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Dear friends,

I am reporting with a brief comment concerning the integral int(1/(1+x^a), x=0..infinity) with a>=2 a real number. This was evaluated here.

Now Maple 15 (X86 64 LINUX) will quite happily compute this in its most simple form involving the sine when a is not a positive integer or a rational number. If it is, however, a beta function term results,...

Dear friends,

this is to share with you what a joy it was to work with Maple on the problem of enumerating non-isomorphic graphs. This problem goes back to Polya and Harary and it is a beautiful example of Polya counting, while also being of notable simplicity, so that a high school student or an undergraduate can follow it easily.

I have worked on this problem over the years, adapting my solutions in Cocoa and Lisp as I gained insights. My first attempt used...

I want to find a triangle with the vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) knowing that the point G(1,1,1) is centroid of the triagle ABC and x1, y1, z1, x2, y2, z2, x3, y3, z3 are integer numbers, but I can not find. How do I tell Maple to do that? 

Hello, I have a set of equations, and I want to solve it over the integers (mod 13). But msolve command fails. Here is my code:

eq:={a4*a1-11, a4*a2-12, a4*a3-6, a5*a1-3, a5*a2-8, a5*a3-4, a6*a1-2, a6*a2-1, a6*a3-7};
msolve(eq,13);

It fails, but I can solve it manually. Here is a solution:

a4*a1 = 11 => a4:=1 and a1:=11

a4*a2 = 12 => a2:=12

a4*a3 = 6 => a3:=6

a5*a1 = 3 => a5:=3/11 mod 13 = 5

a5*a2 = 8 => a2:=8/5 mod 13 = 12

Fine $a_{n}$ of sequence  2 , 3 , 6 , 7 , 10 , 11 , 14 , 15,...

Hi All, I have the following procedure to compute the gcd between two integers:

Egcd := proc(a, b)
while b != 0 do
temp := b;
b := a mod b;
a := temp;
od;
return a;  
end proc;

Why does it simply return the value of a when the function was called? (i.e my statements inside the procedure do nothing

I created a triangle whose length of the medians are integral numbers. Please comment to me about my code.

> resrart:

ListTools[Categorize]:

N:=5:

L:=[]:

for x1 from -N to N do

for y1 from x1 to N do

for z1 from y1 to N do

for x2 from -N to N do

for y2 from -N to N do

In geom3d, how can i make a triangle which coordinates of the center circle circumscribed triangle are integer numbers? Please help me. Thank you.

What is the next two terms for this pattern...

1, -2, 2, -4, 0, ...

Suppose I have a matrix M, with rational entries. I need to go row by row and scale each individually by the smallest possible integer such that the entries in each row are integer-valued - anyone have an easy way to do this? It's a large matrix.

We have 1^2 + 2^2 +2^2 = 3^2; 2^2 + 3^2 + 6^2 = 7^2; 2^2 + 10^2 +11^2 = 15^2. Please write for me a command solve the equation a^2 + b^2 + c^2 = d^2, where, a, b, c, d are integer numbers, and d <=100. Thank you very much.

Dear friends, I encounter the following problem. Let k and i be two given positive integers such that k>=2i>=4. How to find all nonnegative integers tuples (x_1, x_2, ..., x_i) such that k-2i<=x_1+2x_2+...+ix_i<=k-i? The buildin command "isolve" seems not work. Thanks.

 

with(numtheory):

f := proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = x .. infinity) end proc

proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = x .. infinity) end proc

(1)

What are the quotients  ot the  continued fration of the sum of f(1)+f(2)+f(3)+f(4)+...

Here are the  quotients  of some partial sums.

``

cfrac(evalf(sum(f(x), x = 1 .. 2)), 'quotients')

[0, 2, 1, 1, 1, 21, 10, 4, 1, 4, 8, `...`]

(2)

cfrac(evalf(sum(f(x), x = 1 .. 3)), 'quotients')

[0, 6, 1, 2, 3, 1, 1, 2, 3, 3, 24, `...`]

(3)

cfrac(evalf(sum(f(x), x = 1 .. 4)), 'quotients')

[0, 2, 1, 2, 1, 4, 2, 1, 3, 1, 1, `...`]

(4)

cfrac(evalf(sum(f(x), x = 1 .. 5)), 'quotients')

[0, 5, 1, 99, 1, 1, 1, 6, 1, 3, 1, `...`]

(5)

cfrac(evalf(sum(f(x), x = 1 .. 6)), 'quotients')

[0, 2, 1, 6, 1, 2, 1, 2, 2, 1, 1, `...`]

(6)

cfrac(evalf(sum(f(x), x = 1 .. 7)), 'quotients')

[0, 5, 1, 1, 142, 1, 1, 1, 1, 19, 1, `...`]

(7)

cfrac(evalf(sum(f(x), x = 1 .. 8)), 'quotients')

[0, 2, 1, 47, 1, 1, 1, 1, 27, 4, 1, `...`]

(8)

cfrac(evalf(sum(f(x), x = 1 .. 9)), 'quotients')

[0, 5, 5, 3, 1, 7, 1, 1, 1, 2, 1, `...`]

(9)

cfrac(evalf(sum(f(x), x = 1 .. 100)), 'quotients')

[0, 3, 1, 1, 1, 11, 2, 2, 1, 1, 4, `...`]

(10)

cfrac(evalf(sum(f(x), x = 1 .. 200)), 'quotients')

[0, 3, 1, 2, 1, 1, 1, 11, 3, 4, 6, `...`]

(11)

cfrac(evalf(sum(f(x), x = 1 .. 400)), 'quotients')

[0, 3, 1, 3, 3, 3, 1, 18, 1, 2, 1, `...`]

(12)

cfrac(evalf(sum(f(x), x = 1 .. 800)), 'quotients')

[0, 3, 1, 3, 1, 4, 16, 14, 3, 23, 2, `...`]

(13)

cfrac(evalf(sum(f(x), x = 1 .. 1600)), 'quotients')

[0, 3, 1, 4, 7, 4, 436, 1, 1, 1, 2, `...`]

(14)

``

Here are the quotients of the  continued fration  of the sum. 

cfrac(evalf(sum(f(x), x = 1 .. infinity)), 'quotients')

[0, 3, 1, 4, 1, 1, 1, 1, 1, 9, 1, `...`]

(15)

With the exception of the leading 0, that is close to the integer squence of pi.

``evalf((65241/65251)*Pi)

3.141111191

(16)

The exponents of 2 that sum the numerator and denominator, in the following way, of that multiple of pi give rise to the integer sequences {0,1,2,3,8,16},numbers such that floor[a(n)^2 / 7] is a square, and {0,2,3,4,8,16},{0,3} union powers of 2.

evalf((2^16-2^8-2^5-2^2-2-2^0)*Pi/(2^16-2^8-2^4-2^3-2^2-2^0))

3.141111191

(17)

We can do the same thing for the first 20 quotients giving rise to the integer sequences {0,1,2,5,6,8,10,13,17,19,22,23,24,28,31} and {0,4,6,9,12, 14,15,16,18,22, 23,24,28,31}. What can be said of these sequences?

cfrac(evalf(sum(f(x), x = 1 .. infinity), 20), 20, 'quotients')``

[0, 3, 1, 4, 1, 1, 1, 1, 1, 9, 1, 3, 1, 2, 1, 1, 1, 5, 1, 3, 11, `...`]

(18)

evalf((1849023129/1849306543)*Pi, 20)

3.1411111913121115131

(19)

````

evalf((2^31-2^28-2^24-2^23-2^22-2^19-2^17-2^13-2^10-2^8-2^6-2^5-2^2-2-2^0)*Pi/(2^31-2^28-2^24-2^23-2^22-2^18-2^16-2^15-2^14-2^12-2^9-2^6-2^4-2^0), 20)

3.1411111913121115131

(20)

``


 

Let a sequence( an  ) defined by an=1,3,6,12,33,51,...and n=1,2,3,.... Find formula an?

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