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A335368
Harmonic numbers k with a record number of primes p not dividing k such that k*p is also a harmonic number.
3
1, 28, 1638, 30240, 2178540, 2457000, 32997888, 142990848, 1307124000, 71271827200, 547929930240, 2198278051200, 2567400675840, 54409216942080
OFFSET
1,2
COMMENTS
The corresponding record values are 0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 17.
If k is a harmonic number (A001599) and p is a prime that does not divide k, then k*p is a harmonic number if and only if (p+1)/2 is a divisor of the harmonic mean of the divisors of k, k*tau(k)/sigma(k) = k*A000005(k)/A000203(k).
LINKS
Mariano Garcia, On numbers with integral harmonic mean, The American Mathematical Monthly, Vol. 61, No. 2 (1954), pp. 89-96. See page 95.
EXAMPLE
1 is the first harmonic number, and it has 0 primes p such that 1*p = p is a harmonic number, since a prime number cannot be harmonic. The next harmonic number k with at least one prime p such that k*p is also a harmonic number is 28, since 28 * 5 = 140 is a harmonic number.
MATHEMATICA
harmNums = Cases[Import["https://oeis.org/A001599/b001599.txt", "Table"], {_, _}][[;; , 2]]; harMean[n_] := n*DivisorSigma[0, n]/DivisorSigma[1, n]; primeCount[n_] := Module[{d = Divisors[harMean[n]]}, Length @ Select[2*d - 1, PrimeQ[#] && ! Divisible[n, #] &]]; primeCountMax = -1; seq = {}; Do[If[(pc = primeCount[harmNums[[k]]]) > primeCountMax, primeCountMax = pc; AppendTo[seq, harmNums[[k]]]], {k, 1, Length[harmNums]}]; seq
KEYWORD
nonn,more
AUTHOR
Amiram Eldar, Jun 03 2020
STATUS
approved