Problems & Puzzles: Puzzles

Puzzle 952. Consecutive self primes Carlos Rivera asks for sequences of K consecutive self & prime integers. His findings are: K=3, starts with the self & prime 3 K=4, starts with the self & prime 2877817 K=5, starts with the self & prime 12277039 K=6, starts with the self & prime 800974417 Q1. Send your solutions for K>6? Q2. BTW, can you prove that there are not consecutive self integers? Q3. Can you explain why the most frequent difference between two consecutive self integers is 11*?_____________ *See a list of the first 10000 self integers here.

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Contributions came from Oscar Volpatti and Emmanuel Vantieghem.

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Oscar wrote on May 10, 2019

My findings are:
K = 7, starts with the self & prime 35323185061 (or 39397696397, or 39400561237);
K = 8, starts with the self & prime 48126138299 (and no further solutions with K > 7 up to 1e11).

The sequence of (non-negative) integers doesn't contain two consecutive self numbers.
An integer y is not self if it is generated by some integer x, such that y = x + digitsum(x) = g(x).
Such function is piecewise linear:
if x ≠ 9 mod 10, then g(x+1) = g(x) + 2 > g(x);
if x = 9 mod 10, then x+1 = q*(10^m) with q ≠ 0 mod 10, and g(x+1) = g(x) + 2 - 9*m < g(x).

Base step:
The 100 integers in [0,99] generate 100 numbers in [0,117], with no repetitions but with 18 holes (candidate self numbers):
h ∈ {1,3,5,7,9,20,31,42,53,64,75,86,97,108,110,112,114,116}.
The holes are not consecutive integers: the gap between successive holes is either 2 (8 times) or 11 (9 times).
So, the theorem holds within the interval [0,g(99)].

Induction step:
The theorem holds within the interval [0,g(100*Q-1)]
As g(100*Q+R) = g(100*Q) + g(R), the 100 integers in [100*Q,100*Q+99] generate 100 numbers in [g(100*Q),g(100*Q) + 117], with no repetitions but with 18 non-consecutive holes g(100*Q) + h.
So, the theorem also holds within the union of the overlapping intervals [0,g(100*Q-1)] and [g(100*Q),g(100*Q+99)].

Moreover, such overlapping fills many holes resposible for gaps with length 2, so the most common gap length is 11.

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Emmanuel wrote on May 10, 2019

Q1 : The earliest set of seven consecutive primes that are all self is :

  { 35323185061, 35323185083, 35323185151, 35323185217, 35323185239, 35323185241, 35323185263 }.

 

Q2 : This has been settled by Firoozbakht in puzzle 374.

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