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A120110
Diagonal sums of number triangle A120108.
3
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
0,2
LINKS
FORMULA
a(n) = Sum_{k=0..floor(n/2)} lcm(1,..,n-k+1)/lcm(1,..,k+1).
MATHEMATICA
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 *)
PROG
(GAP) 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
(PARI) a(n) = sum(k=0, n\2, lcm([1..n-k+1])/lcm([1..k+1])); \\ Michel Marcus, Mar 04 2019
(Magma)
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
(SageMath)
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
CROSSREFS
Sequence in context: A279286 A282197 A050612 * A047694 A338399 A262016
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jun 09 2006
STATUS
approved