A018892 -id:A018892 - cp's OEIS frontend

A070304 a(n) = number of times n^2*k^2/(n^2+k^2) is an integer as k ranges over 0, 1, 2, ...

1, 2, 1, 2, 2, 2, 1, 2, 1, 6, 1, 2, 1, 2, 3, 2, 2, 2, 1, 7, 1, 2, 1, 2, 2, 4, 1, 2, 1, 8, 1, 2, 1, 3, 2, 2, 2, 2, 2, 7, 1, 2, 1, 2, 3, 2, 1, 2, 1, 7, 2, 4, 1, 2, 2, 2, 1, 4, 1, 9, 1, 2, 1, 2, 5, 2, 1, 4, 1, 7, 1, 2, 1, 3, 3, 2, 1, 5, 1, 7, 1, 3, 1, 2, 4, 2, 1, 2, 1, 8, 1, 2, 1, 2, 2, 2, 1, 2, 1, 8, 2, 4, 1, 4, 3

Offset: 1

Benoit Cloitre, May 10 2002

A018892(n) gives the number of integers of the form nk/(n+k).

Cf. A066451.

for(n=1,150,print1(sum(i=0,n^2,if((n^2*i^2)%(n^2+i^2),0,1)),","))

A182082 Number of pairs, (x,y), with x >= y, whose LCM does not exceed n.

Original entry on oeis.org

1, 3, 5, 8, 10, 15, 17, 21, 24, 29, 31, 39, 41, 46, 51, 56, 58, 66, 68, 76, 81, 86, 88, 99, 102, 107, 111, 119, 121, 135, 137, 143, 148, 153, 158, 171, 173, 178, 183, 194, 196, 210, 212, 220, 228, 233, 235, 249, 252, 260, 265, 273, 275, 286, 291, 302, 307, 312

Offset: 1

Walt Rorie-Baety, Apr 10 2012

nonn

Note that this is the asymmetric count. If all pairs (x,y) are counted, A061503 is obtained. - T. D. Noe, Apr 10 2012

			a(1000000) = 37429395, according to Project Euler problem #379.

Walt Rorie-Baety, Table of n, a(n) for n = 1..1000
Project Euler, Problem 379: Least common multiple count.

Partial sums of A018892.

Cf. A000005, A007875, A013661, A061503 (symmetric case).

a n = length [(x,y)| x <- [1..n], y <- [x..n], lcm x y <= n]

Table[Count[Flatten[Table[LCM[i, j], {i, n}, {j, i, n}]], ?(# <= n &)], {n, 60}] (* _T. D. Noe, Apr 10 2012 *)
nn = 100; (Accumulate[Table[DivisorSigma[0, n^2], {n, nn}]] + Range[nn])/2 (* T. D. Noe, Apr 10 2012 *)

a(n)=(sum(k=1,n,numdiv(k^2))+n)/2 \\ Charles R Greathouse IV, Apr 10 2012

a(n) = Sum_{k=1..n} (d(k^2)+1)/2, where d is the number of divisors function (A000005). - Charles R Greathouse IV, Apr 10 2012
a(n) = Sum_{k=1..n} A007875(k) * floor(n/k). - Daniel Suteu, Jan 08 2021
a(n) ~ n * log(n)^2 /(4*zeta(2)) (see A018892 for a more accurate asymptotic formula). - Amiram Eldar, Feb 01 2025

A309149 Number of solutions of the Diophantine equation 1/n + 1/x = 1/y + 1/z, where n >= 1, x >= n, y > n and z >= y.

Original entry on oeis.org

0, 3, 9, 14, 22, 33, 29, 44, 56, 65, 50, 108, 53, 101, 141, 97, 64, 150, 73, 216, 184, 119, 81, 274, 138, 141, 182, 263, 101, 378, 90, 222, 270, 183, 375, 419, 102, 185, 319, 479, 121, 486, 115, 394, 520, 187, 118, 565, 227, 362

Offset: 1

S. Nazardonyavi, Jul 14 2019

nonn

			n=2: 1/2 + 1/6 = 1/3 + 1/3, 1/2 + 1/12 = 1/3 + 1/4, 1/2 + 1/30 = 1/3 + 1/5.

Cf. A309150, A063647, A018892.

a[n_]:=Length@Solve[1/(n)+1/(x)==1/y+1/z&&x>=n&&z>=y&&y>n,{x,y,z},Integers];
Array[a,50]

A309150 Number of solutions of the Diophantine equation 1/n + 1/x = 1/y + 1/z, where n >= 1, x > n, y > n and z > y.

Original entry on oeis.org

0, 2, 7, 12, 20, 29, 27, 41, 52, 60, 48, 101, 51, 96, 134, 93, 62, 142, 71, 209, 176, 114, 79, 264, 134, 136, 176, 256, 99, 363, 88, 217, 262, 178, 368, 406, 100, 180, 311, 469, 119, 471, 113, 386, 508, 182, 116, 552, 223, 353

Offset: 1

S. Nazardonyavi, Jul 14 2019

nonn

			n=2:  1/2 + 1/12 = 1/3 + 1/4, 1/2 + 1/30 = 1/3 + 1/5.

Cf. A309149, A063647, A018892.

a[n_]:=Length@Solve[1/(n)+1/(x)==1/y+1/z&&x>n&&z>y&&y>n,{x,y,z},Integers];
Array[a,50]

A337298 Sum of the coordinates of all relatively prime pairs of divisors of n, (d1,d2), such that d1 <= d2.

Original entry on oeis.org

2, 5, 6, 10, 8, 21, 10, 19, 16, 29, 14, 46, 16, 37, 36, 36, 20, 61, 22, 64, 46, 53, 26, 91, 34, 61, 44, 82, 32, 141, 34, 69, 66, 77, 64, 136, 40, 85, 76, 127, 44, 181, 46, 118, 106, 101, 50, 176, 60, 133, 96, 136, 56, 173, 92, 163, 106, 125, 62, 316, 64, 133, 136, 134, 106, 261, 70

Offset: 1

Wesley Ivan Hurt, Aug 21 2020

			a(4) = 10; There are 3 divisors of 4: {1,2,4}. If we list the relatively prime pairs (d1,d2), where d1 <= d2, we get (1,1), (1,2), (1,4). The sum of the coordinates from all pairs is 1+1+1+2+1+4 = 10.
a(5) = 8; There are 2 divisors of 5: {1,5}. The relatively prime pairs (d1,d2), where d1 <= d2 are: (1,1) and (1,5). The sum of the coordinates is then 1+1+1+5 = 8.

Cf. A018892, A337246.

Table[Sum[Sum[(i + k) 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) = my(d = divisors(n)); sum(i=1, #d, sum(j=1, i, if (gcd(d[i],d[j])==1, d[i]+d[j])));  \\ Michel Marcus, Aug 22 2020

a(n) = Sum_{i|n, k|n, i<=k, gcd(i,k)=1} (i+k).

A352881 a(n) is the minimal number z having the largest number of solutions to the Diophantine equation 1/z = 1/x + 1/y such that 1 <= x <= y <= 10^n.

Original entry on oeis.org

2, 12, 60, 840, 9240, 55440, 720720, 6126120, 116396280, 232792560, 5354228880, 26771144400, 465817912560, 4813451763120, 24067258815600, 144403552893600, 2671465728531600, 36510031623265200, 219060189739591200, 4709794079401210800, 18839176317604843200, 221360321731856907600

Offset: 1

Darío Clavijo, Apr 06 2022

nonn

Solving for z gives z = (x*y) / (x+y), so x*y == 0 (mod x+y).
All known terms are from A025487:
  a(1) =    2 = 2;
  a(2) =   12 = 2^2 * 3;
  a(3) =   60 = 2^2 * 3 * 5;
  a(4) =  840 = 2^3 * 3 * 5 * 7;
  a(5) = 9240 = 2^3 * 3 * 5 * 7 * 11.
If a solution to the equation 1/z = 1/x + 1/y is found such that gcd(x,y,z) is a square, then x+y, x*y*z, and (x-y)^2 + (2*z)^2 are also squares.
For all solutions, x^2 + y^2 + z^2 is a square.
The sequence is indeed a subsequence of A025487, and likely of A126098 as well. - Max Alekseyev, Mar 01 2023
a(n) < 5*10^(n-1). - Max Alekseyev, Mar 01 2023

			For n=1, we have the following, where r = (x*y) mod (x+y). (In the last four columns, each number marked by an asterisk is a square.)
.
  r  z  x  y  x*y  x+y  x*y*z  x^2+y^2+z^2
  -  -  -  -  ---  ---  -----  -----------
  0  1  2  2    4*   4*     4*           9* (solution)
  2  1  2  4    8    6      8           21
  4  1  2  6   12    8     12           41
  6  1  2  8   16*  10     16*          69
  3  1  3  3    9    6      9*          19
  0  2  3  6   18*   9*    36*          49* (solution)
  3  2  3  9   27   12     54           94
  0  2  4  4   16*   8     32           36* (solution)
  8  2  4  8   32   12     64*          84
  5  2  5  5   25*  10     50           54
  0  3  6  6   36*  12    108           81* (solution)
  7  3  7  7   49*  14    147          107
  0  4  8  8   64*  16*   256*         144* (solution)
  9  4  9  9   81*  18    324*         178
.
z = 2 has the largest number of solutions, so a(1) = 2.
The number of solutions for the resulting z cannot exceed A018892(z).

Max Alekseyev, Table of n, a(n) for n = 1..30
Project Euler, Diophantine reciprocals I, Problem 108.

Cf. A025487, A099829, A126098, A067128, A018892, A126098, A225206.

a(n)=my(bc=0,bk=0,L=10^n);for(k=1,L-1,my(c=0,k2=k^2);for(d=max(1,k2\(L-k)+1),k,if(k2%d==0,c++););if(c>bc,bc=c;bk=k););return(bk); \\ Darío Clavijo, Mar 03 2025

def a(n):
    # k=x*y and d=x+y
    bc, bk, L = 0, None, 10**n
    for k in range(1, L):
        c, k2 = 0, k * k
        for d in range(max(1, k2 // (L - k) + 1), k + 1):
            if k2 % d == 0: c += 1
        if c > bc:
            bc, bk = c, k
    return bk

a(6) from Chai Wah Wu, Apr 10 2022

a(7)-a(22) from Max Alekseyev, Mar 01 2023

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

cp's OEIS Frontend

A070304 a(n) = number of times n^2*k^2/(n^2+k^2) is an integer as k ranges over 0, 1, 2, ...

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A182082 Number of pairs, (x,y), with x >= y, whose LCM does not exceed n.

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A309149 Number of solutions of the Diophantine equation 1/n + 1/x = 1/y + 1/z, where n >= 1, x >= n, y > n and z >= y.

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A309150 Number of solutions of the Diophantine equation 1/n + 1/x = 1/y + 1/z, where n >= 1, x > n, y > n and z > y.

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A337298 Sum of the coordinates of all relatively prime pairs of divisors of n, (d1,d2), such that d1 <= d2.

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A352881 a(n) is the minimal number z having the largest number of solutions to the Diophantine equation 1/z = 1/x + 1/y such that 1 <= x <= y <= 10^n.

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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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