A079427 - OEIS

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A079427

Least m > n having the same number of divisors as n, a(1) = 1.

1, 3, 5, 9, 7, 8, 11, 10, 25, 14, 13, 18, 17, 15, 21, 81, 19, 20, 23, 28, 22, 26, 29, 30, 49, 27, 33, 32, 31, 40, 37, 44, 34, 35, 38, 100, 41, 39, 46, 42, 43, 54, 47, 45, 50, 51, 53, 80, 121, 52, 55, 63, 59, 56, 57, 66, 58, 62, 61, 72, 67, 65, 68, 729, 69, 70, 71, 75, 74, 78, 73

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OFFSET

1,2

COMMENTS

tau(a(n)) = tau(n) and tau(i) <> tau(n), n < i < a(n) (tau = A000005);

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000

FORMULA

a(A000040(k)) = A079428(A000040(k)) = A000040(k+1), as A000005(p)=2 for primes p.

a(n) = A171937(n) + n. - Ridouane Oudra, Sep 14 2021

EXAMPLE

Sets of divisors for n=10,11,12,13 and 14: D(10)={1,2,5,10}, D(11)={1,11}, D(12)={1,2,3,4,6,12}, D(13)={1,13}, D(14)={1,2,7,14}: therefore a(10)=14 (#D(10)=#D(14)).

MATHEMATICA

a[1] = 1; a[n_] := Module[{m = n+1, d=DivisorSigma[0, n]}, While[DivisorSigma[0, m] != d, m++]; m]; Array[a, 100] (* Amiram Eldar, Feb 03 2020 *)

PROG

(PARI) a(n) = if (n==1, 1, my(m=n+1, nd=numdiv(n)); while(numdiv(m) != nd, m++); m); \\ Michel Marcus, Sep 14 2021

(Python)

from sympy import divisors

def a(n):

if n == 1: return 1

divisorsn, m = len(divisors(n)), n + 1

while len(divisors(m)) != divisorsn: m += 1

return m

print([a(n) for n in range(1, 72)]) # Michael S. Branicky, Sep 14 2021