|
|
A204315
|
|
Numbers j such that floor(j^(1/4)) divides j.
|
|
1
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
LINKS
|
|
|
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
|
|
EXAMPLE
|
|
|
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;
|
|
MATHEMATICA
|
Select[Range[150], Mod[#, Floor[Surd[#, 4]]]==0&] (* Harvey P. Dale, Oct 04 2023 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|