77th Birthday

Today I turned 77 years of age and my equivalent diurnal age is 28124 which has the following factorisation:$$28124=2 \times 2 \times 79 \times 89$$Though this number is composite, it has numerous prime number associations. Let's examine some of them beginning with its sum of digits, sum of digits squares and sum of digits cubed:$$ \begin{align} 2 + 8 + 1 + 2 + 4 &=17 \text{ (prime)} \\2^2+8^2+1^2+2^2+8^4 &= 89 \text{ (prime)} \\2^3+8^3+1^3+2^3+8^3 &= 593 \text{ (prime)} \end{align}$$The number is only one step removed from its home prime because:$$28124=2 \times 2 \times 79 \times 89 \rightarrow 227989 \text{ (prime)}$$The number is also a member of OEIS A048381: numbers such that replacing each nonzero digit with the n-th prime (replacing each 0 digit with a 1) yields a prime. Thus:$$28124 \rightarrow 319237 \text{ (prime)}$$The number has a binary complement that is prime. The binary complement of a number is determined by changing the number to binary and swapping any 0's for 1's and vice versa. Thus:$$ \begin{align} 28124_{10} &= 110110111011100_2 \\ &\rightarrow 001001000100011_2 \\ &=4643_{10} \text{ (prime)} \end{align}$$The number is quickly captured by the prime 28109 under the ODD(+) and EVEN(-) algorithm where the sum of the odd digits is added to the number and the sum of the even digits is subtracted recursively until a fixed point is reached or a loop is entered. Here is the trajectory is simply:$$ \begin{align} 28124 &\rightarrow 28124 + 1 -(2 + 8 + 2 + 4) \\ &=28124 + 1 - 16 \\ &=28109 \text{ (prime)} \end{align}$$The number can be considered as a concatenation of powers of the prime 2 because:$$ 28124 = 2^1\, | \,2^3 \,| \,2^0 \,| \,2^1 \,| \, 2^2 $$where | represents concatenation. The number can be generated by adding the prime sum (13) of the digits of the prime 28111 to itself. Thus:$$28111+13=28124$$The digits of the number can be rearranged to form the following primes:$$22481, 24281, 24821, 42281, 42821, 48221, 82241, 82421, 84221$$The position 28124 in the Recaman Sequence is reached after a prime number of iterations:$$0 \rightarrow 28124 \text{ requires } 34183 \text{ (prime) iterations}$$

Encoding Secret Numbers

The previous post about Heinz numbers got me thinking about how secret numbers that we might use as pass numbers for access to various services could be encoded. Let say we have a secret number 659347 that we want to encode using the one to one correspondence between Heinz numbers and integer sequences. Let's treat 659347 as a Heinz number, generate the corresponding integer sequence and then concatenate the numbers in the sequence:$$ \begin{align} \text{Heinz}(659347) &=(134, 19, 6) \\ &\rightarrow 134196 \end{align}$$If we concatenate the numbers in this sequence, there is ambiguity because the sequences (1341, 96) or (13419, 6) give the same result when the members are concatenated. It would thus be necessary to introduce a dot between the sequence members so as to resemble an IP address. An additional dummy 0 could be added so that it does indeed look like an IP address (0 cannot appear in any Heinz integer sequence). Thus we would have (permalink):$$ \begin{align} \text{Heinz}(659347) &= (134, 19, 6) \\ &\rightarrow 134.19.6  \text{ or} \\ &\rightarrow 134.19.6.0 \end{align}$$The point is that 134.19.6 or 134.19.6.0 can be written in plain sight and there's no way to decode the encrypted secret number (unless you know the magic word: Heinz).

An alternative method would be use the Gray Code. I've written about this in a post titled Gray Code in June of 2023. Here the secret number undergoes the following transformation:$$ \begin{align}659347 &= 10100000111110010011_2 \\ &\rightarrow 11110000100001011010_2 \\ &= 985178 \end{align} $$Again this number (985178) can be written in plain sight and there's no way to decode it without the magic phrase: Gray Code. Another encoding mechanism is Binary Complement. See my post Binary Complement from October of 2024. Here the encoding goes as follows:$$ \begin{align} 659347 &= 10100000111110010011_2\\ &\rightarrow 01011111000001101100_2\\ &=389228 \end{align} $$There is a problem once we try to decode from 389228. If we convert this number to binary then we don't have a leading zero anymore, we have 1011111000001101100. We need to add this leading zero because binary complements always have a leading zero. Once we do this, we can flip each 0 to 1 and each 1 to 0 to recover our secret number once we convert from binary. If one wanted to be really devious then two or even three encoding techniques could be used sequentially.