Let {a
n} be a sequence of numbers that satisfy a
0 = 3, and (3 − a
n+1)(6 + a
n) = 18. Find
n 1
∑ ---
i=0 ai
clc,clearvars
a=vpa(3);
for i=2:100
a(end+1)=3-18/(6+a(end));
% disp([i a(end) 1/a(end) sum(1./a)])
fprintf('%3d %15.14f %15.6f %15.6f
',i,a(end),1/a(end),sum(1./a))
end
clc,clearvars
a=vpa(3);
for i=2:100
a(end+1)=3-18/(6+a(end));
% disp([i a(end) 1/a(end) sum(1./a)])
fprintf('%d,',sum(3./a))
end
fprintf('
')
After 1/a(0) = 1/3:
i a(i) 1/a(i) sum so far
2 1.00000000000000 1.000000 1.333333
3 0.42857142857143 2.333333 3.666667
4 0.20000000000000 5.000000 8.666667
5 0.09677419354839 10.333333 19.000000
6 0.04761904761905 21.000000 40.000000
7 0.02362204724409 42.333333 82.333333
8 0.01176470588235 85.000000 167.333333
seems always a multiple of 1/3
The multiples of 1/3 are
4,11,26,57,120,247,502,1013,2036,4083,8178,16369,32752,...
which the OEIS identifies as A000295 Eulerian numbers.
Some formulae given are
a(n) = 2^n - n - 1.
a(n) = Sum_{k=2..n} binomial(n, k)
starting with n=2 giving 1 for the 1/3 for subscript zero
n=3 giving 4 for the 4/3 for subscript 1, etc.
Then these are 2^(n+2) - n - 3.
Dividing by 3:
(2^(n+2) - n - 3) / 3
Edited on March 26, 2025, 10:57 am
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Posted by Charlie
on 2025-03-26 09:48:41 |
computer solution
