Puzzle 994. A puzzle for quarantine days.

Problems & Puzzles: Puzzles

Puzzle 994. A puzzle for quarantine days.

Last week I saw the following curio in a very interesting site in the web:
Returning the Favor

54199³= 159211275242599 and 15921 + 12752 + 42599 = 71272

71272³ = 362040234715648 and 36204 + 02347 + 15648 = 54199

Trying to take advantage of the quarantine covid-19 days that we are experiencing, I started trying to detect more examples related to the curio reported, if using primes, the better.

Property observed: Returning the favor:

"The sum of the k parts of the k-power of A produce B & the sum of the k parts of the k-power of B produce A, k>1"
(or A->B->A)"

Property for this puzzle: Prime-Immediate Return:

"The sum of the k parts of the k-power of P produce P, k>1"
(or P->P)

Soon I got some nice results:

k The smallest P such that P->P

3 77767777

4 7

5 76836761

Example:

76836761⁵ = 2678213815289811000299231341508821319801 and 26782138+15289811+00029923+13415088+21319801 = 76836761

Q1. Send the smallest P for k=2 or show it does not exist.

Q2. Send one example for the property (A->B->A) where A & B are primes, for some k>1.

Q3. Send more interesting examples according your taste.

During the week March 28 to April 3, 2020, contributions came from Jeff Heleen, Emmanuel Vantieghem and Oscar Volpatti.

***

Jeff wrote:

Q3. Searching A^k for A <= 100,000,000 with cycle length <= 20:

k    A     cycle     cycle length     note

2    3552301     3552301 -> 3656485 -> 3892213 -> 3552301     3     first 3-cycle for k=2, also prime

2    5555565     5555565 -> 5555655 -> 5565555 -> 5555565     3

2    5556555     5556555 -> 6555555 -> 5655555 -> 5556555     3

2    41715     41715 -> 58626 -> 42246 -> 42363 -> 41715     4     first 4-cycle for k=2

2    77787     77787 -> 77877 -> 87777 -> 78777 -> 77787     4

3    55330642     55330642 -> 91698220 -> 81343243 -> 73966447 -> 55330642     4     first 4-cycle for k=3

For k>5 I found no solutions <10^8.

***

Emmanuel wrote:

Q1.
The search for primes p with the property that p^2 = a&b with a + b = p is the search for prime Kaprekar numbers.
This was the subject of the puzzles 837, 838, 843.
As far as I know, p = 1111111111111111111 = R(19) is the smallest and there are two other candidates : R(109297) and R(270343).
These are probable primes (has anyone turned some of them into provable primes ?).

Q2.
k = 3
I could find a few examples of type Prime -> Composite -> Prime :
27272827 -> 28272727 -> 27272827
56468807 -> 56469797 -> 56468807
57659779 -> 33001255 -> 57659779

A prime such that the intermediate step is also prime must be bigger than 10^9

k = 5

I think the prime 540539 is the smallest of type P -> P :

540539^5 = 46146117682034200315812026699

and  46146+117682+034200+315812+026699 = 540539

Q3.

I mixed two values for k and found this :
5347^2 = 28590409 and 2859 + 0409 = 3268
3268^3 = 34901664832 and 0349 + 0166 + 4832 = 5347
(5347 is prime but 3268 is not)
and
24391^3 = 14510715208471 and 01451 + 07152 + 08471 = 17074
17074^2 = 291521476 and 02915 + 21476 = 24391
(24394 is prime but 17074 is not).

As you can see, I completed some powers by putting a suitable number of zeros in front.

This was easier for the computation (I wrote the numbers in base 10^k and took the sum of the digits).

I justified this by the fact that in the shown favor numbers with zeros in front were allowed.

***

Oscar wrote:

For some numbers X with n digits, the power Y = X^k has less than k*n digits.

In such cases, before breaking Y into k parts, I assumed to add a suitable number of leading zeros to restore the desired length k*n;

in other words, I always computed the digit sum of Y in base B = 10^n.

For property (A->B->A), I also assumed that A and B must have the same number of digits.

Q1: primes P with property (P->P) and smallest length n, for 2<=k<=10.

k=2, n=19 digits:

1111111111111111111;

k=3, n=8 digits (Rivera):

77767777;

k=4, n=1 digit (Rivera):

7;

k=5, n=6 digits:

540539;

k=6, n=20 digits (4 elements):

60210512756263417123;

...

79121102647615090561;

k=7, n=12 digits:

613801613701;

k=8, n=42 digits (193 elements):

355839956031992243919487769418805047201931;

...

923728645273970952725085655404793995285559;

k=9, n=30 digits:

845694063736573432501140312293;

k=10, n=30 digits;

874976257858102471576213669489.

Q2: prime pairs A,B with property (A->B->A) and smallest length n, for 2<=k<=7.

k=2, n=18 digits:

447368921552631079,

552131579447868421;

k=3, n=15 digits:

413914575696319,

682407697250449;

k=4, n=20 digits (26 pairs):

24373793623946300713,

37263811887044396713;

...

76978828415352912211,

88010744874000180787;

k=5, n=12 digits:

596882270249,

601932265199;

k=6, n=28 digits:

5438891125364615057776655341,

7946923579156953704305635427;

k=7, n=24 digits:

595179647069952543900643,

704269143030762363323611;

Q3: infinite sequences of repdigits X with property (X->X)

For the following repdigits X, the digits of Y = X^k also follow a regular pattern.

If n == 1 mod 9, repunit X=R(n) has property (X->X) for k=2.

X = 111111111,...,111111111,1

Y1= 012345679,...,012345679,0

Y0= 098765432,...,098765432,1

Z = 111111111,...,111111111,1 = X

X is prime for n=19; next candidates are the known PRPs for n=109297 and n=270343.

If n == 1 mod 9, repdigit X=6*R(n)+1 has property (X->X) for k=4.

X = 666666666,...,666666666,7

Y3= 197530864,...,197530864,2
Y2= 370370370,...,370370370,4
Y1= 000000000,...,000000000,0
Y0= 098765432,...,098765432,1

Z = 666666666,...,666666666,7 = X

X is prime for n=1 (Rivera). No more PRPs found so far.

If n == 2 mod 9, repdigit X=6*R(n)+1 has property (X->X) for k=4 too.

X = 666666666,...,666666666,67

Y3= 197530864,...,197530864,20
Y2= 148148148,...,148148148,15
Y1= 111111111,...,111111111,11
Y0= 209876543,...,209876543,21

Z = 666666666,...,666666666,67 = X

X is prime for n in {2,11,20,119,...}

...

Sequence X(i) = 6*R(9*i+2)+1:
the term X(97) = 6*R(875)+1 is prime too.

Sequence X(i) = 6*R(9*i+1)+1:
the only prime term is X(0) = 7.

Proof by induction.
X(0) = 7;
X(1) = 6666666667 = 19*1627*215659;
X(i+2) - X(i) = 6*R(18)*10^(9*i+1).

Both 7 and 19 divide R(18), so 7 divides all the terms with even indexes and 19 divides all the terms with odd indexes.

***

Records | Conjectures | Problems | Puzzles