A051537 -id:A051537 - cp's OEIS frontend

A070262 5th diagonal of triangle defined in A051537.

5, 3, 21, 2, 45, 15, 77, 6, 117, 35, 165, 12, 221, 63, 285, 20, 357, 99, 437, 30, 525, 143, 621, 42, 725, 195, 837, 56, 957, 255, 1085, 72, 1221, 323, 1365, 90, 1517, 399, 1677, 110, 1845, 483, 2021, 132, 2205, 575, 2397, 156, 2597, 675, 2805, 182, 3021, 783

Offset: 1

Amarnath Murthy, May 09 2002

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A061037. [From R. J. Mathar, Sep 29 2008]
Cf. A010873, A051537.
Cf. A000466, A002378, A003185, A085027.

[LCM(n + 4, n)/GCD(n + 4, n): n in [1..50]]; // G. C. Greubel, Sep 20 2018

Table[ LCM[i + 4, i] / GCD[i + 4, i], {i, 1, 60}]
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{5,3,21,2,45,15,77,6,117,35,165,12},90] (* Harvey P. Dale, Jul 13 2019 *)

Vec(x*(5 + 3*x + 21*x^2 + 2*x^3 + 30*x^4 + 6*x^5 + 14*x^6 - 3*x^8 - x^9 - 3*x^10) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3) + O(x^60)) \\ Colin Barker, Mar 27 2017

a(n) = lcm(n+4,n)/gcd(n+4,n); \\ Altug Alkan, Sep 20 2018

a(n) = lcm(n + 4, n) / gcd(n + 4, n).
From Colin Barker, Mar 27 2017: (Start)
G.f.: x*(5 + 3*x + 21*x^2 + 2*x^3 + 30*x^4 + 6*x^5 + 14*x^6 - 3*x^8 - x^9 - 3*x^10) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n>12. (End)
From Luce ETIENNE, May 10 2018: (Start)
a(n) = n*(n+4)*4^((5*(n mod 4)^3 - 24*(n mod 4)^2 + 31*(n mod 4)-12)/6).
a(n) = n*(n+4)*(37-27*cos(n*Pi)-6*cos(n*Pi/2))/64. (End)
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 11/6.
Sum_{n>=1} (-1)^n/a(n) = 7/6.
Sum_{k=1..n} a(k) ~ (37/192) * n^3. (End)

Edited by Robert G. Wilson v, May 10 2002

A070260 Third diagonal of triangle defined in A051537.

Original entry on oeis.org

3, 2, 15, 6, 35, 12, 63, 20, 99, 30, 143, 42, 195, 56, 255, 72, 323, 90, 399, 110, 483, 132, 575, 156, 675, 182, 783, 210, 899, 240, 1023, 272, 1155, 306, 1295, 342, 1443, 380, 1599, 420, 1763, 462, 1935, 506, 2115, 552, 2303, 600, 2499, 650, 2703, 702

Offset: 1

Amarnath Murthy, May 09 2002

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Bisections: A002378, A000466.

Cf. A051537.

Table[ LCM[i + 2, i] / GCD[i + 2, i], {i, 1, 60}]
LinearRecurrence[{0,3,0,-3,0,1},{3,2,15,6,35,12},60] (* Harvey P. Dale, Sep 14 2019 *)

Vec(x*(3+2*x+6*x^2-x^4) / (1-x^2)^3 + O(x^60)) \\ Colin Barker, Mar 27 2017

From Vladeta Jovovic, May 09 2002: (Start)
a(n) = n*(n+2)/4 if n is even else n*(n+2).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x*(3 + 2*x + 6*x^2 - x^4)/(1 - x^2)^3. (End)
E.g.f.: (x/4)*((12 + x)*cosh(x) + (3 + 4*x)*sinh(x)). - G. C. Greubel, Jul 20 2017
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 3/2.
Sum_{n>=1} (-1)^n/a(n) = 1/2.
Sum_{k=1..n} a(k) ~ (5/24) * n^3. (End)

More terms from Vladeta Jovovic, May 09 2002

A070261 4th diagonal of triangle defined in A051537.

Original entry on oeis.org

4, 10, 2, 28, 40, 6, 70, 88, 12, 130, 154, 20, 208, 238, 30, 304, 340, 42, 418, 460, 56, 550, 598, 72, 700, 754, 90, 868, 928, 110, 1054, 1120, 132, 1258, 1330, 156, 1480, 1558, 182, 1720, 1804, 210, 1978, 2068, 240, 2254, 2350, 272, 2548, 2650, 306, 2860

Offset: 1

Amarnath Murthy, May 09 2002

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Cf. A051537.

Table[ LCM[i + 3, i] / GCD[i + 3, i], {i, 1, 60}]

Vec(2*x*(2 + 5*x + x^2 + 8*x^3 + 5*x^4 - x^6 - x^7) / ((1 - x)^3*(1 + x + x^2)^3) + O(x^60)) \\ Colin Barker, Mar 27 2017

a(n) = lcm(n + 3, n) / gcd(n + 3, n).
From Colin Barker, Mar 27 2017: (Start)
G.f.: 2*x*(2 + 5*x + x^2 + 8*x^3 + 5*x^4 - x^6 - x^7) / ((1 - x)^3*(1 + x + x^2)^3).
a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9) for n>9.
(End)
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 3/2.
Sum_{n>=1} (-1)^n/a(n) = 22*log(2)/9 - 7/6.
Sum_{k=1..n} a(k) ~ (19/81) * n^3. (End)

Edited by Robert G. Wilson v, May 10 2002

A050873 Triangular array T read by rows: T(n,k) = gcd(n,k).

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 2, 1, 4, 1, 1, 1, 1, 5, 1, 2, 3, 2, 1, 6, 1, 1, 1, 1, 1, 1, 7, 1, 2, 1, 4, 1, 2, 1, 8, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 2, 3, 4, 1, 6, 1, 4, 3, 2, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Clark Kimberling

The function T(n,k) = T(k,n) is defined for all integer k,n but only the values for 1 <= k <= n as a triangular array are listed here.
For each divisor d of n, the number of d's in row n is phi(n/d). Furthermore, if {a_1, a_2, ..., a_phi(n/d)} is the set of positive integers <= n/d that are relatively prime to n/d then T(n,a_i * d) = d. - Geoffrey Critzer, Feb 22 2015
Starting with any row n and working downwards, consider the infinite rectangular array with k = 1..n. A repeating pattern occurs every A003418(n) rows. For example, n=3: A003418(3) = 6. The 6-row pattern starting with row 3 is {1,1,3}, {1,2,1}, {1,1,1}, {1,2,3}, {1,1,1}, {1,2,1}, and this pattern repeats every 6 rows, i.e., starting with rows {9,15,21,27,...}. - Bob Selcoe and Jamie Morken, Aug 02 2017

			Rows:
  1;
  1, 2;
  1, 1, 3;
  1, 2, 1, 4;
  1, 1, 1, 1, 5;
  1, 2, 3, 2, 1, 6; ...

T. D. Noe, Rows n=1..100, flattened
Marcelo Polezzi, A Geometrical Method for Finding an Explicit Formula for the Greatest Common Divisor, The American Mathematical Monthly, Vol. 104, No. 5 (May, 1997), pp. 445-446.
Eric Weisstein's World of Mathematics, Greatest Common Divisor
Wikipedia, Greatest Common Divisor

Cf. A003989.
Cf. A002262, A054531, A226314.
Cf. A018804 (row sums), A245717.
Cf. A132442 (sums of divisors).
Cf. A003418.

a050873 = gcd
a050873_row n = a050873_tabl !! (n-1)
a050873_tabl = zipWith (map . gcd ) [1..] a002260_tabl
-- Reinhard Zumkeller, Dec 12 2015, Aug 13 2013, Jun 10 2013

ColumnForm[Table[GCD[n, k], {k, 12}, {n, k}], Center] (* Alonso del Arte, Jan 14 2011 *)

{T(n, k) = gcd(n, k)} /* Michael Somos, Jul 18 2011 */

a(n) = gcd(A002260(n), A002024(n)); A054521(n) = A000007(a(n)). - Reinhard Zumkeller, Dec 02 2009
T(n,k) = A075362(n,k)/A051173(n,k), 1 <= k <= n. - Reinhard Zumkeller, Apr 25 2011
T(n, k) = T(k, n) = T(-n, k) = T(n, -k) = T(n, n+k) = T(n+k, k). - Michael Somos, Jul 18 2011
T(n,k) = A051173(n,k) / A051537(n,k). - Reinhard Zumkeller, Jul 07 2013

A051173 Triangle read by rows: T(n, k) = lcm(n, k).

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 4, 12, 4, 5, 10, 15, 20, 5, 6, 6, 6, 12, 30, 6, 7, 14, 21, 28, 35, 42, 7, 8, 8, 24, 8, 40, 24, 56, 8, 9, 18, 9, 36, 45, 18, 63, 72, 9, 10, 10, 30, 20, 10, 30, 70, 40, 90, 10, 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 11, 12, 12, 12, 12, 60, 12, 84, 24, 36, 60, 132, 12

Offset: 1

Clark Kimberling

			Triangle begins (for the full array see A109042):
  [1]  1;
  [2]  2,  2;
  [3]  3,  6,  3;
  [4]  4,  4, 12,  4;
  [5]  5, 10, 15, 20,  5;
  [6]  6,  6,  6, 12, 30,  6;
  [7]  7, 14, 21, 28, 35, 42,  7;
  [8]  8,  8, 24,  8, 40, 24, 56,  8;

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
Keith Bourque and Steve Ligh, Matrices associated with multiplicative functions, Linear Algebra and its Applications 216 (1995) 267-275.
Shaofang Hong and Raphael Loewy, Asymptotic behavior of eigenvalues of greatest common divisor matrices, Glasgow Mathematical Journal, Vol. 46, No. 3 (2004) 551-569.
Eric Weisstein's World of Mathematics, Least Common Multiple
Wikipedia, Least Common Multiple
Index entries for sequences related to lcm's

Cf. A109043 (column 2), A051193 (row sums), A000384 (central terms).

Cf. A109042 (variant), A003990, A002260, A075362, A051537, A050873.

a051173 = lcm
a051173_row n = a051173_tabl !! (n-1)
a051173_tabl = map (\x -> map (lcm x) [1..x]) [1..]
-- Reinhard Zumkeller, Aug 13 2013, Jul 07 2013

A051173 := proc(u,v) ilcm(u,v) ; end proc: # R. J. Mathar, Apr 07 2011

Table[LCM[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

T(n,k) = lcm(n,k);
tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n,k), ", ")); print;) \\ Michel Marcus, Jul 10 2017

T(n, 1) = T(n, n) = n. T(n, 2) = A109043(n). - R. J. Mathar, Apr 07 2011
T(n, k) = A075362(n, k)/A050873(n, k), 1 <= k <= n. - Reinhard Zumkeller, Apr 25 2011
T(n, k) = A051537(n, k) * A050873(n, k). - Reinhard Zumkeller, Jul 07 2013

A056789 a(n) = Sum_{k=1..n} lcm(k,n)/gcd(k,n).

Original entry on oeis.org

1, 3, 10, 19, 51, 48, 148, 147, 253, 253, 606, 352, 1015, 738, 960, 1171, 2313, 1263, 3250, 1869, 2803, 3028, 5820, 2784, 6301, 5073, 6814, 5458, 11775, 4798, 14416, 9363, 11505, 11563, 14898, 9343, 24643, 16248, 19276, 14797, 33621, 14013, 38830

Offset: 1

Leroy Quet, Aug 20 2000

For prime p, a(p) = 1 + p^2*(p-1)/2.
We note lcm(k,n) = k*n iff gcd(k,n) = 1 (and in general lcm(k,n) equals k*n/gcd(k,n)), and so for these values LCM/GCD = k*n. From A023896, we have that Sum_{k=1..n-1: gcd(k,n)=1} k = n*phi(n)/2, and so Sum_{k=1..n-1: gcd(k,n)=1} k*n = n * Sum_{k=1..n-1: gcd(k,n)=1} k = n^2*phi(n)/2. As this is true, certainly Sum_{k=1..n} lcm(k,n)/gcd(k,n) > n^2*phi(n)/2. - Jon Perry, Nov 09 2014 [Edited by Petros Hadjicostas, May 27 2020]
Conjecture: for prime p, a(p^n) = 1 + (1/2)*(p - 1)*p^2*(p^(3*n) - 1)/(p^3 - 1) for n = 1,2,3,.... Cf. A339384. - Peter Bala, Dec 04 2020
The conjecture can be proven by splitting up the sum like this: a(p^n) = 1 + Sum_{1 <= r < p^n if gcd(p,r) = 1} lcm(p^n,r)/gcd(p^n,r) + Sum_{1 <= r < p^(n-1) if gcd(p,r) = 1} lcm(p^n,p*r)/gcd(p^n,p*r) + … + Sum_{1 <= r < p if gcd(p,r) = 1} lcm(p^n,p^(n-1)*r)/gcd(p^n,p^(n-1)*r) = 1 + Sum_{1 <= r < p^n if gcd(p,r) = 1} p^n*r + Sum_{1 <= r < p^(n-1) if gcd(p,r) = 1} p^(n-1)*r + … + Sum_{1 <= r < p if gcd(p,r) = 1} p*r = 1 + p^n*(1/2)*p^n*phi(p^n) +  p^(n-1)*(1/2)*p^(n-1)*phi(p^(n-1)) + … + p*(1/2)*p*phi(p) = 1 + (1/2)*(p-1)*Sum_{k=1..n} p^(3k-1) = 1 + (1/2)*(p-1)*p^2*(p^(3*n)-1)/(p^3-1). - Sebastian Karlsson, Dec 07 2020

			a(6) = 6/1 + 6/2 + 6/3 + 12/2 + 30/1 + 6/6 = 48.

T. D. Noe, Table of n, a(n) for n = 1..1000

Row sums of triangle in A051537.

Cf. A023896, A068963, A339384.

a056789 = sum . a051537_row  -- Reinhard Zumkeller, Jul 07 2013

Table[ Sum[ LCM[k, n] / GCD[k, n], {k, 1, n}], {n, 1, 50}]
f[p_, e_] := p^2*(p-1)*(p^(3*e)-1)/(p^3-1)+1; a[1] = 1; a[n_] := (1 + Times @@ f @@@ FactorInteger[n])/2; Array[a, 40] (* Amiram Eldar, Oct 05 2023 *)

vector(50, n, sum(k=1, n, lcm(k,n)/gcd(k,n))) \\ Michel Marcus, Nov 08 2014

a(n) = sumdiv(n, d, if(d>1, d^2*eulerphi(d)/2, 1)); \\ Daniel Suteu, Dec 10 2020

a(n) > n^2*phi(n)/2. - Thomas Ordowski, Nov 08 2014
a(n) = Sum_{k=1..n} k*n/gcd(k,n)^2. - Thomas Ordowski, Nov 08 2014
a(n) = (1/2)*Sum_{d|n} d^2*(d+1) Sum_{j|n/d} mu(j)*j^2. - Felix A. Pahl, Nov 23 2019
a(n) = 1 + Sum_{d|n, d > 1} phi(d^3)/2. - Daniel Suteu, Dec 10 2020
From Amiram Eldar, Oct 05 2023: (Start)
a(n) = (A068963(n)+1)/2.
Sum_{k=1..n} a(k) ~ (Pi^2/120) * n^4. (End)
a(n) < n^3 / 2, n > 1. - Bill McEachen, Jul 18 2024
Hence n^3/log log n << a(n) << n^3. - Charles R Greathouse IV, Jul 25 2024

Additional comments from Amarnath Murthy, May 09 2002

A083555 Quotient of LCM of prime(n+1)-1 and prime(n)-1 and GCD of the same two numbers.

Original entry on oeis.org

2, 2, 6, 15, 30, 12, 72, 99, 154, 210, 30, 90, 420, 483, 598, 754, 870, 110, 1155, 1260, 156, 1599, 1804, 132, 600, 2550, 2703, 2862, 756, 72, 4095, 4420, 4692, 5106, 5550, 650, 702, 6723, 7138, 7654, 8010, 342, 9120, 2352, 9702, 1155, 1295, 12543, 12882

Offset: 1

Labos Elemer, May 22 2003

nonn

			n=25: prime(25)=97, prime(26)=101; a(25) = lcm(96,100)/gcd(96,100) = 2400/4 = 600.

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

Cf. A006093, A083538..A083554, A051537, A058263.

P:= seq(ithprime(i),i=1..100):
seq(ilcm(P[i+1]-1,P[i]-1)/igcd(P[i+1]-1,P[i]-1),i=1..99); # Robert Israel, Jun 11 2017

f[x_] := Prime[x]-1 Table[LCM[f[w+1], f[w]]/GCD[f[w+1], f[w]], {w, 1, 128}]
(* Second program: *)
Table[Apply[LCM[#1, #2]/GCD[#1, #2] &, Prime[n + {1, 0}] - 1], {n, 49}] (* Michael De Vlieger, Jun 11 2017 *)

first(n)=my(v=vector(n),p=2,k,g); forprime(q=3,, g=gcd(p-1,q-1); v[k++]=(p-1)*(q-1)/g^2; p=q; if(k==n, break)); v \\ Charles R Greathouse IV, Jun 11 2017

a(n) = lcm(A006093(n+1), A006093(n))/gcd(A006093(n+1), A006093(n));
a(n) = A083554(n)/A058263(n).
a(n) = A051537(A006093(n+1), A006093(n)). - Robert Israel, Jun 11 2017

A164306 Triangle read by rows: T(n, k) = k / gcd(k, n), 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 3, 4, 1, 1, 1, 1, 2, 5, 1, 1, 2, 3, 4, 5, 6, 1, 1, 1, 3, 1, 5, 3, 7, 1, 1, 2, 1, 4, 5, 2, 7, 8, 1, 1, 1, 3, 2, 1, 3, 7, 4, 9, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 1, 1, 1, 1, 5, 1, 7, 2, 3, 5, 11, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1

Offset: 1

Reinhard Zumkeller, Aug 12 2009

Also the gcd of the coefficients of the partition polynomials (called 'De Moivre polynomials' by O'Sullivan, see link, Theorem 4.1). - Peter Luschny, Sep 20 2022

			From _Indranil Ghosh_, Feb 14 2017: (Start)
Triangle begins:
1,
1, 1,
1, 2, 1,
1, 1, 3, 1,
1, 2, 3, 4, 1,
1, 1, 1, 2, 5, 1,
1, 2, 3, 4, 5, 6, 1,
. . .
T(4,3) = 3 / gcd(3,4) = 3 / 1 = 3. (End)

Indranil Ghosh, Rows 1..120 of triangle, flattened.
Cormac O'Sullivan, De Moivre and Bell polynomials, arXiv:2203.02868 [math.CO], 2022.

Cf. A051537, A054531, A057661, A167192.

seq(seq(k / igcd(n, k), k = 1..n), n = 1..13); # Peter Luschny, Sep 20 2022

Flatten[Table[k/GCD[k,n],{n,20},{k,n}]] (* Harvey P. Dale, Jul 21 2013 *)

for(n=0,10, for(k=1,n, print1(k/gcd(k,n), ", "))) \\ G. C. Greubel, Sep 13 2017

Sum of n-th row = A057661(n).

T(n, k) = A051537(n, k)/A054531(n, k). - Reinhard Zumkeller, Oct 30 2009

A338797 Triangle read by rows: T(n,k) is the least m such that there exist positive integers x, y and z satisfying x/n + y/k = z/m where all fractions are reduced; 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 4, 12, 1, 5, 10, 15, 20, 1, 6, 3, 2, 12, 30, 1, 7, 14, 21, 28, 35, 42, 1, 8, 8, 24, 8, 40, 24, 56, 1, 9, 18, 9, 36, 45, 18, 63, 72, 1, 10, 5, 30, 20, 2, 15, 70, 40, 90, 1, 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 1

Offset: 1

Peter Kagey, Nov 09 2020

			Table begins:
  n\k|  1   2   3   4   5   6   7   8   9  10   11 12
  ---+-----------------------------------------------
   1 |  1,
   2 |  2,  1,
   3 |  3,  6,  1,
   4 |  4,  4, 12,  1,
   5 |  5, 10, 15, 20,  1,
   6 |  6,  3,  2, 12, 30,  1,
   7 |  7, 14, 21, 28, 35, 42,  1,
   8 |  8,  8, 24,  8, 40, 24, 56,  1,
   9 |  9, 18,  9, 36, 45, 18, 63, 72,  1,
  10 | 10,  5, 30, 20,  2, 15, 70, 40, 90,   1,
  11 | 11, 22, 33, 44, 55, 66, 77, 88, 99, 110,  1,
  12 | 12, 12,  4,  3, 60,  4, 84, 24, 36, 60, 132, 1.
T(20,10) = 4 because 1/20 + 7/10 = 3/4, and there is no choice of numerators on the left that results in a smaller denominator on the right.

Peter Kagey, Table of n, a(n) for n = 1..10011 (first 141 rows, flattened)

Cf. A051173, A051537, A221918.

import Data.Ratio ((%), denominator)
farey n = [k % n | k <- [1..n], gcd n k == 1]
a338797T n k = minimum [denominator $ a + b | a <- farey n, b <- farey k]

A051537(n,k) <= T(n,k) <= A221918(n,k) <= lcm(n,k) = A051173(n,k).

T(n,k) = lcm(n,k) when gcd(n,k) = 1.

A333493 a(n) = Sum_{k=1..n} (-1)^(k+1) * lcm(n,k) / gcd(n,k).

Original entry on oeis.org

1, 1, -2, 13, -9, 28, -20, 109, -11, 151, -54, 256, -77, 442, 48, 877, -135, 757, -170, 1363, 103, 1816, -252, 2080, -59, 3043, -38, 3982, -405, 2878, -464, 7021, 273, 6937, 390, 6817, -665, 9748, 388, 11059, -819, 8407, -902, 16348, 219, 17458, -1080, 16672, -167

Offset: 1

Ilya Gutkovskiy, Mar 24 2020

sign

Alternating row sums of A051537.

Cf. A056789, A089913, A199084, A199806.

Table[Sum[(-1)^(k + 1) LCM[n, k]/GCD[n, k], {k, 1, n}], {n, 1, 49}]

a(n) = sum(k=1, n, (-1)^(k+1)*lcm(n, k)/gcd(n, k)); \\ Michel Marcus, Mar 24 2020

If n odd, a(n) = (1/2) * n * Sum_{d|n} Sum_{j|d} (-1)^(j + 1) * mu(d/j) * (n + d) / j^2.

If n even, a(n) = (1/2) * n^2 * Sum_{d|n} Sum_{j|d} (-1)^(j + 1) * mu(d/j) * (n + d) / (d * j^2).

cp's OEIS Frontend

A070262 5th diagonal of triangle defined in A051537.

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A070260 Third diagonal of triangle defined in A051537.

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A070261 4th diagonal of triangle defined in A051537.

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A050873 Triangular array T read by rows: T(n,k) = gcd(n,k).

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Haskell

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PARI

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A051173 Triangle read by rows: T(n, k) = lcm(n, k).

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Haskell

Maple

Mathematica

PARI

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A056789 a(n) = Sum_{k=1..n} lcm(k,n)/gcd(k,n).

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Haskell

Mathematica

PARI

PARI

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A083555 Quotient of LCM of prime(n+1)-1 and prime(n)-1 and GCD of the same two numbers.

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