A recent number, 27960, associated with my diurnal age has an interesting property if you look at the prime factors more closely:$$27960=2^3 \times 3 \times 5 \times 233$$Let's ignore multiplicity and look at the sum of digits of each distinct prime factor. The 2, 3 and 5 remain the same but 233 becomes 8 and the sequence of prime factors then becomes:$$2, 3, 5, 8$$This is the Fibonacci sequence. How many numbers with four prime factors (ignoring multiplicity) have this property. Well, in the range up to 40000, it turns out that there are 115. These numbers are (permalink):
510, 1020, 1530, 1590, 2040, 2130, 2550, 3060, 3180, 3210, 4080, 4260, 4590, 4770, 5100, 6120, 6360, 6390, 6420, 6990, 7314, 7530, 7650, 7950, 8160, 8520, 8670, 9180, 9540, 9630, 9798, 10200, 10650, 12240, 12720, 12750, 12780, 12840, 12930, 13038, 13770, 13980, 14310, 14628, 14766, 15060, 15090, 15300, 15630, 15900, 16050, 16320, 17040, 17085, 17340, 17466, 18360, 19080, 19170, 19260, 19596, 20400, 20970, 21030, 21300, 21942, 22590, 22950, 23850, 24480, 25440, 25500, 25560, 25680, 25860, 26010, 26076, 26322, 27540, 27960, 28620, 28890, 29256, 29394, 29532, 30120, 30180, 30600, 31260, 31800, 31830, 31950, 32100, 32154, 32637, 32640, 34080, 34530, 34638, 34680, 34932, 34950, 35445, 35511, 36690, 36720, 37650, 38160, 38250, 38340, 38520, 38790, 39114, 39192, 39750
Lets consider the last number in this list: 39750. In this case we have:$$ \begin{align} 39750 &= 2 \times 3 \times 5^3 \times 53 \\ & \rightarrow 2, 3, 5, 8 \end{align}$$If we extend the number of prime factors to five, then only one number satisfies in the range up to 40000:$$ \begin{align}34170 &= 2 \times 3 \times 5 \times 17 \times 67 \\ &\rightarrow 2, 3, 5, 8, 13 \end{align}$$However, there are 160 numbers that satisfy in the range up to one million (permalink). The algorithm actually looks for generalised Fibonacci sequences:$$a,b,c,d,e, \dots \text{ such that } c = a+b, d=b+c, e=c+d \dots$$However, all of the 160 numbers begin with 2. For example, the last of the numbers is 9988520:$$ \begin{align} 998520 &= 2^3 \times 3 \times 5 \times 53 \times 157 \\ &\rightarrow 2, 3, 5, 8, 13 \end{align} $$I tried with six prime factors but without success up to 100 million. Finally I realised that any suitable number must add to 21 and thus be divisible by 3. Such a number can never be a prime factor and so the prime factor sequence must end at 13 and can never progress further.





