1,24

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 element 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(n) = 16 for n = 24, 40, 56, 135, 189, 297, 351, 459, ... (numbers of the form p^3*q, p and q primes with q > p).

a(n) = 17 for n = 54, 88, 104, 136, 152, 184, 232, 248, ... (numbers of the form p^i*q^j, p and q primes and (i, j) = (3, 1) or (1, 3).

a(n) = 30 for n = 36, 225, 441, 1225, 3025, 4225, 5929, ... (numbers of the form p^2*q^2, p and q primes.

It is possible to continue with a(n) = 74, 75, 78, 107, 110, 112, 114, ...

Table of n, a(n) for n=1..103.

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 4 following quadruples of divisors: (2,6,10,30), (3,6,15,30) and (5,10,15,30).

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 *)

(PARI) 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

(PARI) 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

Cf. A275387, A336530.

nonn,new

Michel Lagneau, Oct 05 2020

Terms corrected by David A. Corneth, Oct 31 2020

approved

1,1

N. J. A. Sloane, Table of n, a(n) for n = 1..7096

Cf. A280985.

nonn,new

N. J. A. Sloane, Nov 01 2020

approved

1,1

Does any number occur more than twice in A338529?

Table of n, a(n) for n=1..34.

a(3) = 301310 is in the sequence because A338529(1618) = A338529(2414) = 301310.

A338529:= [seq(ithprime(n+2)*ithprime(n+3)-ithprime(n)*ithprime(n+1), n=1..100000)]):

M:= 12*ithprime(100000):

V:= Vector(M, datatype=integer[4]):

R:= NULL:

for i from 1 to 100000 do

  v:= A338529[i];

  if v <= M then V[v]:= V[v]+1; if V[v] > 1 then R:= R, v fi fi

od:

sort([R]);

Cf. A338529.

nonn,new

J. M. Bergot and Robert Israel, Nov 01 2020

approved

1,1

The primes are listed in the order in which they appear in A338529.  Some appear more than once, e.g. a(481) = a(669) = 2427137 because 2427137 = A338529(11966)/2 = A338529(16893)/2.

Robert Israel, Table of n, a(n) for n = 1..10000

A338529(2)/2 = 31 is prime, so a(1) = 31.

A338529(38)/2 = 1873 is prime, so a(2) = 1873.

A338529(65)/2 = 6163 is prime, so a(3) = 6163.

select(isprime, [seq(ithprime(n+2)*ithprime(n+3)-ithprime(n)*ithprime(n+1), n=2..10000)]/2);

Cf. A338529.

nonn,new

J. M. Bergot and Robert Israel, Nov 01 2020

approved

1,1

a(n) > (A001223(n)+2*A001223(n+1)+A001223(n+2))*prime(n).

In particular, a(n) > 12*prime(n).

Robert Israel, Table of n, a(n) for n = 1..10000

Robert Israel, Plot of a(n)/n for 1 <= n <= 30000

The third through sixth primes are 5,7,11,13, so a(3) = 11*13-5*7 = 108.

seq(ithprime(n+2)*ithprime(n+3)-ithprime(n)*ithprime(n+1), n=1..1000);

Cf. A338533, A338537.

nonn,look,new

J. M. Bergot and Robert Israel, Nov 01 2020

approved

1,1

Numbers k such that A056170(k) = A001221(A057521(k)) = 5.

Numbers divisible by the squares of exactly five distinct primes.

The asymptotic density of this sequence is (eta_1^5 - 10*eta_1^3*eta_2 + 15*eta_1*eta_2^2 + 20*eta_1^2*eta_3 - 20*eta_2*eta_3 - 30*eta_1*eta_4 + 24*eta_5)/(20*Pi^2) = 0.0000015673..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

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

Carl Pomerance and Andrzej Schinzel, Multiplicative Properties of Sets of Residues, Moscow Journal of Combinatorics and Number Theory, Vol. 1, No. 1 (2011), pp. 52-66. See pp. 61-62.

5336100 = 2^2 * 3^2 * 5^2 * 7^2 * 11^2 is a term since it has exactly 5 prime factors, 2, 3, 5, 7 and 11, that are non-unitary.

Select[Range[2*10^7], Count[FactorInteger[#][[;; , 2]], _?(#1 > 1 &)] == 5 &]

Subsequence of A013929, A318720 and A327877.

Cf. A001221, A056170, A057521, A190641, A338539, A338540, A338541.

Cf. A154945 (eta_1), A324833 (eta_2), A324834 (eta_3), A324835 (eta_4), A324836 (eta_5).

nonn,new

Amiram Eldar, Nov 01 2020

approved

1,1

Numbers k such that A056170(k) = A001221(A057521(k)) = 4.

Numbers divisible by the squares of exactly four distinct primes.

The asymptotic density of this sequence is (eta_1^4 - 6*eta_1^2*eta_2 + 3*eta_2^2 + 8*eta_1*eta_3 - 6*eta_4)/(4*Pi^2) = 0.0000970457..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

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

Carl Pomerance and Andrzej Schinzel, Multiplicative Properties of Sets of Residues, Moscow Journal of Combinatorics and Number Theory, Vol. 1, No. 1 (2011), pp. 52-66. See pp. 61-62.

44100 = 2^2 * 3^2 * 5^2 * 7^2 is a term since it has exactly 4 prime factors, 2, 3, 5 and 7, that are non-unitary.

Select[Range[620000], Count[FactorInteger[#][[;; , 2]], _?(#1 > 1 &)] == 4 &]

Subsequence of A013929 and A318720.

Cf. A001221, A056170, A057521, A190641, A338539, A338540, A338542.

Cf. A154945 (eta_1), A324833 (eta_2), A324834 (eta_3), A324835 (eta_4).

nonn,new

Amiram Eldar, Nov 01 2020

approved

1,1

Numbers k such that A056170(k) = A001221(A057521(k)) = 3.

Numbers divisible by the squares of exactly three distinct primes.

Subsequence of A318720 and first differs from it at n = 123.

The asymptotic density of this sequence is (eta_1^3 - 3*eta_1*eta_2 + 2*eta_3)/Pi^2 = 0.0032920755..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

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

Carl Pomerance and Andrzej Schinzel, Multiplicative Properties of Sets of Residues, Moscow Journal of Combinatorics and Number Theory, Vol. 1, No. 1 (2011), pp. 52-66. See pp. 61-62.

900 = 2^2 * 3^2 * 5^2 is a term since it has exactly 3 prime factors, 2, 3 and 5, that are non-unitary.

Select[Range[17000], Count[FactorInteger[#][[;; , 2]], _?(#1 > 1 &)] == 3 &]

Subsequence of A013929, A318720 and A327877.

Cf. A001221, A056170, A057521, A190641, A338539, A338541, A338542.

Cf. A154945 (eta_1), A324833 (eta_2), A324834 (eta_3).

nonn,new

Amiram Eldar, Nov 01 2020

approved

1,1

Numbers k such that A056170(k) = A001221(A057521(k)) = 2.

Numbers divisible by the squares of exactly two distinct primes.

Subsequence of A036785 and first differs from it at n = 44.

The asymptotic density of this sequence is (3/Pi^2)*(eta_1^2 - eta_2) = 0.0532928864..., where eta_j = Sum_{p prime} 1/(p^2-1)^j (Pomerance and Schinzel, 2011).

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

Carl Pomerance and Andrzej Schinzel, Multiplicative Properties of Sets of Residues, Moscow Journal of Combinatorics and Number Theory, Vol. 1, No. 1 (2011), pp. 52-66. See pp. 61-62.

36 = 2^2 * 3^2 is a term since it has exactly 2 prime factors, 2 and 3, that are non-unitary.

Select[Range[1000], Count[FactorInteger[#][[;; , 2]], _?(#1 > 1 &)] == 2 &]

Subsequence of A013929 and A036785.

Cf. A001221, A056170, A057521, A190641, A338540, A338541, A338542.

Cf. A154945 (eta_1), A324833 (eta_2).

nonn,new

Amiram Eldar, Nov 01 2020

approved

0,3

T(n, k) is divisible by A099398(n) for all 0 <= k <= n.

Table of n, a(n) for n=0..44.

Let t(n) denote the triangular numbers and C(n) the Catalan numbers.

T(n, k) = k*(2*n - 1)*(t(2*k + 1)/t(n + 1))*C(n - 1) for n, k > 0.

T(n, k) = k^n if k = 0; if k = n then C(n+1)*t(n+1); else T(n-1, k)*(4-10/(n+2)).

Triangle starts:

[0] 1

[1] 0, 2

[2] 0, 3,    15

[3] 0, 6,    30,    84

[4] 0, 14,   70,    196,   420

[5] 0, 36,   180,   504,   1080,  1980

[6] 0, 99,   495,   1386,  2970,  5445,   9009

[7] 0, 286,  1430,  4004,  8580,  15730,  26026,  40040

[8] 0, 858,  4290,  12012, 25740, 47190,  78078,  120120, 175032

[9] 0, 2652, 13260, 37128, 79560, 145860, 241332, 371280, 541008, 755820

T := proc(n, k) if n = 0 then 1 else

(k*(2*k+2)*(2*k+1)*(2*n-1)*binomial(2*(n-1), n-1))/(n*(n+1)*(n+2)) fi end:

# Recursive:

CatalanNumber := n -> binomial(2*n, n)/(n+1):

T := proc(n, k) option remember; if k=0 then k^n elif k=n then CatalanNumber(n+1)* binomial(n+1, 2) else (4 - 10/(n + 2))*T(n-1, k) fi end:

seq(seq(T(n, k), k=0..n), n=0..9);

T[n_, k_] := If[n == 0, 1, (k (2k + 2)(2k + 1)(2n - 1) CatalanNumber[n-1])/((n + 1) (n + 2))]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten

Cf. A119578 (row sums), (-1)^n*A005430 (alternating row sums), A002740 (main diagonal), A007054 (col 1), A099398 (universal divisor), A000108 (Catalan).

nonn,tabl,new

Peter Luschny, Nov 01 2020

approved