Computing only Prime Powers with NestList

Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:

{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}

up to some specified prime power. Thanks so much!

asked 3 hours ago

user413587

675

add a comment |

Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:

{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}

up to some specified prime power. Thanks so much!

asked 3 hours ago

user413587

675

add a comment |

Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:

{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}

up to some specified prime power. Thanks so much!

asked 3 hours ago

user413587

675

Is there a simple way to compute only prime powers of a function f with NestList? That is, I want to compute:

{f[x_0], f^2[x_0]=f[f[x_0]], f^3[x_0], f^5[x_0]...}

up to some specified prime power. Thanks so much!

recursion

asked 3 hours ago

user413587

675

asked 3 hours ago

user413587

675

asked 3 hours ago

user413587

675

asked 3 hours ago

user413587

675

asked 3 hours ago

user413587

675

add a comment |

6 Answers
6

active

oldest

votes

Lets say you have f(x) and you want results up to 10

n=10;    

NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]

edited 2 hours ago

answered 2 hours ago

J42161217

3,767321

add a comment |

Here's a way using Compose:

n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]



{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

answered 1 hour ago

bill s

53.4k376152

add a comment |

Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:

primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]



primeNest[f, x, 5]

{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], 

  f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

(Depth /@ primeNest[f, x, 20]) - 1

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}

answered 34 mins ago

Chip Hurst

21.3k15790

add a comment |

This may not be an elegant and smart way to do it but here is one way.

  exp = NestList[g, x, 12];

  selector = PrimeQ[LeafCount /@ exp - 1];

  Pick[exp, selector] /. {g -> f, x -> x0}

edited 2 hours ago

answered 2 hours ago

Okkes Dulgerci

5,0021817

add a comment |

Well, maybe not as elegant as with NestList, but for the first five primes (n=5), for example, here is another way

n = 5;

ls = {};

For [k = 0, k < n, k++,

  (

    Q = f[x];

    For [j = 1, j < Prime[k + 1], j++,

       Q = Map[f, Q];

    ];

    ls = Join[ls, {Q}]

  )

 ]

 Print[ls]

Here is the result:

{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

edited 1 hour ago

answered 2 hours ago

mjw

1065

New contributor

1

$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 hours ago

$begingroup$
Yes, I've seen that! And I've got two loops! If n>10^4, great point! But for n=5 or n=10, do we care?, this code ran pretty quickly.
$endgroup$
– mjw
1 hour ago

$begingroup$
Thank you for the tip, here it is with For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
1 hour ago

add a comment |

ClearAll[primesNest0]

primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]

primesNest0[f, x, 10]

Also

ClearAll[primesNest1]

primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]



primesNest1[f, x, 10]

and

ClearAll[primesNest2]

primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]

primesNest2[f, x, 10]

and

ClearAll[primesNest3]

primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@ 

  Prime[Range@PrimePi@n]

primesNest3[f, x, 10]

. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

edited 46 mins ago

answered 1 hour ago

kglr

184k10202419

add a comment |

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

Lets say you have f(x) and you want results up to 10

n=10;    

NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]

edited 2 hours ago

answered 2 hours ago

J42161217

3,767321

add a comment |

Lets say you have f(x) and you want results up to 10

n=10;    

NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]

edited 2 hours ago

answered 2 hours ago

J42161217

3,767321

add a comment |

Lets say you have f(x) and you want results up to 10

n=10;    

NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]

edited 2 hours ago

answered 2 hours ago

J42161217

3,767321

Lets say you have f(x) and you want results up to 10

n=10;    

NestList[f,x,n][[#+1]]&/@Prime[Range@PrimePi@n]

edited 2 hours ago

answered 2 hours ago

J42161217

3,767321

edited 2 hours ago

answered 2 hours ago

J42161217

3,767321

answered 2 hours ago

J42161217

3,767321

answered 2 hours ago

J42161217

3,767321

add a comment |

Here's a way using Compose:

n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]



{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

answered 1 hour ago

bill s

53.4k376152

add a comment |

Here's a way using Compose:

n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]



{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

answered 1 hour ago

bill s

53.4k376152

add a comment |

Here's a way using Compose:

n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]



{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

answered 1 hour ago

bill s

53.4k376152

Here's a way using Compose:

n = 10; ComposeList[ConstantArray[f, n], f[x]][[#]] & /@ Prime[Range@PrimePi@n]



{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

answered 1 hour ago

bill s

53.4k376152

answered 1 hour ago

bill s

53.4k376152

answered 1 hour ago

bill s

53.4k376152

answered 1 hour ago

bill s

53.4k376152

add a comment |

Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:

primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]



primeNest[f, x, 5]

{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], 

  f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

(Depth /@ primeNest[f, x, 20]) - 1

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}

answered 34 mins ago

Chip Hurst

21.3k15790

add a comment |

Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:

primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]



primeNest[f, x, 5]

{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], 

  f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

(Depth /@ primeNest[f, x, 20]) - 1

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}

answered 34 mins ago

Chip Hurst

21.3k15790

add a comment |

Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:

primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]



primeNest[f, x, 5]

{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], 

  f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

(Depth /@ primeNest[f, x, 20]) - 1

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}

answered 34 mins ago

Chip Hurst

21.3k15790

Here's a way to accumulate the nestings, i.e. use the previous nesting as the starting point for the next one:

primeNest[f_, x_, n_] := FoldList[Nest[f, ##] &, f[f[x]], Differences[Prime[Range[n]]]]



primeNest[f, x, 5]

{f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], 

  f[f[f[f[f[f[f[x]]]]]]], f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

(Depth /@ primeNest[f, x, 20]) - 1

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71}

answered 34 mins ago

Chip Hurst

21.3k15790

answered 34 mins ago

Chip Hurst

21.3k15790

answered 34 mins ago

Chip Hurst

21.3k15790

answered 34 mins ago

Chip Hurst

21.3k15790

add a comment |

This may not be an elegant and smart way to do it but here is one way.

  exp = NestList[g, x, 12];

  selector = PrimeQ[LeafCount /@ exp - 1];

  Pick[exp, selector] /. {g -> f, x -> x0}

edited 2 hours ago

answered 2 hours ago

Okkes Dulgerci

5,0021817

add a comment |

This may not be an elegant and smart way to do it but here is one way.

  exp = NestList[g, x, 12];

  selector = PrimeQ[LeafCount /@ exp - 1];

  Pick[exp, selector] /. {g -> f, x -> x0}

edited 2 hours ago

answered 2 hours ago

Okkes Dulgerci

5,0021817

add a comment |

This may not be an elegant and smart way to do it but here is one way.

  exp = NestList[g, x, 12];

  selector = PrimeQ[LeafCount /@ exp - 1];

  Pick[exp, selector] /. {g -> f, x -> x0}

edited 2 hours ago

answered 2 hours ago

Okkes Dulgerci

5,0021817

This may not be an elegant and smart way to do it but here is one way.

  exp = NestList[g, x, 12];

  selector = PrimeQ[LeafCount /@ exp - 1];

  Pick[exp, selector] /. {g -> f, x -> x0}

edited 2 hours ago

answered 2 hours ago

Okkes Dulgerci

5,0021817

edited 2 hours ago

answered 2 hours ago

Okkes Dulgerci

5,0021817

answered 2 hours ago

Okkes Dulgerci

5,0021817

answered 2 hours ago

Okkes Dulgerci

5,0021817

add a comment |

Well, maybe not as elegant as with NestList, but for the first five primes (n=5), for example, here is another way

n = 5;

ls = {};

For [k = 0, k < n, k++,

  (

    Q = f[x];

    For [j = 1, j < Prime[k + 1], j++,

       Q = Map[f, Q];

    ];

    ls = Join[ls, {Q}]

  )

 ]

 Print[ls]

Here is the result:

{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

edited 1 hour ago

answered 2 hours ago

mjw

1065

New contributor

1

$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 hours ago

$begingroup$
Yes, I've seen that! And I've got two loops! If n>10^4, great point! But for n=5 or n=10, do we care?, this code ran pretty quickly.
$endgroup$
– mjw
1 hour ago

$begingroup$
Thank you for the tip, here it is with For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
1 hour ago

add a comment |

Well, maybe not as elegant as with NestList, but for the first five primes (n=5), for example, here is another way

n = 5;

ls = {};

For [k = 0, k < n, k++,

  (

    Q = f[x];

    For [j = 1, j < Prime[k + 1], j++,

       Q = Map[f, Q];

    ];

    ls = Join[ls, {Q}]

  )

 ]

 Print[ls]

Here is the result:

{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

edited 1 hour ago

answered 2 hours ago

mjw

1065

New contributor

1

$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 hours ago

$begingroup$
Yes, I've seen that! And I've got two loops! If n>10^4, great point! But for n=5 or n=10, do we care?, this code ran pretty quickly.
$endgroup$
– mjw
1 hour ago

$begingroup$
Thank you for the tip, here it is with For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
1 hour ago

add a comment |

Well, maybe not as elegant as with NestList, but for the first five primes (n=5), for example, here is another way

n = 5;

ls = {};

For [k = 0, k < n, k++,

  (

    Q = f[x];

    For [j = 1, j < Prime[k + 1], j++,

       Q = Map[f, Q];

    ];

    ls = Join[ls, {Q}]

  )

 ]

 Print[ls]

Here is the result:

{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

edited 1 hour ago

answered 2 hours ago

mjw

1065

New contributor

Well, maybe not as elegant as with NestList, but for the first five primes (n=5), for example, here is another way

n = 5;

ls = {};

For [k = 0, k < n, k++,

  (

    Q = f[x];

    For [j = 1, j < Prime[k + 1], j++,

       Q = Map[f, Q];

    ];

    ls = Join[ls, {Q}]

  )

 ]

 Print[ls]

Here is the result:

{f[f[x]],f[f[f[x]]],f[f[f[f[f[x]]]]],f[f[f[f[f[f[f[x]]]]]]],f[f[f[f[f[f[f[f[f[f[f[x]]]]]]]]]]]}

edited 1 hour ago

answered 2 hours ago

mjw

1065

New contributor

edited 1 hour ago

answered 2 hours ago

mjw

1065

New contributor

answered 2 hours ago

mjw

1065

answered 2 hours ago

mjw

1065

New contributor

mjw is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1

$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 hours ago

$begingroup$
Yes, I've seen that! And I've got two loops! If n>10^4, great point! But for n=5 or n=10, do we care?, this code ran pretty quickly.
$endgroup$
– mjw
1 hour ago

$begingroup$
Thank you for the tip, here it is with For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
1 hour ago

add a comment |

1

$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 hours ago

$begingroup$
Yes, I've seen that! And I've got two loops! If n>10^4, great point! But for n=5 or n=10, do we care?, this code ran pretty quickly.
$endgroup$
– mjw
1 hour ago

$begingroup$
Thank you for the tip, here it is with For replaced by Do: n = 5; ls = {}; Do [ ( Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}]; ls = Join[ls, {Q}] ), {k, Range[n]}] Print[ls]
$endgroup$
– mjw
1 hour ago

Why should I avoid the For loop in Mathematica?

– corey979
2 hours ago

Yes, I've seen that! And I've got two loops! If n>10^4, great point! But for n=5 or n=10, do we care?, this code ran pretty quickly.

– mjw
1 hour ago

Thank you for the tip, here it is with For replaced by Do:

n = 5; ls = {}; Do [  (   Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];   ls = Join[ls, {Q}]   ), {k, Range[n]}] Print[ls]

– mjw
1 hour ago

Thank you for the tip, here it is with For replaced by Do:

n = 5; ls = {}; Do [  (   Q = f[x]; Do [Q = Map[f, Q], {j, Range[Prime[k] - 1]}];   ls = Join[ls, {Q}]   ), {k, Range[n]}] Print[ls]

– mjw
1 hour ago

add a comment |

ClearAll[primesNest0]

primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]

primesNest0[f, x, 10]

Also

ClearAll[primesNest1]

primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]



primesNest1[f, x, 10]

and

ClearAll[primesNest2]

primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]

primesNest2[f, x, 10]

and

ClearAll[primesNest3]

primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@ 

  Prime[Range@PrimePi@n]

primesNest3[f, x, 10]

. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

edited 46 mins ago

answered 1 hour ago

kglr

184k10202419

add a comment |

ClearAll[primesNest0]

primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]

primesNest0[f, x, 10]

Also

ClearAll[primesNest1]

primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]



primesNest1[f, x, 10]

and

ClearAll[primesNest2]

primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]

primesNest2[f, x, 10]

and

ClearAll[primesNest3]

primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@ 

  Prime[Range@PrimePi@n]

primesNest3[f, x, 10]

. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

edited 46 mins ago

answered 1 hour ago

kglr

184k10202419

add a comment |

ClearAll[primesNest0]

primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]

primesNest0[f, x, 10]

Also

ClearAll[primesNest1]

primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]



primesNest1[f, x, 10]

and

ClearAll[primesNest2]

primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]

primesNest2[f, x, 10]

and

ClearAll[primesNest3]

primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@ 

  Prime[Range@PrimePi@n]

primesNest3[f, x, 10]

. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

edited 46 mins ago

answered 1 hour ago

kglr

184k10202419

ClearAll[primesNest0]

primesNest0[f_, x_, n_] := Nest[f, x, #] & /@ Prime[Range@PrimePi@n]

primesNest0[f, x, 10]

Also

ClearAll[primesNest1]

primesNest1[f_, x_, n_] := Fold[#2@# &, x, ConstantArray[f, #]] & /@ Prime[Range@PrimePi@n]



primesNest1[f, x, 10]

and

ClearAll[primesNest2]

primesNest2[f_, x_, n_] := Compose[##&@@ConstantArray[f, #], x] & /@ Prime[Range@PrimePi@n]

primesNest2[f, x, 10]

and

ClearAll[primesNest3]

primesNest3[f_, x_, n_] := Composition[## & @@ ConstantArray[f, #]][x] & /@ 

  Prime[Range@PrimePi@n]

primesNest3[f, x, 10]

. {f[f[x]], f[f[f[x]]], f[f[f[f[f[x]]]]], f[f[f[f[f[f[f[x]]]]]]]}

edited 46 mins ago

answered 1 hour ago

kglr

184k10202419

edited 46 mins ago

answered 1 hour ago

kglr

184k10202419

answered 1 hour ago

kglr

184k10202419

answered 1 hour ago

kglr

184k10202419

add a comment |

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