A380023 a(n) = a(n-1) * a(n-2) * (1 - 2 / (n * (n-1))), with a(1) = 2, a(2) = 3.

2, 3, 4, 10, 36, 336, 11520, 3732480, 41803776000, 152564385447936000, 6261807987664209366220800000, 940854207318376503485146088437972992000000000, 5815917000990435607656487842294594291938222391518950745702400000000000000

Offset: 1

Author

Hamza K. Özer, Jan 09 2025

Keywords

• nonn

Comments

This sequence is produced when Moessner's Magic is applied to the Fibonacci sequence.

Examples

			For n = 6, a(6) = 336 solutions are 36 * 10 * (1 - 2/(6*5)) = 336 or 7 * 5^0 * 4^1 * 3^1 * 2^2 = 336.
		

References

• J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, page 63

Links

• Wikipedia, Moessner's Theorem

Crossrefs

Cf. A000045, A230053.

Programs

Mathematica

RecurrenceTable[{a[n] == a[n-1]*a[n-2]*(1 - 2/(n*(n-1))), a[1] == 2, a[2] == 3}, a, {n, 1, 14}] (* Amiram Eldar, Jan 09 2025 *)

Python

import math
def A380023(n): return None if n < 1 else (n + 1) * math.prod([(n - k - 1) ** fibo(k) for k in range(0, n - 1)])
# fibo(k) is the function for calculating the k-th Fibonacci number

Formula

a(n) = (n+1) * A230053(n-2), for n >= 2.

a(n) = (n+1) * Product_{k=0..n-2} (n-k-1)^Fibonacci(k), for n >= 2.

Using Moessner's Magic (page 63 of The Book of Numbers), start with writing the counting numbers and cross out each Fibonacci number. Underneath, write the partial sums of the uncrossed terms in the first row and cross out the terms which are offset one place to left of the crossed out terms in the first row. Continuing this process for successive rows produces this sequence:

1 2 3 4 5 6 7 8 9 10 11 12 13

* * * 4 * 10 17 * 26 36 45 57 *

* * * * * 10 * * 36 72 117 * *

* * * * * * * * 36 108 * * *

* * * * * * * * 36 * * * *