A066743 -id:A066743 - cp's OEIS frontend

A069930 Number of integers of the form (n+k)/(n-k) with 1 <= k <= n-1.

0, 1, 2, 2, 2, 4, 2, 3, 4, 4, 2, 6, 2, 4, 6, 4, 2, 7, 2, 6, 6, 4, 2, 8, 4, 4, 6, 6, 2, 10, 2, 5, 6, 4, 6, 10, 2, 4, 6, 8, 2, 10, 2, 6, 10, 4, 2, 10, 4, 7, 6, 6, 2, 10, 6, 8, 6, 4, 2, 14, 2, 4, 10, 6, 6, 10, 2, 6, 6, 10, 2, 13, 2, 4, 10, 6, 6, 10, 2, 10, 8, 4, 2, 14, 6, 4, 6, 8, 2, 16, 6, 6, 6, 4, 6

Offset: 1

Benoit Cloitre, May 05 2002

Number of r X s integer-sided rectangles such that r < s, r + s = 2n and r | s. - Wesley Ivan Hurt, Apr 24 2020

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A032741, A069283, A000005 (tau), A001227, A006218, A060831, A066743.

A066660(n) - 1.

Table[Count[Table[(n+k)/(n-k),{k,n-1}],?IntegerQ],{n,100}] (* _Harvey P. Dale, Jun 04 2019 *)

for(n=1,150,print1(sum(i=1,n-1,if((n+i)%(n-i),0,1)),","))

{a(n) = if( n<1, 0, numdiv(2*n) - 2)} /* Michael Somos, Aug 30 2012 */

a(n) = A032741(n) + A069283(n) = A000005(n) - 1 + A001227(n) - 1 = tau(n) + A001227(n) - 2. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jul 13 2002
Asymptotic formula: since sum(k=1, n, a(k)) = sum(k=1, n, tau(k)) + sum(k=1, n, A001227(k)) - 2*n = A006218(n) + A060831(n) - 2*n = 2*A006218(n) - A006218(floor(n/2)) - 2*n with A006218(0) = 0, A006218(n) = sum(k=1, n, tau(k)) and now, by Dirichlet's asymptotic expression A006218(n) = n*log(n) + n*(2*gamma-1) + O(n^theta) (gamma = 0.57721..; 1/4 <= theta < 1/2), we have sum(k=1, n, a(k)) = 2*n*log(n) - (n/2)*log(n) + o(n*log(n)) = 1.5*n*log(n) + o(n*log(n)) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jul 13 2002
a(n) = tau(2*n) - 2. - Michael Somos, Aug 30 2012
Sum_{k=1..n} a(k) ~ n/2 * (3*log(n) + log(2) + 6*gamma - 7), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 13 2019

A066755 Numbers m such that m^2 + 1 is not divisible by k^2 + 1 for any k in [1,m-1].

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 16, 20, 24, 26, 34, 36, 40, 44, 46, 50, 54, 56, 60, 66, 70, 74, 76, 84, 86, 90, 94, 96, 100, 104, 110, 114, 116, 120, 124, 126, 130, 134, 136, 144, 146, 150, 156, 160, 164, 170, 176, 180, 184, 186, 190, 194, 196, 204, 206, 210, 214, 220, 224

Offset: 1

Benoit Cloitre, Jan 16 2002

nonn

Equivalently, A066743(m)=1.
If m^2 + 1 is prime, m is in the sequence; i.e., the sequence contains A005574. But so are many other values of m: 34, 44, 46, 50, 60, 70, 76, 86, 96, ...
The numbers of terms that do not exceed 10^k, for k = 1, 2, ..., are , 5, 29, 247, 2354, 23329, 232646, 2324131, ... . Apparently, the asymptotic density of this sequence exists and equals 0.232... . - Amiram Eldar, May 17 2025

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)

Cf. A066743, A005574.

a:= proc(n) option remember; local k; for k from 1+
      `if`(n=1, 0, a(n-1)) while ormap(t->
      irem(k^2+1, t)=0, [(j^2+1)$j=1..k-1]) do od; k
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Sep 18 2019

a66743[ n_ ] := Length[ Select[ Range[ 1, n ], IntegerQ[ (n^2+1)/(#^2+1) ]& ] ]; Select[ Range[ 1, 300 ], a66743[ # ]==1& ]

{ n=0; for (m=1, 10^10, k=1; b=1; t=m^2 + 1; while (k < m - 1, if (t%(k^2 + 1)==0, b=0; break); k++); if (b, write("b066755.txt", n++, " ", m); if (n==1000, return)) ) } \\ Harry J. Smith, Mar 23 2010

Edited by Dean Hickerson, Jan 20 2002

A069929 Number of k, 1 <= k <= n, such that k^3+1 divides n^3+1.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 3, 2, 2, 1, 3, 1, 2, 3, 3, 1, 3, 1, 3, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 2, 2, 2, 1, 5, 1, 3, 2, 2, 1, 3, 1, 3, 2, 2, 1, 5, 1, 3, 2, 2, 2, 3, 1, 2, 3, 4, 1, 3, 1, 2, 2, 4, 1, 3, 1, 2, 2, 2, 1, 5, 1, 2, 2, 3, 1, 4, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 4, 2, 2

Offset: 1

Benoit Cloitre, May 05 2002

Record values are a(1) = 1, a(3) = 2, a(5) = 3, a(17) = 4, a(47) = 5, a(251) = 6, a(467) = 7, a(719) = 9, a(9299) = 10, a(30203) = 12, a(166319) = 14, a(364979) = 15, a(3080159) = 16. - Charles R Greathouse IV, Nov 30 2024

			a(5) = 3 because among the numbers 1^3+1 = 2, 2^3+1 = 9, 3^3+1 = 28, 4^3+1 = 65, and 5^3 + 1 = 126, only 3 of them (2, 9, 126) divide 5^3+1 = 126. - _Petros Hadjicostas_, Sep 18 2019

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A066743.

a:= n-> add(`if`(irem(n^3+1, k^3+1)=0, 1, 0), k=1..n):
seq(a(n), n=1..120);  # Alois P. Heinz, Sep 18 2019

for(n=1,150,print1(sum(i=1,n,if((n^3+1)%(i^3+1),0,1)),","))

a(n)=sumdiv(n^3+1,d, ispower(d-1,3))-1 \\ Charles R Greathouse IV, Nov 30 2024

Conjecture: (1/n)*Sum_{k=1..n} a(k) = C*log(log(n)) + o(log(log(n))) with 1 < C < 3/2.

a(n) < d(n^3+1). - Charles R Greathouse IV, Nov 29 2024

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A069930 Number of integers of the form (n+k)/(n-k) with 1 <= k <= n-1.

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A066755 Numbers m such that m^2 + 1 is not divisible by k^2 + 1 for any k in [1,m-1].

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A069929 Number of k, 1 <= k <= n, such that k^3+1 divides n^3+1.

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