A396766 Numbers whose number of distinct prime factors equals their digital root.

19, 20, 30, 37, 38, 56, 64, 65, 66, 73, 74, 84, 92, 102, 109, 119, 120, 127, 138, 146, 155, 156, 163, 164, 165, 174, 181, 199, 200, 209, 218, 228, 236, 245, 246, 254, 255, 264, 271, 272, 273, 282, 289, 299, 300, 307, 318, 326, 335, 336, 343, 344, 345, 354, 361, 362, 371, 372, 379, 397

Offset

1,1

Comments

Since the digital root of n lies in {1..9}, every term has between 1 and 9 distinct prime factors. The sequence is infinite: by Dirichlet's theorem there are infinitely many primes p == 1 (mod 9), and each such prime has exactly one distinct prime factor and digital root 1, hence belongs to the sequence.

For each d from 1 to 9, there are infinitely many terms k with A001221(k) = A010888(k) = d. - Robert Israel, Jul 08 2026

Links

Table of n, a(n) for n=1..60.

Formula

{ k : A001221(k) = A010888(k) }.

Example

56 = 2^3 * 7 has 2 distinct prime factors {2, 7}; digital root of 56 is 5+6 = 11, then 1+1 = 2; since 2 = 2, 56 is a term.
120 = 2^3 * 3 * 5 has 3 distinct prime factors; digital root is 1+2+0 = 3; since 3 = 3, 120 is a term.

Maple

q:= x-> is(((x-1) mod 9)+1=nops(ifactors(x)[2])):
select(q, [$2..397])[];  # Alois P. Heinz, Jul 07 2026

Mathematica

Select[Range[400], PrimeNu[#] == (Mod[# - 1, 9] + 1) &]

Prog

(PARI) select(n->omega(n)==(n-1)%9+1, [1..400])
(Python) from sympy import primefactors
print([n for n in range(1, 400) if len(primefactors(n)) == 1+(n-1)%9])

Crossrefs

Cf. A001221, A010888, A061237 (subsequence).

Sequence in context: A274340 A241849 A054304 * A151979 A022109 A041730

Adjacent sequences: A396763 A396764 A396765 * A396767 A396768 A396769

Keyword

nonn,base,easy,new

Author

Alexander Pokorny, Jul 07 2026

Status

approved