Puzzle 1255 Sum of prime factors of (17 +71) is 17.

Puzzle 1255 Sum of prime factors of (17 +71) is 17.

On October 5, 2022, J. M. Bergot (†), reported a curio:"Calculate the sum of the emirp 17 and its reversal. The sum of all prime factors of this number (88) is 17."

Q1. Send in order all the primes/emirps with the same property than 17, that you can find?

Q2. Send the largest prime/emirp of this type, that you can find.

From Jan 24-31, 2026 contributions came from Michael Branicky, Gennady Gusev, Oscar Volpatti, Emmanul Vantieghem

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Michael wrote:

I found one more: 107 + 701 = 808 = 2*2*2*101 and 2+2+2+101 = 107

There were no others < 10^11.

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Gennady wrote:

Q1 & Q2.

If all the numbers p1=10^n+7, p2=7*10^n+1, d=10^n+1 are prime numbers,

then p1 and p2 are emirp and its reversal, p1 + p2 = d*2*2*2 and p1 = d+2+2+2 satisfy the puzzle condition.

It's true for n=1 (17, 71, 11) and n=2 (107, 701, 101 -> 107+701=808=101*8 and 107=101+3*2).

I have not found any other such sets for n<30000.

I have not found any other variants of solutions.

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Oscar wrote:

Next solution is emirp p = 107.
It is likely that there are no more solutions.
For compactness, let's define some functions.
rev(n): digit reversal of n, in base 10, OEIS A004086.
sopfr(n): sum of prime factors of n, with repetition, OEIS A001414.
Solutions p = 17 and p = 107 belong to an infinite family of candidates.
For positive exponent m, define integers f,p,q, as follows:
f = 10^m+1;
p = 10^m+7 = f+6;
q = 7*10^m+1 = rev(p).
If they are simultaneously prime, then p is a solution:
x = p+q = 8*10^m+8 = 2^3*f;
sopfr(x) = 2+2+2+f = p.
If exponent m admits an odd prime factor d, then integer 10^d+1 divides f itself.
So we must restrict our attention to exponents m = 2^k.
The "generalized Fermat numbers" GF_k = 10^(2^k)+1 have been extensively studied.
GF_0 = 11 is prime; choice f = 11 produces solution p=17.

GF_1 = 101 is prime; choice f = 101 produces solution p=107.

GF_k is known to be composite for 2<=k<=30.
Next candidate is GF_31 = 10^2147483648+1, whose status is still unknown due to its huge size.
However, as for "standard" Fermat numbers, a heuristic argument suggests that no number GF_k is prime for k>30.

Maybe more solutions can be found elsewhere?

The answer is negative: every solution necessarily belongs to the infinite family mentioned above.

I had no time to write down a complete, readable proof, so consider it just a claim for now, I'm sending a full proof this weekend.

Consequences of such claim:

there are definitely no solutions within interval 107 < p < 10^2147483648+7;

it is likely that there are no more solutions at all.

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Emmanuel wrote: