Re: [math-fun] Mystery tables
Gareth McCaughan <gareth.mccaughan@pobox.com> wrote [about my first table]:
I have been thinking half-heartedly (and apparently half-brainedly) about this one from time to time ... and in the process of writing down my incoherent thoughts I realise that I have solved it.
I often do the same. See https://en.wikipedia.org/wiki/Rubber_duck_debugging
The value in row m, column n is the k such that m^k = 2n^k-1. For the (1,1) entry any k will do, hence the slightly curious "all" in that spot.
Right. My intention was for k to be the power necessary for the power mean of 1 and m to equal n. For those not familiar with the concept, the xth power mean of a and b is defined as (a^x + b^x)/2 = m^x. Common examples include: -1: Harmonic mean 0: Geometric mean 1: Arithmetic mean 2: Quadratic mean, aka Root Mean Square (RMS) But x can be any real number, not just an integer. For my first table I chose values of x that made the power mean of 1 and an integer (the row number) another integer (the column number). Most such values of x are transcendental numbers. But not all. What rational values can x have? Can it have an irrational algebraic value? (I'm speaking of the row and column both being positive integers, and the column number being less than the row number.) [About my second table:]
But it turns out that the entries in the table satisfy the following (defining) equation: if the entry in position (x,y) is z, then x^z + y^z = 2z^z.
Right.
It is not obvious to me whether this is actually equivalent to some sort of generalized mean as defined above. I'd guess not. What of course _is_ true is that z = ((x^z+y^z)/2)^1/z, so each entry is _a_ "power mean" of its row and column positions ... but with an exponent that equals the entry itself :-).
Yes, that was my intention. For every set of positive real numbers x<z1<y, z1 is some power mean of x and y. And for every positive real numbers x, y, and real number z2, there's a z2 power mean z1 that is between x and y. Since, given fixed x and y, z1 and z2 are monotonic functions of each other, it necessarily follows that for every x and y there must be a unique z1 that equals z2. I got curious what they looked like. Hence my second table. I assume they're all transcendental except of course where x=y, in which case z=x=y. Can anyone prove this? Or find a counterexample?
On 07/02/2020 04:06, Keith F. Lynch wrote:
Most such values of x are transcendental numbers. But not all. What rational values can x have? Can it have an irrational algebraic value?
(This is values of x where m^x - 2n^x = 1, with m,n positive integers.) I betcha x is never an irrational algebraic number, and I betcha there will not in our lifetimes be sufficient mathematical technology to prove this.
Hence my second table. I assume they're all transcendental except of course where x=y, in which case z=x=y. Can anyone prove this? Or find a counterexample?
(This is values of x where m^x + n^x = 2x^x, with m,n positive integers.) I betcha x is never algebraic unless m=n, and I betcha there will not in our lifetimes be sufficient mathematical technology to prove this. (I could be wrong about either; I'm not a transcendence expert; but this sort of thing is usually Really Hard.) -- g
Hello Math-Fun, Those 12 terms on a line share a property impossible to spot, I guess: 1, 10, 8, 15, 23, 25, 40, 39, 52, 62, 66, 71. Each of them is the sum of the digits on its left + the sum of the 2 digits on its right. 1 is the sum of "left of 1" = 0 + "sum of the 2 digits to the right of 1" = 1+0 = 1 => 0 + 1 = 1 10 is the sum of "left of 10" = 1 + "sum of the 2 digits to the right of 10" = 8+1 = 9 => 1 + 9 = 10 8 is the sum of "left of 8" = 1+1+0 = 2 + "sum of the 2 digits to the right of 8" = 1+5 = 6 => 2 + 6 = 8 15 is the sum of "left of 15" = 1+1+0+8 = 10 + "sum of the 2 digits to the right of 15" = 2+3 = 5 => 10 + 5 = 15 23 is the sum of "left of 23" = 1+1+0+8+1+5 = 16 + "sum of the 2 digits to the right of 23" = 2+5 = 7 => 16 + 7 = 23 25 is the sum of "left of 25" = 1+1+0+8+1+5+2+3 = 21 + "sum of the 2 digits to the right of 25" = 4+0 = 4 => 21 + 4 = 25 ... 71 is the sum of "left of 71" = 1+1+0+8+1+5+2+3+2+5+4+0+3+9+5+2+6+2+6+6 = 71 + "sum of the 2 digits to the right of 71" = 0 => 71 + 0 = 71. Question: Is it possible to build the lexico-first infinite seq having this property -- or does such a seq inexist? Would it then exist if we add the _three_ digits following a(n) to the sum of the digits preceding a(n)? Best, É.
Correction and extension: 1, 10, 8, 15, 23, 25, 40, 39, 52, 62, 74, 87, 89, 112, 118, 129, 139, 155, 167, 177,... Best, É.
1, 10, 8, 15, 23, 25, 40, 39, 52, 62, 66, 71.
Each of them is the sum of the digits on its left + the sum of the 2 digits on its right.
1 is the sum of "left of 1" = 0 + "sum of the 2 digits to the right of 1" = 1+0 = 1 => 0 + 1 = 1 10 is the sum of "left of 10" = 1 + "sum of the 2 digits to the right of 10" = 8+1 = 9 => 1 + 9 = 10 8 is the sum of "left of 8" = 1+1+0 = 2 + "sum of the 2 digits to the right of 8" = 1+5 = 6 => 2 + 6 = 8 15 is the sum of "left of 15" = 1+1+0+8 = 10 + "sum of the 2 digits to the right of 15" = 2+3 = 5 => 10 + 5 = 15 23 is the sum of "left of 23" = 1+1+0+8+1+5 = 16 + "sum of the 2 digits to the right of 23" = 2+5 = 7 => 16 + 7 = 23 25 is the sum of "left of 25" = 1+1+0+8+1+5+2+3 = 21 + "sum of the 2 digits to the right of 25" = 4+0 = 4 => 21 + 4 = 25 ... 71 is the sum of "left of 71" = 1+1+0+8+1+5+2+3+2+5+4+0+3+9+5+2+6+2+6+6 = 71 + "sum of the 2 digits to the right of 71" = 0 => 71 + 0 = 71.
Question: Is it possible to build the lexico-first infinite seq having this property -- or does such a seq inexist?
Would it then exist if we add the _three_ digits following a(n) to the sum of the digits preceding a(n)? Best, É.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Gareth McCaughan -
Keith F. Lynch -
Éric Angelini -
Éric Angelini