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

%I A380023 #36 Mar 16 2025 22:56:15
%S A380023 2,3,4,10,36,336,11520,3732480,41803776000,152564385447936000,
%T A380023 6261807987664209366220800000,
%U A380023 940854207318376503485146088437972992000000000,5815917000990435607656487842294594291938222391518950745702400000000000000
%N A380023 a(n) = a(n-1) * a(n-2) * (1 - 2 / (n * (n-1))), with a(1) = 2, a(2) = 3.
%C A380023 This sequence is produced when Moessner's Magic is applied to the Fibonacci sequence.
%D A380023 J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, page 63
%H A380023 Wikipedia, <a href="https://en.wikipedia.org/wiki/Moessner%27s_theorem">Moessner's Theorem</a>
%F A380023 a(n) = (n+1) * A230053(n-2), for n >= 2.
%F A380023 a(n) = (n+1) * Product_{k=0..n-2} (n-k-1)^Fibonacci(k), for n >= 2.
%F A380023 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:
%F A380023 1 2 3 4 5 6  7  8 9  10  11  12 13
%F A380023 * * * 4 * 10 17 * 26 36  45  57 *
%F A380023 * * * * * 10 *  * 36 72  117 *  *
%F A380023 * * * * * *  *  * 36 108 *   *  *
%F A380023 * * * * * *  *  * 36 *   *   *  *
%e A380023 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.
%t A380023 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 *)
%o A380023 (Python)
%o A380023 import math
%o A380023 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)])
%o A380023 # fibo(k) is the function for calculating the k-th Fibonacci number
%Y A380023 Cf. A000045, A230053.
%K A380023 nonn
%O A380023 1,1
%A A380023 _Hamza K. Özer_, Jan 09 2025