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
Keywords
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
Programs
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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 * * * *
Comments