Make 2023 with math: PowerMod[7, 7, 7!] = 2023. Your turn!

What is your one-liner for getting 2023? Comment below!

DOWNLOAD-DESKTOP-NOTEBOOK.nb (31.6 KB)

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A few more

LinearRecurrence[{3, -3, 1}, {7, 34, 79}, 15] // Last 
LinearRecurrence[{0, 1, 1, 2}, {0, 2, 3, 10}, 17] // Last 
LinearRecurrence[{4, 0, -5, -2}, {0, 2, 10, 41}, 7] // Last  
N[FromContinuedFraction[Flatten[{44, Table[{1, 43, 1, 88}, 5]}]]^2, 40]
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Also can be checked in Wolfram|Alpha: 7^7 mod 7! - Wolfram|Alpha

BTW - Cool idea! Here is something very simple:

2^{11}-5^2 = 2023
In[]:= 2^11-5^2
Out[]= 2023

And thanks to John Conway’s discovery of 1988 that e^{\pi }-\pi \approx 20 we have

\text{Round}\left[101 e^{\pi }-100 \pi \right]= 2023
In[]:= Round[101*E^Pi-100*Pi]
Out[]= 2023

I am trying to find some exact ones based on Golden Ratio, Pi, E, other irrational and transcendental numbers. Similar to Fibonacci sequence formula, but much neater than this (I hope someone will figure this out):

\frac{\phi ^{18}-(1-\phi) ^{18}}{\sqrt{5}}-3 \cdot 11 \cdot 17=2023
In[]:= FullSimplify[(GoldenRatio^18-(1-GoldenRatio)^18)/Sqrt[5]-3 11 17]
Out[]= 2023
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Going into algebraic number theory

-NumberFieldDiscriminant[x /. Solve[x^3 == x^2 + 6 x + 5][[1]]]
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With this:

I thought it would be interesting to see what could be one in base 7 to get to 2023.

https://www.wolframalpha.com/input?i=decimal+form+of+7*23_7%5E2

Or in WL:
7 7^^23^2

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There are some good answers in comments on Reddit Math: Reddit

Also there is this link with the properties: Properties of number 2023

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I prefer to look forward to what 2023 might bring to us!
2023 is a very interesting 3-color totalistic cellular automaton:

ArrayPlot[CellularAutomaton[{2023,{3,1}},{{1},0},500],ColorRules->{0->White,1->Gray,2->Red}]
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3^2+7^2+11^2+15^2+19^2+23^2+27^2 = 2023

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Not a formal per se but 2023 is a multiple of 7 whose digits also add to 7. In other words 2023 is the numbers congruent to 7 mod 63.

Looks like we got another interesting postcard post by Fermat’s Library on LinkedIn:

which is in Wolfram:

In[]:= (10+(9+8*7)*6)*5+4*3*2-1
Out[]= 2023

I wonder is it easy to find such combination? How many numbers we can get out of specific range using a specific set of operations? And how many combinations we have for a single number on average?

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From a Tweet by Srinivasa Raghava (he also shared a number of other interesting results of 2023):

𝐒𝐫𝐢𝐧𝐢𝐯𝐚𝐬𝐚 𝐑𝐚𝐠𝐡𝐚𝐯𝐚 ζ(1/2 + i σₙ )=0 on X: "Happy new year wishes to all my friends and well-wishers. Expressing the New year number 2022 in terms of the factorials of numbers 1 to 9. #math #Maths #HappyNewYear2023 #NewYear #NewYear2023" / X

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If we allow concatenation.

1*2*3*456+7-8*9*10
1+2*34*5*6-7+8-9-10
1/2*345*6+78+910
12+3/4*5*67*8-9+10
12/34*5678+9+10
12+345*6-78+9+10
10-9*8-76+5*432+1
10-9*87+65*43+2-1
10*98-7*6+543*2-1
10*987/6+54/3*21
1 Like

[ (2023)^2 - 2023 ] / [6* (2023)] = 337

New year : 337 *(6) + 1 = 2023
or [337 (7) (2023)] / 2023] - 338 = 2023

Last year : 337 (6) = 2022
or 337(7) - 337 = 2022

Moving along : 337 + 1 = 338

Finding the next primes using the formula above :

[ [2023 * ( 338)* (7) - 2023] / 2023] - 338 = 2027 is prime

[ [2023 ( 338) (7) + 2023] / 2023] - 338 = 2029 is prime

[(2029+338) 2023 - 2023 / (2366)] = 2023

[(2027+338) 2023 -+2023 / (2366)] = 2023

2023 within a solution using the next twin primes

= (11/x+1 )^-1
which = 184

184 * 11 - 1 = 2023