A382257 a(n) is the numerator of tanh(Sum_{k=1..n-1} artanh(k/n)), where artanh is the inverse hyperbolic tangent function.
0, 1, 9, 17, 125, 461, 1715, 3217, 24309, 92377, 352715, 1352077, 5200299, 20058299, 77558759, 150270097, 1166803109, 4537567649, 17672631899, 68923264409, 269128937219, 1052049481859, 4116715363799, 16123801841549, 63205303218875, 247959266474051, 973469712824055, 3824345300380219, 15033633249770519
Offset: 1
Examples
For n=2, tanh(artanh(1/2)) = 1/2, so a(2) = numerator(1/2) = 1. For n=3, tanh(artanh(1/3) + artanh(2/3)) = (1/3 + 2/3) / (1 + 1/3 * 2/3) = 9/11, so a(3) = 9. Numerators of 0, 1/2, 9/11, 17/18, 125/127, 461/463, 1715/1717, 3217/3218, ...
Links
- Wikipedia contributors, Area function (inverse hyperbolic function), in: Wikipedia, the free encyclopedia. As of April 7, 2025.
Programs
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PARI
apply( {A382257(n)=my(s=0); for(i=1, n-1, s=(s*n+i)/(n+s*i));numerator(s)}, [1..30]) -
Python
from sympy import S,expand_trig as ET tanh,artanh = S("tanh, artanh") def A382257_test(n): # for illustration only -- slow for n >= 19 n=S(n); return ET(tanh(sum(artanh(k/n) for k in range(1,n)))).numerator def A382257(n): s=0; i=S.One/n for k in range(1,n): s = (s + i*k)/(1 + s*k*i) return s.numerator -
Python
from functools import reduce from fractions import Fraction def A382257(n): return reduce(lambda x,y: Fraction(x+y,1+x*y),(Fraction(i,n) for i in range(1,n)),0).numerator # Chai Wah Wu, Apr 23 2025
Formula
a(n) = (binomial(2n-1, n-1) - 1)/2 if n = 2^m or a(n) = binomial(2n-1, n-1) - 1 = A010763(n-1) otherwise, since tanh(Sum_{k=1..n-1} artanh(k/n)) = (binomial(2n-1, n-1) - 1)/(binomial(2n-1, n-1) + 1) reduced. - Thomas Ordowski, Apr 27 2025
Comments