A275387 -id:A275387 - cp's OEIS frontend

A336530 Number of triples of divisors d_i < d_j < d_k of n such that gcd(d_i, d_j, d_k) > 1.

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 0, 4, 0, 5, 0, 5, 0, 0, 0, 23, 0, 0, 1, 5, 0, 12, 0, 10, 0, 0, 0, 36, 0, 0, 0, 23, 0, 12, 0, 5, 5, 0, 0, 62, 0, 5, 0, 5, 0, 23, 0, 23, 0, 0, 0, 87, 0, 0, 5, 20, 0, 12, 0, 5, 0, 12, 0, 120, 0, 0, 5, 5, 0, 12, 0, 62, 4

Offset: 1

Michel Lagneau, Oct 04 2020

nonn

Number of elements in the set {(x, y, z): x|n, y|n, z|n, x < y < z, GCD(x, y, z) > 1}.
Every element of the sequence is repeated indefinitely, for instance:
a(n) = 0 for n = 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, ... (Numbers with at most 2 prime factors (counted with multiplicity). See A037143);
a(n) = 1 for n = 8, 27, 125, 343, 1331, 2197, 4913,... (cubes of primes. See A030078);
a(n) = 4 for n = 16, 81, 625, 2401, 14641, 28561, ... (prime(n)^4. See A030514);
a(n) = 5 for n = 12, 18, 20, 28, 44, 45, ... (Numbers which are the product of a prime and the square of a different prime (p^2 * q). See A054753);
a(n) = 12 for n = 30, 42, 66, 70, 78, 102, 105, 110,... (Sphenic numbers: products of 3 distinct primes. See A007304);
a(n) = 20 for n = 64, 729, 15625, 117649, ... (Numbers with 7 divisors. 6th powers of primes. See A030516);
a(n) = 23 for n = 24, 40, 54, 56, 88, 104, 135, 136, ... (Product of the cube of a prime (A030078) and a different prime. See A065036);
a(n) = 36 for n = 36, 100, 196, 225, 441, 484, 676,... (Squares of the squarefree semiprimes (p^2*q^2). See A085986);
a(n) = 62 for n = 48, 80, 112, 162, 176, 208, 272, ... (Product of the 4th power of a prime (A030514) and a different prime (p^4*q). See A178739);
a(n) = 87 for n = 60, 84, 90, 126, 132, 140, 150, 156, ... (Product of exactly four primes, three of which are distinct (p^2*q*r). See A085987);
a(n) = 120 for n = 72, 108, 200, 392, 500, 675, 968, ... (Numbers of the form p^2*q^3, where p,q are distinct primes. See A143610);
It is possible to continue with a(n) = 130, 235, 284, 289, 356, ...

			a(12) = 5 because the divisors of 12 are {1, 2, 3, 4, 6, 12} and GCD(d_i, d_j, d_k) > 1 for the 5 following triples of divisors: (2,4,6), (2,4,12), (2,6,12), (3,6,12) and (4,6,12).

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A275387.

with(numtheory):nn:=100:
for n from 1 to nn do:
it:=0:d:=divisors(n):n0:=nops(d):
  for i from 1 to n0-2 do:
   for j from i+1 to n0-1 do:
     for k from j+1 to n0 do:
    if igcd(d[i],d[j],d[k])> 1
       then
       it:=it+1:
       else
      fi:
     od:
     od:
     od:
    printf(`%d, `,it):
   od:

Array[Count[GCD @@ # & /@ Subsets[Divisors[#], {3}], ?(# > 1 &)] &, 81] (* _Michael De Vlieger, Oct 05 2020 *)

a(n) = my(d=divisors(n)); sum(i=1, #d-2, sum (j=i+1, #d-1, sum (k=j+1, #d, gcd([d[i], d[j], d[k]]) > 1))); \\ Michel Marcus, Oct 31 2020

a(n) = {my(f = factor(n), vp = vecprod(f[,1]), d = divisors(vp), res = 0);
for(i = 2, #d, res-=binomial(numdiv(n/d[i]), 3)*(-1)^omega(d[i])); res} \\ David A. Corneth, Nov 01 2020

Name clarified by editors, Oct 31 2020

A333976 Number of pairs of divisors of n, (d1,d2), such that d1 <= d2 and gcd(d1,d2) > 1.

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 1, 6, 3, 5, 1, 13, 1, 5, 5, 10, 1, 13, 1, 13, 5, 5, 1, 25, 3, 5, 6, 13, 1, 22, 1, 15, 5, 5, 5, 32, 1, 5, 5, 25, 1, 22, 1, 13, 13, 5, 1, 41, 3, 13, 5, 13, 1, 25, 5, 25, 5, 5, 1, 55, 1, 5, 13, 21, 5, 22, 1, 13, 5, 22, 1, 60, 1, 5, 13, 13, 5, 22, 1, 41, 10

Offset: 1

Wesley Ivan Hurt, Jan 01 2021

nonn

			a(7) = 1; (7,7)
a(8) = 6; (2,2), (2,4), (2,8), (4,4), (4,8), (8,8)
a(9) = 3; (3,3), (3,9), (9,9)
a(10) = 5; (2,2), (2,10), (5,5), (5,10), (10,10)

Cf. A275387 (same with d1A337298.

Table[Sum[Sum[(1 - KroneckerDelta[GCD[i, k], 1]) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}]

a(n) = Sum_{d1|n, d2|n, d1<=d2} (1-[gcd(d1,d2) = 1]), where [ ] is the Iverson bracket.

A336432 Number of ordered quadruples of divisors (d_i, d_j, d_k, d_m) of n such that GCD(d_i, d_j, d_k, d_m) > 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 16, 0, 0, 0, 1, 0, 3, 0, 5, 0, 0, 0, 29, 0, 0, 0, 16, 0, 3, 0, 1, 1, 0, 0, 74, 0, 1, 0, 1, 0, 16, 0, 16, 0, 0, 0, 98, 0, 0, 1, 15, 0, 3, 0, 1, 0, 3, 0, 181, 0, 0, 1, 1, 0, 3, 0, 74, 1, 0, 0, 98, 0, 0, 0, 16, 0, 98, 0, 1, 0, 0, 0, 220, 0, 1, 1, 29, 0, 3, 0

Offset: 1

Michel Lagneau, Oct 05 2020

nonn

Number of elements in the set {(x, y, z, w): x|n, y|n, z|n, w|n, x < y < z < w, GCD(x, y, z, w) > 1}.
Every term of the sequence is repeated indefinitely; for instance:
a(n) = 0 for n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, ... (numbers k such that product of proper divisors of k is <= k; i.e., product of divisors of k is <= k^2; see A007964).
a(n) = 1 for n = 12, 16, 18, 20, 28, 44, 45, 50, 52, 63, 68, 75, 76, 81, 92, 98, 99, ... (either 4th power of a prime, or product of a prime and the square of a different prime; see A080258).
a(n) = 5 for n = 32, 243, 3125, 16807, ... (fifth powers of primes; see A050997).
a(n) = 15 for n = 64, 729, 15625, 117649, ... (numbers with 7 divisors: 6th powers of primes; see A030516).

			a(30) = 3 because the divisors of 30 are {1, 2, 3, 5, 6, 10, 15, 30} and GCD(d_i, d_j, d_k, d_m) > 1 for the following 3 quadruples of divisors: (2,6,10,30), (3,6,15,30) and (5,10,15,30).

Antti Karttunen, Table of n, a(n) for n = 1..10000
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A275387, A336530.

with(numtheory):nn:=100:
for n from 1 to nn do:
it:=0:d:=divisors(n):n0:=nops(d):
  for i from 1 to n0-3 do:
   for j from i+1 to n0-2 do:
     for k from j+1 to n0-1 do:
       for l from k+1 to n0 do:
    if igcd(d[i],d[j],d[k],d[l])> 1
       then
       it:=it+1:
       else
      fi:
     od:
    od:
   od:
  od:
    printf(`%d, `,it):
od:

Array[Count[GCD @@ # & /@ Subsets[Divisors[#], {4}], ?(# > 1 &)] &, 100] (* _Amiram Eldar, Oct 31 2020 after Michael De Vlieger at A336530 *)

a(n) = my(d=divisors(n)); sum(i=1, #d-3, sum (j=i+1, #d-2, sum (k=j+1, #d-1, sum (m=k+1, #d, gcd([d[i], d[j], d[k], d[m]]) > 1)))); \\ Michel Marcus, Oct 31 2020

a(n) = {my(f = factor(n), vp = vecprod(f[,1]), d = divisors(vp), res = 0); for(i = 2, #d, res-=binomial(numdiv(n/d[i]), 4)*(-1)^omega(d[i])); res} \\ David A. Corneth, Oct 31 2020

Terms corrected by David A. Corneth, Oct 31 2020

A284167 a(n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the i-th divisor of n.

Original entry on oeis.org

2, 5, 7, 10, 8, 15, 8, 18, 16, 18, 10, 29, 8, 19, 25, 28, 10, 33, 10, 35, 26, 20, 12, 50, 18, 20, 31, 36, 12, 51, 10, 42, 27, 23, 33, 62, 8, 22, 30, 60, 12, 53, 10, 40, 52, 22, 14, 78, 20, 41, 28, 38, 12, 63, 36, 63, 30, 24, 16, 95, 8, 23, 59, 60, 32, 54, 10

Offset: 1

Ctibor O. Zizka, Mar 21 2017

nonn

Let S(n,n) be the number of solutions of the equation n/x + n/y = c where n, c, x, and y are positive integers. Then S(n,n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the i-th divisor of n.
For c = 1 , S(n,n) = A000005(n).
Let S(n,m) be the number of solutions of the equation n/x + m/y = c where n, m, c, x, and y are positive integers, n not equal to m. Let k(i) be the i-th divisor of n, and k(j) the j-th divisor of m. Let d(t) be the number of divisors of t. Let R = d(k(i) + k(j)). Then S(n,m) = Sum_{i=1..A000005(n)} Sum_{j=1..A000005(m)} [R*1 if gcd(k(i),k(j)) = 1 , R*0 else].
For c = 1 , S(n,m) = A000005(n) * A000005(m) - P, where P is the number of divisor pairs such that gcd(k(i),k(j)) >= 2.

			For n = 4, divisors of 4 are 1, 2, 4; thus a(4) = d(4+1) + d(4+2) + d(4+4) = d(5) + d(6) + d(8) = 2 + 4 + 4 = 10.

R. L. Graham, Paul Erdos and Egyptian Fractions, Bolyai Society Mathematical Studies 25, pp 289-309, 2013.

Cf. A000005, A018892, A063647, A089233, A275387.

a[n_] := Sum[DivisorSigma[0, d + n], {d, Divisors@n}]; Array[a, 67] (* Giovanni Resta, Mar 21 2017 *)

for(n=1, 101, print1(sumdiv(n, d, numdiv(d + n)),", ")) \\ Indranil Ghosh, Mar 22 2017

from sympy import divisor_count, divisors
def a(n):
    return sum(divisor_count(n + d) for d in divisors(n)) # Indranil Ghosh, Mar 22 2017

a(21)-a(67) from Giovanni Resta, Mar 21 2017

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A336530 Number of triples of divisors d_i < d_j < d_k of n such that gcd(d_i, d_j, d_k) > 1.

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A333976 Number of pairs of divisors of n, (d1,d2), such that d1 <= d2 and gcd(d1,d2) > 1.

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A336432 Number of ordered quadruples of divisors (d_i, d_j, d_k, d_m) of n such that GCD(d_i, d_j, d_k, d_m) > 1.

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A284167 a(n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the i-th divisor of n.

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