A297446 a(1) = 1; a(n) = (2^n - 1)*((3^n - 1)/(2^n - 1) mod 1), n >= 2. Unreduced numerators of fractional parts of (3^n - 1)/(2^n - 1).

1, 2, 5, 5, 25, 35, 27, 185, 264, 737, 1104, 3185, 5268, 15515, 29727, 55760, 35227, 235277, 441474, 272525, 1861165, 3478865, 6231072, 1899170, 5672261, 50533340, 17325481, 186108950, 21328108, 63792575, 1264831924, 3794064335, 7086578553

Offset: 1

Author

Fred Daniel Kline, Dec 30 2017

Keywords

• nonn

Comments

An easy way to get the numerator of the fractional part of the proper fraction (3/2)^n is (3^n - 2^n) (mod 2^n), which is not considered an elementary function. So, we created a function that subtracted the denominator from this difference until we got a sign change from positive to negative. I asked if this might be considered elementary at the Kline-Iwaniuk link. Mariusz Iwaniuk noticed the similarity to the sawtooth wave, and crafted a closed form for the floor of (3/2)^n from which we can get the modulus value for the numerator.

A back-of-the-envelope proof sketch of Waring's Problem.

We start with the original Diophantine equation from A060692, which we designate as x(n)+y(n), and substitute it into the "if statement" from Wikipedia Waring's Problem link: "if x(n) + y(n) <= 2^n." This has had no proof because we need more information.

So we extend the expression to three variables, (x,y,z), with z as the numerator of the fractional part of (3^n-1)/(2^n-1), and add the restriction that x is the common floor of (3^n - 1) / (2^n - 1) and 3^n / 2^n.

We find an identity for n >= 2, x(n) + y(n) == z(n) + 1, and substitute it into the if statement: "if x(n) + y(n) == z(n) + 1 <= 2^n."

Since the numerator of the fractional part must be within the bounds, 1 < z < 2^n -1, we determine that the greatest possible value of z is 2^n -2. Substituting for z(n), "if 2^n - 2 + 1 <= 2^n," shows it is always True. And more importantly, the Diophantine equation is always less than 2^n.

Inspection of z[1] shows it is also always True, with and without the anomaly. So, Waring is shown for n >= 1.

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..3280
• Fred Daniel Kline and Mariusz Iwaniuk, Is this a closed form?, Mathematica StackExchange, 2017.
• Wikipedia, Waring's problem

Crossrefs

Cf. A000225, A002379, A002380, A060692.

Programs

GAP

Concatenation([1],List([2..35],n->(3^n-1) mod (2^n-1))); # Muniru A Asiru, Dec 19 2018

Magma

[1] cat [(3^n-1) mod (2^n -1): n in [2..30]]; // G. C. Greubel, Dec 16 2018

Maple

a:=n->`if`(n=1,1,modp(3^n-1,2^n-1)): seq(a(n),n=1..35); # Muniru A Asiru, Dec 19 2018

Mathematica

x[n_] := -(1/2) + (3/2)^n + ArcTan[Cot[(3/2)^n Pi]]/Pi;
y[n_] := 3^n - 2^n * x[n];
z[n_] := x[n] + y[n] - 1;
Array[z, {33}]
f[n_] := PowerMod[3, n, 2^n -1] -1; f[1] = 1; f[2] = 2; Array[f, 33] (* Robert G. Wilson v, Jan 05 2018 *)

PARI

a(n) = if (n==1, 1, (3^n-1) % (2^n-1)); \\ Michel Marcus, Jan 02 2018

Python

def A297446(n): return pow(3,n,(1<2 else n # Chai Wah Wu, Jun 25 2024

Sage

[1] + [mod(3^n-1, 2^n-1) for n in (2..30)] # G. C. Greubel, Dec 16 2018

Formula

a(1) = 1; a(n) = (2^n - 1)*((3^n - 1)/(2^n - 1) mod 1), n >= 2, is the conventional way to describe the sequence. z(n) is the closed form which includes the anomaly.

a(n) = z(n).

x(n) := (3/2)^n + ( tan^-1 ( cot( Pi * (3/2)^n ) ) ) / Pi - 1/2;

y(n) := 3^n - 2^n * x(n);

z(n) := x(n) + y(n) - 1.

a(n) = A060692(n) - 1. - Fred Daniel Kline, Dec 13 2018