A204315 Numbers j such that floor(j^(1/4)) divides j.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126
Offset: 1
Keywords
Examples
26 is a term as floor(26^(1/4)) = 2 divides 26. - _David A. Corneth_, Oct 04 2023
Links
- David A. Corneth, Table of n, a(n) for n = 1..11115
Programs
-
Maple
isA204315 := proc(n) if modp(n,floor(root[4](n))) = 0 then true ; else false ; fi ; end proc: for n from 1 to 130 do if isA204315(n) then printf("%d,",n) ; end if; end do: # R. J. Mathar, Sep 10 2017
-
Mathematica
Select[Range[150],Mod[#,Floor[Surd[#,4]]]==0&] (* Harvey P. Dale, Oct 04 2023 *)
-
PARI
a(n) = {my(k = 0, t = 0); while(t < n, k++; t = 4*k^3/3 + 5*k^2 + 26*k/3); (k+1)^4 - 1 - k * (t - n)} \\ David A. Corneth, Oct 06 2023
-
PARI
first(n) = {my(res = vector(n), t = 0); for(i = 1, oo, forstep(j = i^4, (i + 1)^4 - 1, i, t++; if(t > n, return(res)); res[t] = j))} \\ David A. Corneth, Oct 06 2023
Formula
Let f(x) = 4*x^3/3 + 5*x^2 + 26*x/3 and let k be the smallest integer x such that f(x) >= n. Then a(n) = (k+1)^4 - 1 - k * (f(k) - n). - David A. Corneth, Oct 06 2023
Comments