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This sequence works as a "sentinel" for the inverse of EKG-sequence by matching to any other sequence that is obtained as f(A064664(n)), where f(n) is any function that depends only on the prime signature of n (see the index entry for "sequences computed from exponents in ..."). As of Nov 11 2016 no such sequences were present in the database.

Terms and b-file computed from b-file of A064664 provided by T. D. Noe and Ray Chandler.

As the sequence A064664 has no S-property defined in the comments of A368900, therefore this is not equal to A014963(A064664(n)).

Locally, the graph looks like an EKG (American English) or ECG (British English).

Calculating the square of A064413 and plotting the results shows the EKG behavior even more dramatically - see A104125. - Parthasarathy Nambi, Jan 27 2005

Theorem: (1) Every number appears exactly once: this is a permutation of the positive numbers. - J. C. Lagarias, E. M. Rains, N. J. A. Sloane, Oct 03 2001

The permutation has cycles (1) (2) (3, 4, 6, 9, 10, 5) (..., 20, 18, 12, 7, 14, 13, 28, 26, ...) (8) ...

Theorem: (2) The primes appear in increasing order. - J. C. Lagarias, E. M. Rains, N. J. A. Sloane, Oct 03 2001

Theorem: (3) When an odd prime p appears it is immediately preceded by 2p and followed by 3p. - Conjectured by Lagarias-Rains-Sloane, proved by Hofman-Pilipczuk.

Theorem: (4) Let a'(n) be the same sequence but with all terms p and 3p (p prime) changed to 2p (see A256417). Then lim a'(n)/n = 1, i.e., a(n) ~ n except for the values p and 3p for p prime. - Conjectured by Lagarias-Rains-Sloane, proved by Hofman-Pilipczuk.

Conjecture: If a(n) != p, then almost everywhere a(n) > n. - Thomas Ordowski, Jan 23 2009

Conjecture: lim #(a_n > n) / n = 1, i.e., #(a_n > n) ~ n. - Thomas Ordowski, Jan 23 2009

Conjecture: A term p^2, p a prime, is immediately preceded by p*(p+1) and followed by p*(p+2). - Vladimir Baltic, Oct 03 2001. This is false, for example the sequence contains the 3 terms p*(p+2), p^2, p*(p+3) for p = 157. - Eric Rains

Theorem: If a(k) = 3p, then |{a(m) : a(m>k) < 3p}| = 3p - k. Proof: If a(k) = 3p, then all a(mk) > p and |{a(m) : a(m>k) < 3p}| = 3p - k. - Thomas Ordowski, Jan 22 2009

Let ...,a_i,...,2p,p,3p,...,a_j,... There does not exist a_i > 3p. There does not exist a_j < p. - Thomas Ordowski, Jan 20 2009

Let...,a,...,2p,p,3p,...,b,... All a<3p and b>p. #(a>2p) <= #(b<2p). - Thomas Ordowski, Jan 21 2009

If a(k)=3p then |{a(m):a(m>k)<3p}|=3p-k. - Thomas Ordowski, Jan 22 2009

GCD(a(n),n) = A247379(n). - Reinhard Zumkeller, Sep 16 2014

If the definition is changed to require that the GCD of successive terms be a prime power > 1, the sequence stays the same until a(578)=620, at which point a(579)=610 has GCD = 10 with the previous term. - N. J. A. Sloane, Mar 30 2015

From Michael De Vlieger, Dec 06 2021: (Start)

For prime p > 2, we have the chain {j : 2|j} -> 2p -> p -> 3p -> {k : 3|k}. The term j introducing 2p must be even, since 2p is an even squarefree semiprime proved by Hofman-Pilipczuk to introduce p itself. Hence no term a(i) such that p | a(i) exists in the sequence for i < n-1, where a(n) = p, leaving 2|j. Similarly, the term k following 3p must be divisible by 3 since the terms mp that are not coprime to p (thus implying p | mp) have m >= 4, thereby large compared to numbers k such that 3|k that belong to the cototient of 3p. For the chain {4, 6, 3, 9, 12}, the term 12 following 3p indeed is 4p, but p = 3; this is the only case of 4p following 3p in the sequence. As a consequence, for i > 1, A073734(A064955(i)-1) = 2 and A073734(A064955(i)+2) = 3.

For Fermat primes p, we have the chain {j : 2|j} -> 2^e-> {2p = 2^e + 2} -> {p = 2^(e-1) + 1} -> 3p -> {k : 3|k}.

a(3) = 4 = 2^2, a(5) = 3 = 2^1 + 1;

a(8) = 8 = 2^3, a(10) = 5 = 2^2 + 1;

a(31) = 32 = 2^5, a(33) = 17 = 2^4 + 1;

a(485) = 512 = 2^9, a(487) = 257 = 2^8 + 1;

a(127354) = 131072 = 2^17, a(127356) = 65537 = 2^16 + 1.

(End)

It can be shown that this sequence is monotonic.

A073734(a(n)) = A000040(n) for n > 1. - Reinhard Zumkeller, Sep 17 2001

This is a permutation of the natural numbers: the proof for A098550 applies with essentially no changes. [Confirmed by N. J. A. Sloane, Feb 27 2015]

For n > 3, all primes first appear in order as composites with one smaller prime (proof similar to that in A098550).

For any given set S of primes, the subsequence consisting of numbers whose prime factors are exactly the primes in S appears in increasing order. For example, if S = {2,3}, 6 appears first, followed by 12, 18, 24, 36, 48, 54, 72, etc.

Appears to be very similar to A064413. Compare the respective inverses A255479 and A064664; see also A255482. Speaking very loosely, the ratio a(n)/n seems to be about 1/2, 1, or 3/2, just as for A064413, although this is a long way from being proved for either sequence. David Applegate points out that this is (presumably) because primes p >= 13 always occur as part of a subsequence 2p X p Y 3p, and subsequences 2p X p Y 5p, 2p X p Y 7p, etc. that produced the extra curves in the graph of A098550 just do not happen. - N. J. A. Sloane, Feb 27 2015, Mar 05 2015.

First differs from A254077 at a(29). - Omar E. Pol, May 21 2015