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I have a question. I can’t wrap my head around it. It is suppose to be done on Mathematica.

We have a sequence formula x^2+y^2 Where x is positive and y is greater than once. There is a different sequence for each value of y from 1 to 100. For the first 10,000 terms of the sequence((for a given y), we need to find the gcd of the two consecutive terms of the sequence.

Then I’m suppose to make aa list with the max value for the GCD for each of the 100 sequences

this is what i did so far. I did GCD[a, a+1] so I can get the consecutive elements from the list, but how do I make it continue down the list? Thank you!

    f = x^2 + y^2
    l1 = f /. y -> Range[1, 100];
    l2 = GCD[l1 /. x -> Range[1, 10000]];
    a = Simplify[list2]
    GCD[a,a+1]
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When y is fixed,for example y=50 we get the result as below.

Clear["`*"];
f = x^2 + y^2;
y = 50;
GCD @@@ Partition[f /. x -> Range[10000], 2,1] 
%// Counts
<|1 -> 9790, 73 -> 136, 137 -> 72, 10001 -> 1|>
(* when x,y range from 1 to 10 *)
Clear["`*"];
f = x^2 + y^2;
(GCD @@@ Partition[f /. x -> Range[10], 2,1] /. y -> Range[10])
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  • $\begingroup$ what does the 2 mean in Range[10], 2]?''thank you for help! $\endgroup$
    – Aran
    Sep 29 '20 at 2:00
  • $\begingroup$ @Aran Sorry! We should use Partition[#,2,1], instead of Partition[#,2] for example, Partition[{a, b, c, d}, 2, 1] get {{a, b}, {b, c}, {c, d}} $\endgroup$
    – cvgmt
    Sep 29 '20 at 2:03
  • $\begingroup$ thank you so much! $\endgroup$
    – Aran
    Sep 29 '20 at 2:16
  • $\begingroup$ [f /. x -> Range[10000], 2, 1] what is the 2 and 1 for? thanks again! $\endgroup$
    – Aran
    Sep 29 '20 at 2:29
  • $\begingroup$ Partition[f /. x -> Range[10], 2, 1] means that list = f /. x -> Range[10]; Partition[list, 2, 1] . They are equivalent. 2 and 1 is the variables of Partition $\endgroup$
    – cvgmt
    Sep 29 '20 at 2:38

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