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Hi,

 

tmp:=seq([phi[5-j]*(1-p[6-j]),phi[5-j]*p[6-j]],j=1..4);
Vector(8,[seq(op(tmp[i]),i=1..4)]);

"tmp" is basically what I want. Some kind of 'paired' terms, indexed in reverse order.

I wonder if there is a better (perhaps more efficient and "direct") way to do it?

 

Thanks,

casper

Hi all,

seq(a[i],i=1..3);

Works fine.

seq(a[i],i=3..1);

gives nothing.

But if what I really want, is a ordered sequence

for i from 3 by -1 to 1 do

r:=[op(r),a[i]];

end do;

a[3],a[2],a[1]

What's the best way to get it? (apart from writing a loop)

 

Casper

Hey,

I'm currently working on sequences and can't figure out how to get all of the terms to work modp, is there an easy way of doing this instead of having to use mod(a,p) on each separate term?

Thanks

Rach

 

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)

``


 

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