Puzzle 3: Münchhausen numbers of length 3

The number 3435 = 3³+4⁴+3³+5⁵ is well-known as the Münchhausen number in the base 10 and its length (number of digits) is 4.

More examples:

The number 96446 is the Münchhasuen number in the base 9 and its length is 6, because it has 6 digits in the base 9:

96446₁₀  = 156262₉  

digits 1, 5, 6, 2, 6, 2    

96446 = 1¹+5⁵+6⁶+2²+6⁶+2²

The number 20017650854 is the Münchhasuen number in the base 12 and its length is 10, because it has 10 digits in the base 12:

20017650854₁₀  = 3a67a54832₁₂ 

digits 3, a, 6, 7, a, 5, 4, 8, 3, 2  

20017650854 = 3³+10¹⁰+6⁶+7⁷+10¹⁰+5⁵+4⁴+8⁸+3³+2²   (the digit a is ten)

Bases can be considered arbitrary. Then it is not difficult to prove that there are infinitely many Münchhasuen numbers of length 2 in some base. In particular, 1+nⁿ is of length 2 in the base 1+nⁿ-n for each n greater than 1.

But what about length 3? Are there infinitely many Münchhausen numbers of length 3? Up to now, we know only these Münchhasuen numbers of length 3:

29 in base 4

55 in base 4

3153 in base 25

49782 in base 91

46661 in base 215
New: 823545 in base 904 (June 21, 2020; by mersenneforum.org user axn) 

Puzzle 2: Happy New Year 2020 (and the link with 2- 4- 6- concecutive prime gaps)

Gaps between prime numbers represent an interesting and widely studied topic. Taking the first few primes

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...

there are gaps between them

  1, 2, 2, 4, 2, 4, 6, ...

We see that the smallest gap 1 is an exception: regardless of this, gaps are always positive even numbers. The gap 2 is between so-called prime twins, the smallest such twins are 3 and 5. The prime twins may be infinitely many, but proof of this has not yet been given.

The largest gap does not exist because a gap can be of any length. This follows from this finding: if we take any natural number n, the numbers n! +2, n! +3, ..., n! + n are all composed (surely, n! +2 is divisible by two, n! +3 is divisible by three, and so on) and is therefore a part of some gap between prime numbers.

As already mentioned, primes 3, 5 are smallest consecutive primes with gap 2. Furthermore:

· Prime numbers 5, 7, 11 are the smallest consecutive primes with gaps 2, 4.

· Prime numbers 17, 19, 23, 29 are the smallest consecutive prime numbers with gaps 2, 4, 6.

· Prime numbers 347, 349, 353, 359, 367 are the smallest consecutive prime numbers with gaps 2, 4, 6, 8.

· Prime numbers 13901, 13903, 13907, 13913, 13921, 13931 are the smallest consecutive prime numbers with gaps 2, 4, 6, 8, 10.

In this way, we could continue up to a string of fifteen consecutive prime numbers, each with by two increasing gaps from 2 to 28, which is the last result of this kind. The first prime number here is 221860944705726407. Finding the following string of 16 consecutive primes with gaps from 2 to 30 can be a challenge for the reader (Puzzle 2). See OEIS A016045.

Let us look at four consecutive prime numbers 17, 19, 23, 29, which are the smallest with gaps 2, 4 and 6. If we calculate the sum of squares

17² + 19² + 23² + 29²,

the result is 2020. Pour féliciter 2020.

Puzzle 1: Sum of Powers of Primes / The paper

Now, I have written known facts about numbers in question in my paper published in the journal Notes on Number Theory and Discrete Mathematics; not only that, I tried to derive some properties of these numbers. I'll be glad for your eventual response to this paper. The link to it is here.

2019

Sure, the number 2019 has a lot of great properties. For example, 2019 is the least number having six different representations as a sum of three squares of prime numbers.

2019 = 23²+23²+31²

2019 = 17²+19²+37²

2019 = 11²+23²+37²

2019 = 13²+13²+41²

2019 = 7²+17²+41²

2019 = 7²+11²+43²

There a lot of numbers having six different representations as above, not seven or more. They form a sequence 2019, 2091, 2499, 3099, 3219, 3339, 3579, 3939, 4011, ... E.g., 4011 = 31² + 37² + 41² = 29² + 31² + 47² = 11² + 41² + 47² = 19² + 29² + 53² = 13² + 19² + 59² = 11² + 13² + 61².

Puzzle 1: Sum of Powers of Primes / Results - 2

Dana Jacobsen has published a short code for (P2) in the Perl language on mersenneforum. He notes that testing up to 10⁹ takes 90 minutes.

Puzzle 1: Sum of Powers of Primes / Results - 1

The mersenneforum user, nicknamed a1call, has announced the finding of a new number with the (P2) property. It is the number 9634877 (Sep 3rd, 2018). Excellent work!

Puzzle 1: Sum of Powers of Primes

The whole story began around the turn of the millennium when Carlos Rivera  observed that 39 is a number with the following property:

(P1)     A non-prime number equal to the sum of consecutive primes from the least prime factor to the largest prime factor.

Of course, 39 = 3 * 13 = 3 + 5 + 7 + 11 + 13.

It is not difficult to verify that 10, 155 and 371 are other numbers with this property. Later (Jul. 2000), Jud McCranie found the 5th number:

2935561623745 = 5 * 19 * 53 * 61 * 9557877 = 5 + 7 + 11 + ... + 9557877.

On Nov. 2002, Robert Munafo discovered that

454539357304421 = 3536123 * 128541727 = 3536123 + 3536129 + 3536131 + ... + 128541727.

So, six numbers with the property (P1) are known up to now. One can find them in the Online Encyclopedia of Integer Sequences under the code A055233.

We change the property somewhat now:

(P2)     A non-prime-square number equal to the sum of squares of consecutive primes from the least prime factor to the largest prime factor.

The smallest number with this property is 315797.

Indeed,

315797 = 31 * 61 * 167 = 31² + 37² + 41² + ... + 167².

Question A: Find other numbers with the property (P2).

We can continue.

(P3)     A non-prime-cube number equal to the sum of cubes of consecutive primes from the least prime factor to the largest prime factor.

In this case, there is a nice solution: 160.

160 = 2⁵ * 5 = 2³ + 3³ + 5³.

Question B: Find other numbers with the property (P3).

And finally:

Question C: The first, second and third powers of primes can be generalized to any n-th power. Find also any numbers for n>3.