A088301 a(n) = p(n)/p(n-1), where p(n) = ( floor(n*log(n)) / Product_{j=2..pi(floor(n*log(n)))} prime(j) )!.

1, 2, 4, 48, 90, 12, 3360, 18, 9240, 15600, 756, 31680, 42840, 59280, 1848, 99360, 6497400, 2970, 185136, 234360, 18670080, 347760, 421800, 480480, 557928, 55965360, 70073640, 857280, 98960400, 1157520, 11880, 162983520, 190578024

Offset

2,2

Links

G. C. Greubel, Table of n, a(n) for n = 2..1000

Formula

a(n) = p(n)/p(n-1), where p(n) = ( floor(n*log(n)) / Product_{j=2..pi(floor(n*log(n)))} prime(j) )!.

Mathematica

p[n_]:=Factorial[Floor[n*Log[n]]]/ Product[Prime[i], {i, 2, PrimePi[Floor[n*Log[n]]]}];
Table[p[n]/p[n-1], {n, 2, 50}]

Prog

(Magma)
m:=50;
b:= [ #PrimesUpTo(n): n in [1..2+Floor(2*m*Log(2*m))] ];
f:= func< n | Factorial( Floor(n*Log(n)) )/(&*[ NthPrime(j): j in [2..b[Floor(n*Log(n))]] ]) >;
A088301:= func< n | n le 3 select n-1 else f(n)/f(n-1) >;
[A088301(n): n in [2..m]]; // G. C. Greubel, Dec 18 2022
(SageMath)
def p(n): return factorial( floor(n*log(n)) )/product(nth_prime(j) for j in (2..prime_pi(floor(n*log(n)))))
def A088301(n): return p(n)/p(n-1)
[A088301(n) for n in range(2, 50)] # G. C. Greubel, Dec 18 2022

Crossrefs

Cf. A000720, A050504.

Sequence in context: A018325 A099804 A019596 * A212429 A298903 A127211

Adjacent sequences: A088298 A088299 A088300 * A088302 A088303 A088304

Keyword

nonn,less

Author

Roger L. Bagula, Nov 04 2003

Extensions

Edited by G. C. Greubel, Dec 18 2022

Status

approved