A324388 -id:A324388 - cp's OEIS frontend

A306264 a(n) = 1 + d*a(n/d); a(1)=0. If n has only one prime divisor, then d=n, otherwise d is the greatest proper unitary divisor of n.

0, 1, 1, 1, 1, 4, 1, 1, 1, 6, 1, 5, 1, 8, 6, 1, 1, 10, 1, 6, 8, 12, 1, 9, 1, 14, 1, 8, 1, 16, 1, 1, 12, 18, 8, 10, 1, 20, 14, 9, 1, 22, 1, 12, 10, 24, 1, 17, 1, 26, 18, 14, 1, 28, 12, 9, 20, 30, 1, 21, 1, 32, 10, 1, 14, 34, 1, 18, 24, 36, 1, 10, 1, 38, 26, 20, 12, 40, 1, 17

Offset: 1

David James Sycamore, Feb 01 2019

nonn

Name related to recursive formula of A006022.
a(n) = 1 if and only if n is a prime power; p^t; t >= 1.
The sequence of indices k on which a(k) is a record (1,2,6,10,14,18,22,26,30,...), appears to be A111284.

			a(8) = a(25) = 1 because 8 and 25 are prime powers.
a(30) = 16 because 15 is the greatest proper unitary divisor of 30, so a(30) = 1 + 15*a(2) = 1 + 15 = 16.

Antti Karttunen, Table of n, a(n) for n = 1..20000
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Cf. A006022, A077610, A111284, A324388.

f[n_] := If[PrimePowerQ[n], n,
  SelectFirst[Transpose@
    {Reverse@ #[[-Ceiling[Length[#]/2] ;; -2]],
      #[[2 ;; Ceiling[Length[#]/2]]]} &@ Divisors[n],
    CoprimeQ @@ # &][[1]] ]; f[1] = 1;
a[n_] := 1 + #*a[n/#] &[f[n]]; a[1] = 0;
Array[a, 120] (* Michael De Vlieger, Jun 24 2025 *)

d(n) = if (omega(n) == 1, n, my(v=select(x->(gcd(x, n/x)==1), divisors(n))); v[#v-1]);
lista(nn) = {va = vector(nn); va[1] = 0; for (n=2, nn, dn = d(n); va[n] = 1 + dn*va[n/dn];); va;} \\ Michel Marcus, Feb 10 2019

A324388(n) = if(1>=omega(n),n,fordiv(n,d,if((d>1)&&(1==gcd(d,n/d)),return(n/d))));
A306264(n) = if(1==n,0,my(d=A324388(n)); 1+(d*A306264(n/d))); \\ Antti Karttunen, Feb 28 2019

a(1) = 0; for n > 1, a(n) = 1 + (A324388(n) * a(n/A324388(n))). - Antti Karttunen, Feb 28 2019

A346596 Let m = A344005(n) = smallest m such that n divides m*(m+1); a(n) = max(gcd(n,m), gcd(n,m+1)).

Original entry on oeis.org

1, 2, 3, 4, 5, 3, 7, 8, 9, 5, 11, 4, 13, 7, 5, 16, 17, 9, 19, 5, 7, 11, 23, 8, 25, 13, 27, 7, 29, 6, 31, 32, 11, 17, 7, 9, 37, 19, 13, 8, 41, 7, 43, 11, 9, 23, 47, 16, 49, 25, 17, 13, 53, 27, 11, 8, 19, 29, 59, 15, 61, 31, 9, 64, 13, 11, 67, 17, 23, 14, 71, 9, 73, 37, 25, 19

Offset: 1

Robert Dougherty-Bliss and N. J. A. Sloane, Jul 15 2021

nonn

This is the maximum of A345992 and A345993.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Cf. A344005, A345992, A345993, A345994, A345995.

Cf. also A034699, A324388.

f(n) = my(m=1); while ((m*(m+1)) % n, m++); m; \\ A344005
a(n) = my(m=f(n)); max(gcd(n,m), gcd(n,m+1)); \\ Michel Marcus, Aug 06 2021
(Python 3.8+)
from math import gcd, prod
from itertools import combinations
from sympy import factorint
from sympy.ntheory.modular import crt
def A346596(n):
    if n == 1:
        return 1
    plist = tuple(p**q for p, q in factorint(n).items())
    return n if len(plist) == 1 else max(gcd(n,s:=int(min(min(crt((m, n//m), (0, -1))[0], crt((n//m, m), (0, -1))[0]) for m in (prod(d) for l in range(1, len(plist)//2+1) for d in combinations(plist, l))))),gcd(n,s+1)) # Chai Wah Wu, Jun 17 2022

cp's OEIS Frontend

A306264 a(n) = 1 + d*a(n/d); a(1)=0. If n has only one prime divisor, then d=n, otherwise d is the greatest proper unitary divisor of n.

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A346596 Let m = A344005(n) = smallest m such that n divides m*(m+1); a(n) = max(gcd(n,m), gcd(n,m+1)).

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