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i would like to sum n=0 to n=(some value) over all integers except those that are generated by some function with integer inputs and outputs, like f[n]=(n^2+3n)/2

so it would sum when n=0, n=1, skip n=2, n=3, n=4, skip n=5 ...

the best I've been able to do so far is make a range for both values, then delete my special cases from the first list, but I feel there is a better, more mathematical way to do this so i can start playing with limits and other fun stuff.

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    $\begingroup$ Is the resulting list such that you can write a function that gives the $i$-th value? $\endgroup$
    – JimB
    Commented Jan 10, 2023 at 18:40

3 Answers 3

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The comment that you want to sum a convergent function over this domain suggests a different approach. First, an example. Say you want to sum

g[n_] := 1/n^2;

over the domain of f[n] for all n. This can be done

Sum[g[f[n]], {n, 1, Infinity}]

If you want to sum over the complement of the f[n], then:

f[n_] := (n^2 + 3 n)/2;
g[n_] := 1/n^2; 
Sum[g[n], {n, 1, Infinity}] - Sum[g[f[n]], {n, 1, Infinity}]

Your suggested convergent function 1/10^n can be summed over the complementary domain like this:

f1[n_] := n^2/2;
f2[n_] := 3/2 n;
g[n_] := 1/10^n; 
Sum[g[n], {n, 1, Infinity}] - Sum[g[f1[n]], {n, 1, Infinity}] - 
       Sum[g[f2[n]], {n, 1, Infinity}]

Your answer is in terms of the EllipticTheta function.

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  • $\begingroup$ Thank you, this is exactly what i was looking for! $\endgroup$
    – Wombles
    Commented Jan 17, 2023 at 17:53
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One approach is to define your function f

f[n_] := (n^2 + 3 n)/2;

Then

n = 10;
Complement[Range[n], f /@ Range[n]]

gives the set of elements that you want to sum over. Hence, you can sum them using:

Sum[i, {i, Complement[Range[n], f /@ Range[n]]}]
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  • $\begingroup$ Is it possible to take the limit as this goes to infinity? $\endgroup$
    – Wombles
    Commented Jan 12, 2023 at 17:45
  • $\begingroup$ The limit of the sum would be infinite, no? $\endgroup$
    – bill s
    Commented Jan 12, 2023 at 23:15
  • $\begingroup$ well, i want to sum a convergent series over this domain, something like 1/10^(n) $\endgroup$
    – Wombles
    Commented Jan 13, 2023 at 18:01
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Clear[cSum, k, n]
f[n_] := (n^2 + 3 n)/2;
cSum[k_] := 
 Plus @@ Extract[Range[k], 
   Position[(Solve[f[n] == #, n, PositiveIntegers] & /@ 
      Range[k] ), {}]]

cSum /@ Range[100]

$$\{1,1,4,8,8,14,21,29,29,39,50,62,75,75,90,106,123,141,160,160,181,203,226,250,275,301,301,329,358,388,419,451,484,518,518,554,591,629,668,708,749,791,834,834,879,925,972,1020,1069,1119,1170,1222,1275,1275,1330,1386,1443,1501,1560,1620,1681,1743,1806,1870,1870,1936,2003,2071,2140,2210,2281,2353,2426,2500,2575,2651,2651,2729,2808,2888,2969,3051,3134,3218,3303,3389,3476,3564,3653,3653,3744,3836,3929,4023,4118,4214,4311,4409,4508,4608\}$$

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