A271328 -id:A271328 - cp's OEIS frontend

A271471 Values k > 1 in A271328 such that k does not have any nontrivial divisors in A271328.

5, 17, 28, 37, 82, 106, 122, 197, 228, 257, 294, 362, 406, 577, 628, 677, 842, 906, 1161, 1228, 1297, 1376, 1522, 1606, 1682, 1937, 2028, 2117, 2402, 2513, 2606, 2917, 3028, 3142, 3482, 3606, 3722, 4236, 4357, 4629, 4762, 4906, 5042, 5483, 5777, 6242, 6406

Offset: 1

Peter Kagey, Apr 08 2016

nonn

If k is in this sequence, then all multiples of k are in A271468.

			A271328(1) = 1 is not in this sequence because A271328(1) is not greater than 1.
A271328(2) = 5 is in this sequence because 5 does not have any nontrivial divisors.
A271328(3) = 10 is not in this sequence because A271328(2) is a proper divisor of A271328(3).
A271328(4) = 17 is in this sequence because 17 does not have any nontrivial divisors.
A271328(5) = 28 is in this sequence because 2, 4, 7, and 14 are not in A271328.

Peter Kagey, Table of n, a(n) for n = 1..10000

nn = 240; s = {1}; Do[AppendTo[s, Total@ Select[Range[n - 1], Divisible[n, s[[#]]] &]], {n, 2, nn}]; t = Table[s[[3 n]]/3, {n, nn/3}];
Select[Rest@ t, NoneTrue[Most@ Rest@ Divisors@ #, MemberQ[t, #] &] &] (* Michael De Vlieger, Apr 08 2016 *)

A269347 With a(1) = 1, a(n) is the sum of all 0 < m < n for which a(m) divides n.

Original entry on oeis.org

1, 1, 3, 3, 3, 15, 3, 3, 30, 3, 3, 51, 3, 3, 84, 3, 3, 111, 3, 3, 150, 3, 3, 195, 3, 3, 246, 3, 3, 318, 3, 3, 366, 3, 3, 435, 3, 3, 510, 3, 3, 591, 3, 3, 684, 3, 3, 771, 3, 3, 882, 3, 3, 975, 3, 3, 1086, 3, 3, 1218, 3, 3, 1326, 3, 3, 1455

Offset: 1

Alec Jones, Feb 24 2016

For n > 2, I can prove that a(n) = 3 if 3 does not divide n, and in general, 3 divides a(n).
The base case is a(3) = 3. Suppose that the results hold for a(n) over 3 < n < k; we will show that the results hold for a(k) also. In the case that 3 does not divide k, then a(k) = 3, since a(1) and a(2) divide k but no other previous term can. This proves the first claim.
Otherwise, if 3 does divide k, then a(m) divides k for each 0 < m < k not divisible by 3; these numbers can be divided into k/3 pairs so that the sum of each pair is congruent to 0 modulo 3 (for instance, 1 + 2 == 4 + 5 == 7 + 8 == ... == 0 (mod 3)). If a(m) divides k for some 0 < m < k divisible by 3, this m does not change the congruence class of the sum that forms a(k). Thus, a(k) == 0 (mod 3) as required to prove the second claim.

			a(1) = 1;
a(2) = 1 because a(1) divides 2;
a(3) = 3 because a(1) and a(2) divide 3: 1+2=3;
a(4) = 3 because a(1) and a(2) divide 4: 1+2=3;
a(5) = 3 because a(1) and a(2) divide 5: 1+2=3;
a(6) = 15 because a(1), a(2), a(3), a(4), and a(5) divide 6: 1+2+3+4+5=15.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from Alec Jones)

Cf. A088167 which gives the number of m < n for which a(m) divides n.

Cf. A271326, A271328.

a269347 1 = 1
a269347 n = genericIndex a269347_list (n - 1)
a269347_list = map a [1..] where
  a n = sum $ filter ((==) 0 . mod n . a269347) [1..n-1]
-- Peter Kagey, Jun 17 2016

int[] terms = new int[1000];
terms[0] = 1;
for (int i = 1; i < 1000; i++) {
     int count = 0;
     for (int j = 0; j < i; j++) {
          if ((i + 1) % terms[j] == 0) {
               count = count + (j + 1);
          }
     }
     terms[i] = count;
}

a = {1}; Do[AppendTo[a, Total@ Select[Range[n - 1], Divisible[n, a[[#]]] &]], {n, 2, 66}]; a (* Michael De Vlieger, Mar 24 2016 *)

lista(nn) = {va = vector(nn); va[1] = 1; for (n=2, nn, va[n] = sum(k=1, n-1, k*((n % va[k])==0));); va;} \\ Michel Marcus, Feb 24 2016

from itertools import count, islice
from sympy import divisors
def A269347_gen(): # generator of terms
    A268347_dict = {1:1}
    yield 1
    for n in count(2):
        yield (s:=sum(A268347_dict.get(d,0) for d in divisors(n,generator=True)))
        A268347_dict[s] = A268347_dict.get(s,0) + n
A269347_list = list(islice(A269347_gen(),40)) # Chai Wah Wu, Nov 17 2022

def a(n)
  seq = [1]
  (2..Float::INFINITY).each do |i|
    return seq.last[0...n].last if seq.length > n
    indices = seq.each_index.select { |j| i % seq[j] == 0 }
    seq << indices.map(&:next).reduce(:+)
  end
end # Peter Kagey, Feb 25 2016

A271326 a(n) = A269347(3n).

Original entry on oeis.org

3, 15, 30, 51, 84, 111, 150, 195, 246, 318, 366, 435, 510, 591, 684, 771, 882, 975, 1086, 1218, 1326, 1455, 1590, 1731, 1884, 2031, 2190, 2370, 2526, 2718, 2886, 3075, 3270, 3483, 3684, 3891, 4128, 4335, 4566, 4818, 5046, 5295, 5550, 5811, 6084

Offset: 1

Alec Jones, Apr 04 2016

This sequence represents the "peaks" of A269347(n), which occur for that sequence when n is divisible by 3.

Terms of this sequence are equal to 3n^2 + 3 with predictable frequency; see my comment on A271328.

Alec Jones, Table of n, a(n) for n = 1..10001

Cf. A269347, A271328

A271468 List of indices i such that A269347(3i) != 3(i^2 + 1).

Original entry on oeis.org

1, 5, 10, 15, 17, 20, 25, 28, 30, 34, 35, 37, 40, 45, 50, 51, 55, 56, 60, 65, 68, 70, 74, 75, 80, 82, 84, 85, 90, 95, 100, 102, 105, 106, 110, 111, 112, 115, 119, 120, 122, 125, 130, 135, 136, 140, 145, 148, 150, 153, 155, 160, 164, 165, 168, 170, 175, 180

Offset: 1

Peter Kagey, Apr 08 2016

nonn

Equivalently, this sequences is a list of indices i such that A271328(i) != i^2 + 1.
If k is in A271328, then k is in this sequence.
All multiples of a(k) are in the sequence for k > 1.

			A269347(3*1) = 3 != 3*(1^2 + 1) = 6 so 1 is in the sequence.
A269347(3*2) = 15 = 3*(2^2 + 1) so 2 is not in the sequence.
A269347(3*3) = 30 = 3*(3^2 + 1) so 3 is not in the sequence.
A269347(3*4) = 51 = 3*(4^2 + 1) so 4 is not in the sequence.
A269347(3*5) = 84 != 3*(5^2 + 1) = 78 so 5 is in the sequence.

Peter Kagey, Table of n, a(n) for n = 1..10000

cp's OEIS Frontend

A271471 Values k > 1 in A271328 such that k does not have any nontrivial divisors in A271328.

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A269347 With a(1) = 1, a(n) is the sum of all 0 < m < n for which a(m) divides n.

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A271326 a(n) = A269347(3n).

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A271468 List of indices i such that A269347(3i) != 3(i^2 + 1).

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cp's OEIS Frontend

A271471 Values k > 1 in A271328 such that k does not have any nontrivial divisors in A271328.

Views

Author

Keywords

Comments

Examples

Links

Programs

Mathematica

A269347 With a(1) = 1, a(n) is the sum of all 0 < m < n for which a(m) divides n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Java

Mathematica

PARI

Python

Ruby

A271326 a(n) = A269347(3n).

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Author

Keywords

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Crossrefs

A271468 List of indices i such that A269347(3*i) != 3*(i^2 + 1).

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Author

Keywords

Comments

Examples

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A271468 List of indices i such that A269347(3i) != 3(i^2 + 1).