# Nested function with an increasing argument

If, for example, I have a nesting function like

Nest[1 + x^n (#) &, 1, 3]


it gives the result

1 + x^n (1 + x^n (1 + x^n))


I would need the n to increase by one with every step, just like in a sum or product function, so that the result would be

1 + x^1 (1 + x^2 (1 + x^3))


Is there a way to do this in Mathematica? If so, how?

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## 5 Answers

Module[{i = 1}, Nest[1 + x^n (#) &, 1, 3] /. n :> i++] Or (same result)

Fold[1 + x^#2 #1 &, 1, {3, 2, 1}]

• +1 for Module construct. – garej Dec 5 '15 at 11:48

Taking a different approach it appears that you are seeking to apply Horner's Method to the polynomial

$$\underset{i=0}{\overset{n}{\sum }}x^{\underset{j=0}{\overset{i}{\sum }}j}$$

Taking $\underset{j=0}{\overset{i}{\sum }}j = \frac{1}{2}i(1+i)$ and making use of HornerForm a function can be created to construct the factored polynomial.

f[n_Integer?Positive] := HornerForm@Sum[x^(i (1 + i)/2), {i, 0, n}]


You are seeking f

f
(* 1 + x (1 + x^2 (1 + x^3)) *)


Hope this helps.

In case the intermediate result 1 + x^n (1 + x^n (...)) is not required,

n = 3;
First@Nest[{1 + x^#[] #[], #[] - 1} &, {1, n}, n]


Alternative

A simpler formulation under the same assumption as above

n = 3;
Nest[1 + x^(n--) # &, 1, n]

• +1 for (n - -) construct – garej Dec 5 '15 at 11:49

I would use Fold.

Fold[1 + x^#2 (#1) &, 1, Range[3, 1, -1]]
(*  1 + x (1 + x^2 (1 + x^3))  *)


Two approaches with Replacement.

z = Nest[1 + x^n (#) &, 1, 3]
ReplacePart[z, Position[z, n][[#]] -> Defer@# & /@ Range] Also inspired by the @Karsten7's answer and @Eldo's Module version:

Module[{r = 3, i = 1},
Quiet@ReplaceRepeated[(1 + x^n), (1 + x^n) -> 1 + x^n (1 + x^n),
MaxIterations -> r - 1] /. n :> i++]


1 + x (1 + x^2 (1 + x^3))

• Some more inspiration: Module[{n = 3}, ReplaceRepeated[(1 + x^n), p_Plus /; n > 1 :> 1 + x^(--n) p]] – Karsten 7. Dec 5 '15 at 16:58