cp's OEIS Frontend

Let gpf(m) = A006530(m) and let phi(m) = A000010(m) for m in A005117.

Row n contains m in A005117 such that 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} (as A059894) to achieve codes M -> m for each row n.

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.

This sequence interprets the code M as a binary value. The sequence is a permutation of the natural numbers since the ratio phi(m)/m is unique for squarefree m.

This sequence and A059894 are identical for 1 <= n <= 23.

Numbers of terms in rows n of this sequence and A059894 (partitioned by powers of 2) that are coincident: 1, 2, 4, 8, 14, 14, 10, 26, 14, 20, 10, 16, 22, 12, 18, 18, 16, 14, 18, 18, 18, 14, 16, ...}.

The graphs of this sequence and A059894 are similar.

The graph of this sequence feature squares of size 2^(j-1) at (x,y) = (h,h) where h = 2^j, integers, that have pi-radian rotational symmetry.

Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117.

There are 2^(n-1) numbers k with gpf(k) = prime(n), since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which phi(k)/k - 1/2 is positive and minimal.

We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but more than 1/2.

Apart from a(1) = 2, all terms are odd. For n > 1 and k even, phi(k)/k - 1/2 is negative.

Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117. There are 2^(n-1) numbers k with gpf(k) = n, since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which 1/2 - phi(k)/k is positive and minimal.

We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but less than 1/2.

Conjecture: for n > 3, k is always odd. This assertion is reliant upon phi(2 prime(n))/2 prime(n) = phi(2)/2 * phi(prime(n))/prime(n) = 1/2 * (prime(n) - 1)/prime(n), and it is clear that 1/2 is an asymptote for even k.

Permutation of natural numbers.

Once we have a(n) = prime(m), we require a(n+1) = u, the smallest missing number. Thereafter, we find the smallest k new to the sequence such that phi(k)/k > phi(j)/j, where j = a(n-1) until we reach another prime. Primes appear in order, since phi(p)/p = (p-1)/p.

Sequence can be interpreted as an irregular triangle where row 0 = {1} and row m > 1 begins with prime(m). In such a triangle, we observe prime(m) > min(row m) for m > 5, yet we can find prime(m) either less than or exceeding max(row m) for 2^20 terms of this sequence, or m = 1..82032.

Since phi(k)/k ascribes to squarefree kernel K = rad(k) = A007947(k) and K < mK where mK is nonsquarefree yet rad(m) | K, squarefree k appear before mK. For example, a(3) = 6 and a(13) = 12; a(11) = 10 and a(20) = 20, etc.

Odd prime numbers tend to appear early, even numbers tend to appear late.