Devansh Singh - cp's OEIS frontend

A350878 Integers m that divide the sum of values dp < m, where d is a divisor of m, p is a prime, and dp does not divide m.

1, 2, 5, 10, 18, 24, 32, 60, 71, 100, 512, 2990, 9910, 10031, 12618, 32674, 53586, 153878, 223500, 312608, 369119, 386110, 466569, 4491817, 7068356, 8765871, 65311881

Offset: 1

Devansh Singh, Jan 20 2022

Conjecture: The sum of values d*p < m in the definition of the sequence is equal to m for m = 5 only. True for m <= 15000.
A007506 is the subsequence of the prime terms of this sequence. - Amiram Eldar, Jan 20 2022
a(28) > 10^8. - David A. Corneth, Jan 21 2022

David A. Corneth, PARI program
Jon E. Schoenfield, Magma program

Cf. A334800, A000040, A007506.

q[n_] := Module[{ds = Divisors[n], s = 0, r}, Do[r = n/d; ps = Select[Range[2, r], PrimeQ[#] && ! Divisible[n, d*#] &]; s += Total[d*ps], {d, ds}]; Divisible[s, n]]; Select[Range[3000], q] (* Amiram Eldar, Jan 20 2022 *)

isok(m) = {my(d=divisors(m), s=0); forprime(p=2, m, for(k=1, #d, my(x=d[k]*p); if ((x < m) && (m % x), s+=x););); (s % m) == 0;} \\ Michel Marcus, Jan 21 2022

\\ See Corneth link \\ David A. Corneth, Jan 21 2022

import sympy
A350878=[]
for m in range(1,15001):
    sum=0
    primes_lessthan_m_by2 = list(sympy.primerange(2,-(m//-2)))
    primes_between_m_by2_and_m = list(sympy.primerange(m//2+1,m))
    divisors_of_m=sympy.divisors(m,generator=False)
    divisors_of_m.remove(m)
    if m%2==0:
        divisors_of_m.remove(m//2)
    for p in primes_between_m_by2_and_m:
        sum+=p
    for p in primes_lessthan_m_by2:
        for d in divisors_of_m:
            if p< m//d and m%(d*p)!=0:
                sum+=d*p
    if sum%m==0:
       A350878.append(m)
print(A350878)

a(16)-a(20) from Amiram Eldar, Jan 21 2022

a(21)-a(27) from David A. Corneth, Jan 21 2022

A350046 Numbers m such that, in the prime factorization of m! (with the primes in ascending order), no two successive exponents differ by a composite number.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 15, 32, 35, 162, 163

Offset: 1

Devansh Singh, Dec 11 2021

nonn

Is this sequence finite?
After 163, there are no more terms through 10^10. - Jon E. Schoenfield, Dec 15 2021
Sequence is the same as numbers m such that, in the prime factorization of m! (with the exponents in ascending order), no two successive exponents differ by a composite number. This is due to the fact that in the factorization of m! with the primes in ascending order, the corresponding exponents are in descending order. - Chai Wah Wu, Jan 10 2022

			15! = 2^11 * 3^6 * 5^3 * 7^2 * 11^1 * 13^1; the exponents are 11, 6, 3, 2, 1, 1, and no two successive/neighboring exponents differ by a composite number, so 15 is a term of the sequence.

Cf. A000142, A330706.

q[n_] := AllTrue[Differences @ Reverse[FactorInteger[n!][[;; , 2]]], !CompositeQ[#] &]; Select[Range[2, 200], q] (* Amiram Eldar, Dec 11 2021 *)

valp(n,p)=my(s); while(n\=p, s+=n); s
is(n)=my(o=valp(n,2),e); forprime(p=3,, e=valp(n,p); if(o-e<4, return(1)); if(isprime(o-e), o=e, return(0))) \\ Charles R Greathouse IV, Dec 15 2021

import sympy
def last_exponent(c,i):
    if sympy.nextprime(i//2)<=i:
        if c==1 or sympy.isprime(c-1):
            return(True)
        else: return(False)
    else:
        if c==1 or sympy.isprime(c):
            return(True)
        else: return(False)
A350046_n=[2,3]
for i in range(4,1001):
    p_expo=True
    x = list(sympy.primerange(2,i//2+1))
    prime_expo=[]
    for j in (x):
        c=i//j
        s=0
        while c!=0:
            s=s+c
            c=c//j
        prime_expo.append(s)
    c=prime_expo[0]
    l=len(prime_expo)
    for j in range(1,l):
        c=c-prime_expo[j]
        if c!=0:
            if c!=1 and not sympy.isprime(c):
                p_expo=False
                break
        c=prime_expo[j]
    prime_expo=last_exponent(c,i)
    if p_expo==True:
       A350046_n.append(i)
print(A350046_n)

from collections import Counter
from itertools import count, islice
from sympy import factorint, isprime
def A350046_gen(): # generator of terms
    f = Counter()
    for m in count(2):
        f += Counter(factorint(m))
        e = sorted(f.items())
        if all(d <= 1 or isprime(d) for d in (abs(e[i+1][1]-e[i][1]) for i in range(len(e)-1))):
            yield m
A350046_list = list(islice(A350046_gen(),15)) # Chai Wah Wu, Jan 10 2022

A343541 For n > 1, a(n) is the largest base b <= prime(n)-1 such that the digits of prime(n)-1 in base b contain the digit b-1.

Original entry on oeis.org

2, 2, 3, 2, 4, 3, 3, 5, 4, 6, 2, 2, 7, 7, 4, 8, 8, 6, 6, 9, 9, 2, 3, 10, 5, 6, 6, 5, 11, 8, 3, 12, 12, 5, 3, 13, 13, 13, 5, 6, 6, 14, 14, 10, 10, 15, 15, 5, 5, 11, 11, 16, 16, 2, 3, 4, 5, 17, 17, 17, 10, 18, 18, 18, 18, 13, 13, 19, 19, 19

Offset: 2

Devansh Singh, Apr 18 2021

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A000040, A338295.

f:= proc(n) local p,b,L;
    p:= ithprime(n);
    for b from floor((1 + sqrt(4*p - 3))/2) by -1 do
      L:= convert(p-1,base,b);
      if member(b-1,L) then return b fi
    od;
end proc:
map(f, [$2 .. 100]); # Robert Israel, Dec 10 2024

Table[Max@Select[Range[2,Prime@n-1],MemberQ[IntegerDigits[Prime@n-1,#],#-1]&],{n,2,71}] (* Giorgos Kalogeropoulos, Nov 22 2021 *)

a(n) = my(q=prime(n)-1); forstep(b=q, 2, -1, if (vecmax(digits(q, b)) == b-1, return (b))); \\ Michel Marcus, Apr 19 2021

import sympy
def a_n(N):
    a_n=[2]
    for i in sympy.primerange(5, N+1):
        a_n.append(A338295(i-1))
    print(a_n)
def A338295(n):
    checker=0
    for b in range(n//2, 1,-1):
        checker=main_base_check(n, b)
        if checker!=0:
            break
    return checker
def main_base_check(m, b):
    while m!=0:
        if m%b == b-1:
            return b
        m = m//b
    return 0
a_n(500)

a(n) <= (1 + sqrt(4*prime(n) - 3))/2 for all n. Prime(n), which is 111 in some base Q, has a(n) = Q+1. Example: 31 = 6*5 + 1 and it is 111 in base 5. - Devansh Singh, Nov 22 2021

A342679 Number of steps for n to reach 1 or n by repeated application of A037916, or -1 if they are never reached.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 4, 1, 3, 2, 2, 1, 2, 2, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 3, 3, 2, 1, 2, 2, 4, 2, 3, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 3, 3, 2, 3, 1, 3, 2, 3, 1, 3, 1, 2, 4, 3, 2, 3, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 1

Offset: 2

Devansh Singh, Mar 18 2021

Let n = (p1^a1)*(p2^a2)*...*(pk^aj) be the prime-factorization of n >= 2, where primes are in ascending order and ai >= 1 for all i >= 1 and <= k, then form n' = a1 a2 a3 ... ak = A037916(n), the concatenation of the exponents. Repeat this process if n' != 1 and != n, otherwise stop.
When n is prime, a(n) = 1; when n is semiprime, a(n) = 2.
Does every n reach 1 by this process, or does there exist some n whose trajectory enters a cycle, i.e., we reach n again instead of 1?
No cycles for n <= 10^9. - Michael S. Branicky, Mar 21 2021

			3 = 3^1 -> 1, so a(3) = 1;
6 = 2^1 * 3^1 -> 11 = 11^1 -> 1, so a(6) = 2;
16 = 2^4 -> 4 = 2^2 -> 2 = 2^1 -> 1, so a(16) = 3;
50 = 2^1 * 5^2 -> 12 = 2^2 * 3^1 -> 21 = 3^1 * 7^1 -> 11 -> 1, so a(50) = 4.

Cf. A000040, A037916.

Table[Length@Rest@Most@FixedPointList[FromDigits[Last/@FactorInteger@#]&,k],{k,2,100}] (* Giorgos Kalogeropoulos, Apr 01 2021 *)

f(n) = my(f=factor(n)[,2], s=""); for(i=1, #f~, s=concat(s,Str(f[i]))); eval(s); \\ A037916
a(n) = my(k=n, nb=0); while (k != 1, k = f(k); nb++); nb; \\ Michel Marcus, Mar 18 2021

import sympy
N=int(input())
A342679_n=[]
for n in range(2,N+1):
    n_0=n
    steps=0
    while not sympy.isprime(n) :
        exponents=list(sympy.factorint(n).values())
        m=""
        for i in exponents:
            m=m+str(i)
        n=int(m)
        if n==n_0:
            break
        steps+=1
    A342679_n.append(steps+1)
print(A342679_n)

def a(n):
  c, iter = 1, A037916(n)
  while iter != 1 and iter != n: c, iter = c+1, A037916(iter)
  return c
print([a(n) for n in range(2, 89)]) # Michael S. Branicky, Mar 20 2021

A340287 Numbers k for which there are A000005(k+1)-1 bases b such that k in base b contains digit b-1.

Original entry on oeis.org

1, 3, 5, 9, 11, 17, 27, 35, 37, 39, 41, 65, 81, 83, 85, 89, 131, 149, 179, 203, 255, 257, 263, 407, 419, 455, 539, 739, 811, 899, 1031, 1109, 1385, 1619, 1631, 1883, 2819, 3299, 3527, 4133, 4139, 4151, 4919, 5669, 5939, 6299, 7055, 7307, 8303, 9829, 9839, 10661

Offset: 1

Devansh Singh, Jan 03 2021

If k written in some base b has b-1 as its least significant digit, then b is a divisor of k+1 (because k = high*b + b-1 means k+1 = b*(high+1)).  The present sequence is those k where such divisors are in fact the only bases b where digit b-1 occurs.
In terms of the count of bases in A337496, the divisors give a lower bound A337496(k) >= tau(k+1)-1 (number of divisors except 1).  The present sequence is those k at this lower bound.
There are 147 terms to k <= 10^8.

			1 is a term because A000005(1+1)-1 = 1 such that 1 in base 2 (2|2) contains digit 1 and there are no such bases of 1 which are non-divisors of 1+1.
5 is a term because A000005(5+1)-1 = 3 such that 5 in base 2,3,6 (2|6,3|6 and 6|6) contains digit 1,2,5 respectively and there are no such bases of 5 which are non-divisors of 5+1.

Kevin Ryde, C code searching for terms

Cf. A000005, A337496.

baseQ[n_, b_] := MemberQ[IntegerDigits[n, b], b - 1]; q[n_] := Count[Range[2, n + 1], ?(baseQ[n, #] &)] == DivisorSigma[0, n + 1] - 1; Select[Range[1000], q] (* _Amiram Eldar, Jan 03 2021 *)

isok(k) = sum(b=2, k+1, (#select(x->(x==(b-1)), digits(k, b)))>0) == numdiv(k+1)-1; \\ Michel Marcus, Jan 03 2021

def is_n_with_no_nondivisor_baseb(N):
    return list(filter(n_with_no_nondivisor_baseb,range(1,N+1,2)))
def n_with_no_nondivisor_baseb(n):
    main_base_counter=0
    for b in range(3,((n+1)//2) +1):
        if (n+1)%b!=0:
            main_base_counter=main_base_check(n//b,b)+main_base_counter
            if main_base_counter==1:
                break
    return main_base_counter==0
def main_base_check(m,b):
    while m!=0:
        if m%b == b-1:
            return 1
        m = m//b
    if m==0:
        return 0
print(is_n_with_no_nondivisor_baseb(int(input())))

A337162 Numbers m such that Sum_{d|m: 1<=d<=sqrt(m)} m/d-d is a multiple of m.

Original entry on oeis.org

1, 6, 840, 3420

Offset: 1

Devansh Singh, Jan 28 2021

Integers m such that A079667(m) is a multiple of m.
Sum_{d|m: 1<=d<=sqrt(m)} (m/d)-d = 0 only when m=1 and Sum_{d|m: 1<=d<=sqrt(m)} (m/d)-d = m only when m=6.
If m is q+1-perfect and 2*Sum_{d|m: 1<=d<=sqrt(m)} d=m then Sum_{d|m: 1<=d<=sqrt(m)} (m/d)-d = m*q or if m is member of this sequence i.e. Sum_{d|m: 1<=d<=sqrt(m)} (m/d)-d = m*q and 2*Sum_{d|m: 1<=d<=sqrt(m)} d=m then m is q+1-perfect.
Does there exist any m apart from 6 which is q+1-perfect, q>=2 and satisfies 2*Sum_{d|m: 1<=d<=sqrt(m)} d=m? Because if it exists then m is member of this sequence.

Cf. A079667, A007691 (in comment).

Select[Range[10^5], Function[m, Mod[DivisorSum[m, Abs[m/# - #] &, # <= Sqrt[m] &], m] == 0]] (* Michael De Vlieger, Mar 17 2021 *)

isok(m) = !(sumdiv(m, d, if (d^2 <= m, m/d-d)) % m); \\ Michel Marcus, Jan 28 2021

A338420 Numbers k having exactly one base b which is not a divisor of k+1, and k contains the digit b-1 in base b.

Original entry on oeis.org

2, 4, 7, 8, 10, 13, 15, 19, 23, 25, 26, 29, 31, 36, 38, 40, 51, 53, 55, 59, 63, 71, 80, 82, 84, 86, 87, 99, 101, 107, 109, 119, 127, 128, 129, 137, 143, 151, 152, 155, 161, 167, 169, 209, 215, 227, 256, 259, 260, 261, 265, 266, 267, 269, 271

Offset: 1

Devansh Singh, Oct 25 2020

All the terms of A337536 are in this sequence except A337536(2)=3.

There are only 30 terms which are even up to n=124705.

Cf. A337536.

baseCount[n_] := Count[Complement[Range[n + 1], Divisors[n + 1]], ?(MemberQ[ IntegerDigits[n, #], # - 1] &)]; Select[Range[1000], baseCount[#] == 1 &] (* _Amiram Eldar, Oct 25 2020 *)

isok(k) = sum(b=2, k+1, ((k+1) % b) && #select(x->(x==b-1), digits(k, b))) == 1; \\ Michel Marcus, Oct 30 2020

def A338420(N):
    return list(filter(isA338420,range(1,N+1)))
def isA338420(n):
    counter=0
    if n==2 or n==4:
        return True
    if n%2==0:
        counter=1
    for b in range(3,(n//2) +1):
        if (n+1)%b!=0:
            counter=main_base_check(int(n/b),b)+counter
    return counter==1
def main_base_check(m,b):
    while m!=0:
        if m%b == b-1:
            return 1
        m = m//b
    return 0
print(A338420(int(input())))

A333970 Irregular triangle read by rows where the n-th row lists the bases 2<=b<=n+1 where n in base b contains the digit b-1.

Original entry on oeis.org

2, 2, 3, 2, 4, 2, 5, 2, 3, 6, 2, 3, 7, 2, 3, 4, 8, 2, 3, 9, 2, 5, 10, 2, 11, 2, 3, 4, 6, 12, 2, 4, 13, 2, 4, 7, 14, 2, 3, 4, 5, 15, 2, 3, 4, 8, 16, 2, 3, 17, 2, 3, 6, 9, 18, 2, 3, 19, 2, 3, 4, 5, 10, 20, 2, 3, 5, 7, 21, 2, 3, 5, 11, 22, 2, 3, 5, 23, 2, 3, 4, 5, 6, 8, 12, 24

Offset: 1

Devansh Singh, Sep 03 2020

If a number n has base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)) contains digit b-1, where b = q*(k+1)/k, k>=1 , and Sum_{i>=0} ((A(i)(mod b-q))*((b-q)^i)) > 0 then there exists  n' < n such that that n' in base b-q = b' contains digit b'-1 at the same place as n in base b and 0 <= (A(i)-A'(i))/b' <= (k+1)-((A'(i)+1)/b') (A'(i) is digit of n' in base b')for all i>=0.*
This condition is necessary and sufficient.
Proof that Condition is Necessary:
Since b-1 = b-q+q-1 and b' = q/k (as b = q*(k+1)/k). Therefore (b-1) (mod b') = (b'+q-1) (mod b') = (q-1) (mod b') = b'-1 :-(1).
n in base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)).Then n = Sum_{i>=0} (A(i)*(b^i)) = Sum_{i>=0} (A(i)*((b-q+q)^i)).
n = Sum_{i>=0} (A(i)*(b'^i)) +
    Sum_{i>=1} (A(i)*(b^i - b'^i))
   = Sum_{i>=0} (A'(i)*(b'^i)) + Sum_{i>=0} ((A(i)-A'(i))* (b'^i)) + Sum_{i>=1} (A(i)*(b^i - b'^i)),
         where A'(i) = A(i) (mod b').
Now n-Sum_{i>=0} ((A(i)-A'(i))*(b'^i))
       - Sum_{i>=1} (A(i)*(b^i - b'^i))
       = Sum_{i>=0} (A'(i)*(b'^i)).
Since A'(j) = A(j) (mod b') =  (b-1) (mod b') = b'-1(due to equation (1) above and A(j) = b-1.
Hence there exists n'  = Sum_{i>=0} (A'(i)*(b'^i)) > 0 containing digit b'-1 in base b'.
Table of n/b with cell containing T(n, b) = (n', b') for q = b/2. n' = Sum_{i>=0} (A'(i)*(b'^i))
n/b|  4  |  6  |  8  |  10 | 12
3  |(1,2)|     |     |     |
4  |     |     |     |     |
5  |     |(2,3)|     |     |
6  |     |     |     |     |
7  |(3,2)|     |(3,4)|     |
8  |     |     |     |     |
9  |     |     |     |(4,5)|
10 |     |     |     |     |
11 |(1,2)|(5,3)|     |     |(5,6)
Example: For table n/b in comments containing (n',b') in its cells.
For n = 7:
In base b = 4, n = 13 :- q = b' = 4/2 = 2, and n' = (3 mod (2))*(2)^0 + (1 mod(2))*(2)^1 = 1+2 = 3.
In base b = 8, n = 7 :- q = b' = 8/2 = 4, and n' = (7 mod (4))*(4)^0 = 3.
There are no other bases b >= 4 except 4, 8 for n = 7.
(n, b) maps to (0, 1) if b is prime. Following this and comment in A337536 we can say that all of the terms of A337536 will map to (0, 1) only, except A337536(2).
For above (n, b) -> (n', b') one possible (n, b) pair for (n', b') is { Sum_{i>=0} ((A'(i)+b') *((2*b')^i)), 2*b'}.

			Triangle begins
  Row    Bases
  n=1:   2
  n=2:   2  3
  n=3:   2  4
  n=4:   2  5
  n=5:   2  3  6
  n=6:   2  3  7
  n=7:   2  3  4  8
  n=8:   2  3  9
  n=9:   2  5  10
  n=10:  2  11

Rémy Sigrist, Table of n, a(n) for n = 1..10027 (rows for n = 1..1014, flattened)

Cf. A337535 (second column), A338295 (penultimate column), A337496 (row widths), A337536 (width 2), A337143 (width 3).

Rows containing bases 3..11 respectively: A074940, A337250, A337572, A333656, A337141, A337239, A338090, A011539, A095778.

row(n) = {my(list = List()); for (b=2, n+1, if (vecmax(digits(n, b)) == b-1, listput(list, b));); Vec(list);} \\ Michel Marcus, Sep 11 2020

More terms from Michel Marcus, Sep 11 2020

A337355 a(n) is the denominator of Product_{i=0..n-1} (n-i)^((-1)^ceiling(i/2)).

Original entry on oeis.org

1, 1, 2, 3, 6, 5, 5, 63, 56, 4, 225, 77, 1232, 312, 65, 2695, 1408, 14144, 9945, 92169, 53504, 28288, 41769, 370139, 447488, 217600, 502775, 1427679, 2684928, 2524160, 10414625, 132774147, 443908096, 2375680, 1168520925, 3129357, 18644140032, 35160064, 1108596775, 2318853537

Offset: 1

Devansh Singh, Aug 24 2020

			a(5) = denominator of (5*2)/(4*3) = 6.

Cf. A337354 (numerators).

a(n) = {denominator(prod(i=0, n-1, (n-i)^(-1)^((i+1)\2)))} \\ Andrew Howroyd, Aug 25 2020

Terms a(31) and beyond from Andrew Howroyd, Aug 25 2020

A337354 a(n) is the numerator of Product_{i=0..n-1} (n-i)^((-1)^ceiling(i/2)).

Original entry on oeis.org

1, 2, 3, 2, 5, 9, 7, 40, 45, 7, 308, 48, 975, 539, 88, 1664, 1105, 24255, 13376, 56576, 41769, 48279, 55936, 226304, 348075, 370139, 671232, 870400, 2082925, 4283037, 13872128, 80773120, 343682625, 4023459, 1553678336, 1900544, 14411758075, 59457783, 1471905792, 1406402560

Offset: 1

Devansh Singh, Aug 24 2020

a(n) is the numerator of (n/(n-1)) * ((n-3)/(n-2)) * ((n-4)/(n-5)) ...

			a(n)/A337355(n) equals 1, 2, 3/2, 2/3, 5/6, 9/5, 7/5, 40/63, 45/56, 7/4 ...
a(4) = numerator of (4*1)/(3*2) = numerator of 2/3 = 2.
a(5) = numerator of (5*2)/(4*3) = numerator of 5/6 = 5.
                      12  *   9*8  *  5*4  *  1
a(12) = numerator of --------------------------- = 48.
                        11*10  *  7*6  *  3*2

Cf. A337355 (denominators).

Cf. A076390, A085565, A007662.

a(n) = {numerator(prod(i=0, n-1, (n-i)^(-1)^((i+1)\2)))} \\ Andrew Howroyd, Aug 24 2020

a(n) = numerator of (n*A337355(n-2))/(a(n-2)*(n-1)) for n>=3.

Conjecture: a(4*n)/A337355(4*n) ~ 0.5990701173677... (=A076390). - Andrew Howroyd, Aug 25 2020

Terms a(31) and beyond from Andrew Howroyd, Aug 25 2020

cp's OEIS Frontend

User: Devansh Singh

A350878 Integers m that divide the sum of values d*p < m, where d is a divisor of m, p is a prime, and d*p does not divide m.

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A350046 Numbers m such that, in the prime factorization of m! (with the primes in ascending order), no two successive exponents differ by a composite number.

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A343541 For n > 1, a(n) is the largest base b <= prime(n)-1 such that the digits of prime(n)-1 in base b contain the digit b-1.

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A342679 Number of steps for n to reach 1 or n by repeated application of A037916, or -1 if they are never reached.

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A340287 Numbers k for which there are A000005(k+1)-1 bases b such that k in base b contains digit b-1.

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A337162 Numbers m such that Sum_{d|m: 1<=d<=sqrt(m)} m/d-d is a multiple of m.

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A338420 Numbers k having exactly one base b which is not a divisor of k+1, and k contains the digit b-1 in base b.

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A350878 Integers m that divide the sum of values dp < m, where d is a divisor of m, p is a prime, and dp does not divide m.