Pham G. Hoang - cp's OEIS frontend

User: Pham G. Hoang

Pham G. Hoang has authored 2 sequences.

A378960 The "tetrational mean" of 2 and 3 determined as the mutual limit of interdependent sequences.

2, 4, 1, 9, 3, 6, 5, 3, 2, 1, 8, 4, 4, 9, 2, 1, 7, 8, 8, 8, 6, 0, 7, 4, 5, 4, 6, 8, 9, 3, 2, 7, 5, 4, 3, 5, 4, 4, 1, 6, 4, 6, 2, 6, 2, 4, 3, 6, 8, 7, 9, 3, 9, 1, 4, 5, 5, 7, 2, 2, 8, 4, 7, 0, 1, 1, 2, 0, 9, 6, 3, 6, 2, 4, 3, 5, 6, 3, 9, 7, 4, 1, 4, 4, 8, 4, 0, 1, 3, 7, 9, 2, 2, 4, 5, 0, 7, 8, 9, 6, 8, 2, 7, 0, 2, 7, 2, 8, 9, 1, 7, 7, 3, 7, 7

Offset: 1

Pham G. Hoang, Dec 12 2024

In an attempt to generalize the arithmetic mean (sum-based) and the geometric mean (product-based) to a similar construct for exponentiation, one can devise a simple definition using 2 interdependent sequences:
  a_0 = x, b_0 = y,
  a_n = exp(LambertW(log(a_{n-1}^b_{n-1}))),
  b_n = exp(LambertW(log(b_{n-1}^a_{n-1}))), where x and y are the numbers for which we have to determine their "tetrational mean".
The averaging operation is the square super-root of each of the 2 possible exponentiation orders to give out the successive term of each defining sequence. The square super-root of x is exp(LambertW(log(x))) for a particular branch of the LambertW function.
If a_n and b_n converge to a number C then the "tetrational mean" of x and y is C. There may be a need to choose a particular branch of the LambertW function depending on the values of x and y (and that of log(x^y) and log(y^x)). This constant is based on the principal branch of the LambertW function.

			2.419365321844921788860745468932754354...

Pham G. Hoang, Table of n, a(n) for n = 1..1000
Wikipedia, Arithmetic mean
Wikipedia, Geometric mean
Wikipedia, Mean
Wikipedia, Tetration

Cf. A068985, A030798, A010464.

A308508 Numbers (including primorials) that satisfy "three rules that highly composite numbers must have" (see Comments below) but are not highly composite numbers.

Original entry on oeis.org

30, 96, 192, 210, 384, 420, 480, 768, 960, 1080, 1440, 1536, 1920, 2160, 2310, 2880, 3072, 3360, 3840, 4320, 4620, 5760, 6144, 6300, 6480, 6720, 7680, 8640, 9240, 11520, 12288, 12600, 12960, 13440, 13860, 15360, 17280, 18480, 23040, 24576, 25920, 26880, 30030

Offset: 1

Pham G. Hoang, Jun 02 2019

nonn

The three rules that the highly composite numbers (and this sequence too) must have are:
   - The primes in the product have to be consecutive and start with 2,
   - The exponents of the primes must be decreasing,
   - The final exponent of the primes must be 1 (except 4 = 2^2 and 36 = 2^2 * 3^2, but those are highly composite numbers and are excluded).
Except for numbers of the form 2^n * 3, the terms are divisible by 10, because a(n) has the form of 2^n * 3^m * 5^j * c = (2 * 5) * 2^(n - 1) * 3^m * 5^(j - 1) * c.
Except for terms that are primorials, all others are divisible by 12, because a(n) has the form 2^n * 3^m * c = (2^2 * 3) * 2^{n - 2} * 3^(m - 1) * c.
All terms are divisible by 6, because a(n) has the form 2^n * 3^m * c = (2 * 3) * 2^(n - 1) * 3^(m - 1) * c.
It seems that eulerphi(a(n)) is always divisible by 8.

Michael De Vlieger, Table of n, a(n) for n = 1..16384 (first 137 terms from Giovanni Resta).

Contains A002110(n) (primorials).

Does not contain A002182(n) (highly composite numbers).

limit = 2*10^5; hc = {1}; r=1; Do[t = DivisorSigma[0, n]; If[t > r, r=t; AppendTo[hc, n]], {n, 2, limit, 2}]; ok[n_] := Block[{f = FactorInteger[n]}, ! MemberQ[hc, n] && f[[-1, 2]] == 1 && Max[ Differences[Last /@ f]] <= 0 && Union[ Differences[ PrimePi[ First /@ f]]] == {1}]; Select[Range[2, limit, 2], ok] (* Giovanni Resta, Jun 10 2019 *)

a(40)-a(43) from Giovanni Resta, Jun 04 2019

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User: Pham G. Hoang

A378960 The "tetrational mean" of 2 and 3 determined as the mutual limit of interdependent sequences.

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