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 |

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 |

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 48 mins ago

bill s

53.3k376152

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 30 mins ago

answered 52 mins ago

kglr

184k10202419

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 17 mins ago

Chip Hurst

21.3k15790

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 |

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 |

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 48 mins ago

bill s

53.3k376152

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 48 mins ago

bill s

53.3k376152

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 48 mins ago

bill s

53.3k376152

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 48 mins ago

bill s

53.3k376152

answered 48 mins ago

bill s

53.3k376152

answered 48 mins ago

bill s

53.3k376152

answered 48 mins ago

bill s

53.3k376152

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 30 mins ago

answered 52 mins 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 30 mins ago

answered 52 mins 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 30 mins ago

answered 52 mins 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 30 mins ago

answered 52 mins ago

kglr

184k10202419

edited 30 mins ago

answered 52 mins ago

kglr

184k10202419

answered 52 mins ago

kglr

184k10202419

answered 52 mins ago

kglr

184k10202419

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 17 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 17 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 17 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 17 mins ago

Chip Hurst

21.3k15790

answered 17 mins ago

Chip Hurst

21.3k15790

answered 17 mins ago

Chip Hurst

21.3k15790

answered 17 mins ago

Chip Hurst

21.3k15790

add a comment |

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