. (1/2) . . .
. . . . .
. . . . 105
. . . . (16/35)
. . . . 14
. . . 10 (3/7)
. . . (2/5) .
. . . . .
. . . . 70
. . 6 . (12/35)
. . (1/3) . .
. . . . 42
. . . 30 (2/7)
. . . (4/15) .
. . . . 210
. . . . (8/35)
...


MATHEMATICA

Prepend[Array[SortBy[#, Last] &@ Map[{#1, #2, EulerPhi[#1]/#1} & @@ {Times @@ MapIndexed[Prime[First@ #2]^#1 &, Reverse@ #], FromDigits@ #} &, Map[Prepend[Reverse@ #, 1] &, Tuples[{1, 0}, #  1]]] &, 6], {{1, 0, 1}}][[All, All, 1]] // Flatten


CROSSREFS

Cf. A000010, A000040, A000079, A000720, A002110, A005117, A006094, A006530, A007947, A059894, A070826, A077017, A306237.


KEYWORD

nonn,tabf,easy


AUTHOR



STATUS

approved



