Infinitude of super happy primes

Question

Similar to happy primes, I define super happy primes by the following process:

$(1)$ Find the sum of the digits raised to the power of themselves.

Ex. $13$ gives sum $ = 1^1 + 3^3 = 28$

$(2)$ If the digit root of the sum is equal to $1$ then it is a super happy number.

Ex. digitroot(sum)$=$digitroot($28$)$=1$. So $13$ is a super happy prime.

Here are some super happy primes I got by searching using PARI/GP with the help of user Peter.

13 19 31 61 83 89 139 157 163 193 199 313 331 383 389 571 587 613 619 631 
661 68 3 691 751 839 857 863 919 983 991 1087 1117 1171 1187 1277 1399 1567 
1579 1597 1 657 1663 1669 1693 1699 1747 1753 1759 1871 1933 1993 1999 2141 2281 2411 2447 

Note: Since digitroot$(sum)$ is nothing but $sum \mod 9$, the sum will be of the form $9k+1$, for some natural number $k$.

Now coming to the question:

Can we prove the infinitude of super happy primes?

Answer 1

Original Problem

This is not a complete answer but my progress on the conjecture (Infinitude of super happy primes) :

First off, we can clearly see that a number's digital root is just the number itself modulo $9$ since we have $n \equiv S(n) \pmod{9}$ where $S(n)$ represents the sum of the digits of $n$. By repeated iteration, the digital root will also be congruent to the original number modulo $9$ and since it is one digit, it is the least positive residue of $n \pmod{9}$ (we have $9$ instead of $0$). Now, let our number in base $10$ be of the form $n=\overline{d_{[1+\log{n}]} \cdots d_1d_0}$ where the logarithm is taken base $10$ and $[x]$ is the floor function. Now, we have: $$n=\sum_{i=0}^{[1+\log{n}]}10^id_i \implies \sum_{i=0}^{[1+\log{n}]}d_i^{d_i} \equiv 1 \pmod{9} \implies \sum_{i=0}^{[1+\log{n}]}(d_i^{d_i} \bmod{9}) \equiv 1 \pmod{9}$$ $$ \begin{array}{c|c} d & d^d \pmod{9} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 0 \\ 4 & 4 \\ 5 & 2 \\ 6 & 0 \\ 7 & 7 \\ 8 & 1 \\ 9 & 0 \\ \end{array} $$ Let a prime number $n$ have $A_d$ appearances of the digit $d$ base $10$. Thus, for $n$ to be a super happy prime: $$A_1+4A_2+4A_4+2A_5+7A_7+A_8 \equiv 1 \pmod{9}$$ It might be easier to search for super happy primes using the above congruence. Since prime numbers are expected to show no bias in their digital representation in any base, one should expect the above to hold true about $\frac{1}{9}$th of the time for primes. Since there are an infinite number of primes, it is most likely that there are an infinite number of super happy primes. However, I am not sure of how to prove the same.

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It might be helpful to analyze the same problem in smaller bases. For base $b$, the digital root is the least positive residue of the number modulo $(b-1)$, where $0$ is represented by $(b-1)$ itself.

Base $2$

This is trivial in the base $2$ system since every number would generate $1$ as the digital root, since we work with modulo $1$ system.

Result : Every prime is a super happy prime in base $2$

Base $3$

$$ \begin{array}{c|c} d & d^d \pmod{2} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 0 \\ \end{array} $$ Thus, for a prime to be a super happy prime in base $3$, it needs to have an odd number of $1$s in its base $3$ digit expansion i.e. $2 \nmid A_1$ or $A_1 \equiv 1 \pmod{2}$. However: $$n=\overline{d_{[1+\log_3{n}]} \cdots d_1d_0}_{\space 3} \implies n \equiv A_1 \pmod{2}$$ since the other digits ($0$ and $2$) are even. For $n>2$ to be a prime, we must have $2 \nmid n$. Thus, we directly have $2 \nmid A_1$ which shows that every odd prime is a super happy prime base $3$.

Result : Every odd prime is a super happy prime base $3$.

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The Generalized Problem

It can be clearly seen that for any base $b$, we can say that a prime is super happy if: $$C_1A_1 + C_2A_2 + \cdots + C_{b-2}A_{b-2} \equiv 1 \pmod{b-1}$$ for some coefficients $0 \leqslant C_d < b-1$. Define the following sequence: $$t_0=0$$ $$t_{bi+d}=[(t_i+C_d) \bmod{(b-1)}] \quad (0 \leqslant d < b)$$ where one can set $C_0 = C_{b-1} = 0$. By definition, it follows that if $t_i=1$ for prime $i$, then $i$ is also a super happy prime. One needs to prove that there exist infinitely many such $i$.

One of my ideas to approach this problem is by using the Green-Tao theorem. I suspect that for any $(t_i)$, we will always have $M \in \mathbb{N}$ such that for any arithmetic progression $\{x_i \space | \space 0 \leqslant i < M\}$, the set- $$\{t_{x_i} \space |\space 0 \leqslant i < M\}$$ always contains one of its elements as $1$. Since there exist infinitely many arithmetic progressions of length of at least $M$ filled with primes by the Green-Tao theorem, it will follow that there are infinitely many super happy primes in all bases.

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Overview

• All primes are super happy primes in base $2$
• All odd primes are super happy primes in base $3$
• One can prove that there are infinitely many super happy primes for any base $b>3$ by proving the following:

Define the following sequence: $$t_0=0$$ $$t_{bi+d}=[(t_i+C_d) \bmod{(b-1)}] \quad (0 \leqslant d < b)$$ where $C_d \equiv d^d \pmod{b-1}$.

For any arithmetic progression $\{x_i \space | \space 0 \leqslant i < M\}$ (length $M$), we have: $$\{t_{x_i} \space |\space 0 \leqslant i < M\}$$ containing $1$ atleast once where $M$ is a sufficiently large constant.

One can note that this problem has many similarities with the Prouhet-Thue-Morse sequence.