A120109 -id:A120109 - cp's OEIS frontend

A120108 Number triangle T(n,k) = lcm(1,..,n+1)/lcm(1,..,k+1).

1, 2, 1, 6, 3, 1, 12, 6, 2, 1, 60, 30, 10, 5, 1, 60, 30, 10, 5, 1, 1, 420, 210, 70, 35, 7, 7, 1, 840, 420, 140, 70, 14, 14, 2, 1, 2520, 1260, 420, 210, 42, 42, 6, 3, 1, 2520, 1260, 420, 210, 42, 42, 6, 3, 1, 1, 27720, 13860, 4620, 2310, 462, 462, 66, 33, 11, 11, 1

Offset: 0

Paul Barry, Jun 09 2006

			Triangle begins:
    1;
    2,   1;
    6,   3,  1;
   12,   6,  2,  1;
   60,  30, 10,  5, 1;
   60,  30, 10,  5, 1, 1;
  420, 210, 70, 35, 7, 7, 1;

Muniru A Asiru, Rows n=0..100 of triangle, flattened

First column is A003418(n+1). Second column is A025555. Row sums are A120109. Diagonal sums are A120110. Inverse is A120111.

Flat(List([0..10],n->List([0..n],k->Lcm(List([1..n+1],i->i))/Lcm(List([1..k+1],i->i))))); # Muniru A Asiru, Feb 26 2019

A120108:= func< n,k | Lcm([1..n+1])/Lcm([1..k+1]) >;
[A120108(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 04 2023

T:= (n,k)-> ilcm(seq(q,q=1..n+1))/ilcm(seq(r,r=1..k+1)):
seq(seq(T(n,k),k=0..n),n=0..10); # Muniru A Asiru, Feb 26 2019

f[n_] := f[n] = LCM @@ Range[n];
T[n_, k_] := f[n+1]/f[k+1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 01 2021 *)

def f(n): return lcm(range(1,n+2))
def A120108(n,k):
    return f(n)/f(k)
flatten([[A120108(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, May 04 2023

Number triangle T(n,k) = [k<=n]*lcm(1,..,n+1)/lcm(1,..,k+1).

A120110 Diagonal sums of number triangle A120108.

Original entry on oeis.org

1, 2, 7, 15, 67, 92, 461, 1065, 3016, 3956, 29478, 42231, 379107, 547556, 603421, 991923, 12709228, 18540622, 241033695, 352271227, 389226278, 407797820, 5532937710, 8097345425, 30368363481, 41503874738, 98701094676, 127342427241

Offset: 0

Paul Barry, Jun 09 2006

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

Cf. A120108, A120109.

List([0..30],n->Sum([0..Int(n/2)],k->Lcm(List([1..n-k+1],i->i))/Lcm(List([1..k+1],i->i)))); # Muniru A Asiru, Mar 04 2019

A120108:= func< n,k | Lcm([1..n+1])/Lcm([1..k+1]) >;
[(&+[A120108(n-k,k): k in [0..Floor(n/2)]]): n in [0..50]]; // G. C. Greubel, May 04 2023

A120108[n_, k_]:= LCM@@Range[n+1]/(LCM@@Range[k+1]);
A120110[n_]:= Sum[A120108[n-k,k], {k,0,n/2}];
Table[A120110[n], {n,0,50}] (* G. C. Greubel, May 04 2023 *)

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

def f(n): return lcm(range(1,n+2))
def A120110(n):
    return sum(f(n-k)//f(k) for k in range((n//2)+1))
[A120110(n) for n in range(51)] # G. C. Greubel, May 04 2023

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

cp's OEIS Frontend

A120108 Number triangle T(n,k) = lcm(1,..,n+1)/lcm(1,..,k+1).

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