A392819 Numbers k such that p + e divides k for every prime power p^e in the canonical prime factorization of k.

1, 4, 12, 48, 54, 60, 64, 90, 108, 120, 132, 180, 192, 224, 240, 270, 336, 360, 432, 448, 528, 540, 660, 672, 720, 729, 840, 918, 960, 1000, 1080, 1104, 1140, 1188, 1320, 1344, 1400, 1458, 1530, 1680, 1728, 1740, 1836, 1980, 1984, 2112, 2160, 2256, 2280, 2520, 2610

Offset

1,2

Comments

Numbers k such that lcm over p^e || k of (p + e) divides k.

There are no primes and no squarefree numbers > 1 in this sequence: for k = p we would need (p + 1) | p; for squarefree k every e = 1 forces (p + 1) | k for each p | k, which is impossible.

If k is odd and has at least two distinct odd primes, then either some exponent e is odd, in which case p + e is even and the lcm is even (so it cannot divide odd k); or all exponents are even, so k is a perfect square. Write k = Product_{i} p_i^{2*f_i}. Then for each i, p_i + 2*f_i is odd and coprime to p_i, so any prime factor of p_i + 2*f_i must be among the other p_j. For two distinct indices i <> j this would force simultaneous congruences p_i == -2*f_i (mod p_j) and p_j == -2*f_j (mod p_i), which cannot all hold unless k has only one prime factor. Hence k cannot have >= 2 distinct odd primes, so odd terms are prime powers with even exponent (A259417). Smallest odd term > 1: 729 = 3^6.

Conjecture: This sequence has natural density 0: the divisibility conditions are stronger than for refactorable numbers (which already have density 0); in particular, for k = Product_{i} p_i^{e_i}, each (p_i + e_i) must be p_n-smooth and divide k.

Comparison with A033950:

Property A033950 (Refactorable numbers) A392819 (This sequence)

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Divisibility Product_{i} (e_i + 1) divides k Product_{i} (e_i + p_i) divides k

Primes Only prime term is 2 No prime term

Squarefree numbers Only 1 and 2 Only 1

Odd terms Perfect squares Perfect squares that are odd prime powers

Natural density 0 0 (conjectured)

Links

Felix Huber, Table of n, a(n) for n = 1..10000

Example

54 = 2^1*3^3 is a term because (2 + 1) = 3 and (3 + 3) = 6 both divide 54.
840 = 2^3*3^1*5^1*7^1 is a term because (2 + 3) = 5, (3 + 1) = 4, (5 + 1) = 6 and (7 + 1) = 8 all divide 840.

Maple

A392819 := proc(n)
    option remember;
    local i, k;
    if n = 1 then
        1
    else
        for k from A392819(n - 1) + 1 do
            for i in ifactors(k)[2] do
                if k mod (i[1] + i[2]) <> 0 then
                    next k
                end if
            end do;
            return k
        end do
    end if;
end proc:
seq(A392819(n), n = 1 .. 51);

Mathematica

q[k_] := AllTrue[FactorInteger[k], Divisible[k, Total[#]] &]; q[1] = True; Select[Range[3000], q] (* Amiram Eldar, Feb 24 2026 *)

Prog

(PARI) isok(k) = if (k==1, 1, my(f=factor(k)); for (i=1, #f~, if (k % (f[i, 1]+f[i, 2]), return(0))); 1); \\ Michel Marcus, Feb 25 2026

Crossrefs

Cf. A000040, A000290, A005117, A033950, A259417, A390229.

Sequence in context: A035310 A022016 A151441 * A361428 A032380 A057346

Adjacent sequences: A392816 A392817 A392818 * A392820 A392821 A392822

Keyword

nonn

Author

Felix Huber, Feb 23 2026

Status

approved