A120101 -id:A120101 - cp's OEIS frontend

A120113 Bi-diagonal inverse of number triangle A120101.

1, -6, 1, 0, -5, 1, 0, 0, -14, 1, 0, 0, 0, -3, 1, 0, 0, 0, 0, -11, 1, 0, 0, 0, 0, 0, -13, 1, 0, 0, 0, 0, 0, 0, -2, 1, 0, 0, 0, 0, 0, 0, 0, -17, 1, 0, 0, 0, 0, 0, 0, 0, 0, -19, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1

Offset: 0

Paul Barry, Jun 09 2006

Subdiagonal is -A120114(n-1).

			Triangle begins
   1;
  -6,  1;
   0, -5,   1;
   0,  0, -14,  1;
   0,  0,   0, -3,   1;
   0,  0,   0,  0, -11,   1;
   0,  0,   0,  0,   0, -13,  1;
   0,  0,   0,  0,   0,   0, -2,   1;
   0,  0,   0,  0,   0,   0,  0, -17,   1;
   0,  0,   0,  0,   0,   0,  0,   0, -19,  1;
   0,  0,   0,  0,   0,   0,  0,   0,   0, -1,  1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A120101, A120111, A120114.

A120114:= func< n | Lcm([1..2*n+4])/Lcm([1..2*n+2]) >;
A120113:= func< n,k | k eq n select 1 else k eq n-1 select -A120114(n-1) else 0 >;
[A120113(n,k): k in [0..n], n in [0..16]]; // G. C. Greubel, May 05 2023

A120114[n_]:= LCM@@Range[2*n+4]/(LCM@@Range[2*n+2]);
A120113[n_, k_]:= If[k==n, 1, If[k==n-1, -A120114[n-1], 0]];
Table[A120113[n, k], {n,0,16}, {k,0,n}]//Flatten

def A120113(n,k):
    if (kA120113(n,k) for k in range(n+1)] for n in range(17)]) # G. C. Greubel, May 05 2023

T(n, k) = 1 if k = n, T(n, k) = -A120114(n-1) if k = n-1, otherwise 0. - G. C. Greubel, May 05 2023

A051538 Least common multiple of {b(1),...,b(n)}, where b(k) = k(k+1)(2k+1)/6 = A000330(k).

Original entry on oeis.org

1, 5, 70, 210, 2310, 30030, 60060, 1021020, 19399380, 19399380, 446185740, 2230928700, 6692786100, 194090796900, 12033629407800, 12033629407800, 12033629407800, 445244288088600, 445244288088600, 18255015811632600

Offset: 1

Asher Auel

Also a(n) = lcm(1,...,(2n+2))/12. - Paul Barry, Jun 09 2006. Proof that this is the same sequence, from Martin Fuller, May 06 2007: k, k+1, 2k+1 are coprime so their lcm is the same as their product. Hence a(n) = lcm{k, k+1, 2k+1 | k=1..n}/6. {k, k+1, 2k+1 | k=1..n} = {1..2n+2 excluding even numbers >n+1}. Adding the higher even numbers to the set doubles the LCM. Hence lcm{k, k+1, 2k+1 | k=1..n}/6 = lcm{1..2n+2}/12.

			a(4) = lcm(1, 5, 14, 30) = 210.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Second column of A120101.
Cf. A000330.
Cf. A051542 (LCM), A025555.

a051538 n = a051538_list !! (n-1)
a051538_list = scanl1 lcm $ tail a000330_list
-- Reinhard Zumkeller, Mar 12 2014

[Lcm([1..2*n+2])/12: n in [1..30]]; // G. C. Greubel, May 03 2023

Table[LCM@@Range[2n+2]/12,{n,30}] (* Harvey P. Dale, Apr 25 2011 *)

def A051538(n):
    return lcm(range(1,2*n+3))/12
[A051538(n) for n in range(1,31)] # G. C. Greubel, May 03 2023

Corrected by James Sellers

Edited by N. J. A. Sloane, May 06 2007

A120106 a(n) = Sum_{k=0..n} lcm(1..2n+2)/lcm(1..2k+2).

Original entry on oeis.org

1, 7, 36, 505, 1516, 16677, 216802, 433605, 7371286, 140054435, 140054436, 3221252029, 16106260146, 48318780439, 1401244632732, 86877167229385, 86877167229386, 86877167229387, 3214455187487320, 3214455187487321

Offset: 0

Paul Barry, Jun 09 2006

			a(2) = lcm(1,2,3,4,5,6)*(1/lcm(1,2) + 1/lcm(1,2,3,4) + 1/lcm(1,2,3,4,5,6)) = 60 (1/2 + 1/12 + 1/60) = 60 * 3/5 = 36. - _Bernard Schott_, Feb 27 2019

Muniru A Asiru, Table of n, a(n) for n = 0..500

Row sums of number triangle A120101.

List([0..20], n-> Sum([0..n], k-> Lcm(List([1..2*n+2],i->i) )/Lcm(List([1..2*k+2],i->i)))); # Muniru A Asiru, Feb 26 2019

[(&+[ LCM([j: j in [1..2*n+2]])/LCM([j: j in [1..2*k+2]]): k in [0..n]]): n in [0..20]]; // G. C. Greubel, Feb 26 2019

Table[Sum[(LCM@@Range[2n+2])/LCM@@Range[2k+2],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Mar 25 2012 *)

lcv(n) = lcm(vector(2*n+2, j, j));
a(n) = lcv(n)*sum(k=0, n, 1/lcv(k)); \\ Michel Marcus, Feb 27 2019

[sum(lcm(range(1,2*n+3))/lcm(range(1,2*k+3)) for k in (0..n)) for n in (0..20)] # G. C. Greubel, Feb 26 2019

A119634 a(n) = lcm(1,...,2n+2)/2.

Original entry on oeis.org

1, 6, 30, 420, 1260, 13860, 180180, 360360, 6126120, 116396280, 116396280, 2677114440, 13385572200, 40156716600, 1164544781400, 72201776446800, 72201776446800, 72201776446800, 2671465728531600, 2671465728531600

Offset: 0

Paul Barry, Jun 09 2006

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
Vaclav Kotesovec, Plot of log(a(n))/n for n = 1..10000

Cf. A003418.

Column k=0 of A120101.

List([0..20],n->Lcm(List([1..2*n+2]))/2); # Muniru A Asiru, Mar 04 2019

[LCM([1..2*n+2])/2: n in [0..20]]; // G. C. Greubel, Mar 05 2019

Table[LCM @@ Range[2*n + 2]/2, {n, 0, 20}] (* Vaclav Kotesovec, Mar 04 2019 *)

{a(n) = lcm(vector(2*n+2, i, i))/2}; \\ G. C. Greubel, Mar 05 2019

[lcm(range(1,2*n+3))/2 for n in (0..20)] # G. C. Greubel, Mar 05 2019

A120107 a(n) = Sum_{k=0..floor(n/2)} lcm(1,...,2(n-k)+2)/lcm(1,...,2k+2).

Original entry on oeis.org

1, 6, 31, 425, 1331, 14084, 182533, 390855, 6192220, 117429752, 136000866, 2700408581, 13835919839, 42477252404, 1171690228133, 72397239805085, 84274330442804, 86644937313210, 2686078920033439, 3119346038772923

Offset: 0

Paul Barry, Jun 09 2006

Diagonal sums of number triangle A120101.

Muniru A Asiru, Table of n, a(n) for n = 0..1000

Cf. A120101, A120105, A120106.

List([0..20],n->Sum([0..Int(n/2)],k->Lcm(List([1..2*(n-k)+2],i->i))/Lcm(List([1..2*k+2],i->i)))); # Muniru A Asiru, Mar 03 2019

A120105:= func< n,k | Lcm([1..2*n+2])/Lcm([1..2*k+2]) >;
[(&+[A120105(n-k,k): k in [0..Floor(n/2)]]): n in [0..50]]; // G. C. Greubel, May 04 2023

A120105[n_, k_]:= LCM@@Range[2*n+2]/(LCM@@Range[2*k+2]);
A120107[n_]:= Sum[A120105[n-k, k], {k, 0, Floor[n/2]}];
Table[A120107[n], {n,0,50}] (* G. C. Greubel, May 04 2023 *)

a(n) = sum(k=0, n\2, lcm([1..2*(n-k)+2])/lcm([1..2*k+2])); \\ Michel Marcus, Mar 04 2019

def f(n): return lcm(range(1,2*n+3))
def A120107(n):
    return sum(f(n-k)/f(k) for k in range(1+(n//2)))
[A120107(n) for n in range(51)] # G. C. Greubel, May 04 2023

a(n) = Sum_{k=0..floor(n/2)} A120105(n-k, k). - G. C. Greubel, May 04 2023

A119636 a(n) = lcm(1,...,2n+4)/((n+1)*binomial(2n+2, n+1)).

Original entry on oeis.org

6, 5, 14, 9, 22, 65, 30, 119, 532, 126, 690, 825, 594, 4147, 62062, 15015, 3640, 32708, 7956, 79458, 833340, 203490, 2337874, 4004231, 980628, 12738550, 3124550, 766935, 11113830, 166613265, 81940950

Offset: 0

Paul Barry, Jun 09 2006

Subdiagonal of A120101.

Muniru A Asiru, Table of n, a(n) for n = 0..3000

Cf. A068553.

List([0..40],n->Lcm(List([1..2*n+4]))/((n+1)*Binomial(2*n+2,n+1))); # Muniru A Asiru, Mar 04 2019

[Lcm([1..2*n+4])/((n+1)*Binomial(2*n+2, n+1)): n in [0..40]]; // G. C. Greubel, Mar 04 2019

Table[LCM@@Range[2n+4]/((n+1)Binomial[2n+2,n+1]),{n,0,30}] (* Harvey P. Dale, Jun 08 2018 *)

[lcm(range(1, 2*(n+2)+1))/((n+1)*binomial(2*n+2, n+1)) for n in (0..40)] # G. C. Greubel, Mar 04 2019

cp's OEIS Frontend

A120113 Bi-diagonal inverse of number triangle A120101.

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A051538 Least common multiple of {b(1),...,b(n)}, where b(k) = k(k+1)(2k+1)/6 = A000330(k).

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A120106 a(n) = Sum_{k=0..n} lcm(1..2n+2)/lcm(1..2k+2).

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A119634 a(n) = lcm(1,...,2n+2)/2.

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A120107 a(n) = Sum_{k=0..floor(n/2)} lcm(1,...,2*(n-k)+2)/lcm(1,...,2*k+2).

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GAP

Magma

Mathematica

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Formula

A119636 a(n) = lcm(1,...,2n+4)/((n+1)*binomial(2n+2, n+1)).

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GAP

Magma

Mathematica

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A120107 a(n) = Sum_{k=0..floor(n/2)} lcm(1,...,2(n-k)+2)/lcm(1,...,2k+2).