|
|
A340048
|
|
Numbers that are the sum of a cube s and a fourth power t such that 0 < s <= t.
|
|
0
|
|
|
2, 17, 24, 82, 89, 108, 145, 257, 264, 283, 320, 381, 472, 626, 633, 652, 689, 750, 841, 968, 1137, 1297, 1304, 1323, 1360, 1421, 1512, 1639, 1808, 2025, 2296, 2402, 2409, 2428, 2465, 2526, 2617, 2744, 2913, 3130, 3401, 3732, 4097, 4104, 4123, 4129, 4160, 4221, 4312
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Contains the entries of A340050 and numbers like 2, 8192, 1062882,.. which are 2 times 12th powers (A008456). - R. J. Mathar, Jan 05 2021
|
|
LINKS
|
|
|
EXAMPLE
|
24 is in the sequence since 2^3 + 2^4 = 8 + 16 = 24, where 0 < 8 <= 16.
|
|
MAPLE
|
isA340048 := proc(n)
local t, s3 ;
for t from 0 do
s3 := n-t^4 ;
if s3 <= 0 then
return false ;
elif s3 <= t^4 and isA000578(s3) then
return true;
end if;
end do:
end proc:
for n from 1 do
if isA340048(n) then
printf("%d, \n", n) ;
end if;
|
|
MATHEMATICA
|
Table[If[Sum[(Floor[i^(1/3)] - Floor[(i - 1)^(1/3)]) (Floor[(n - i)^(1/4)] - Floor[(n - i - 1)^(1/4)]), {i, Floor[n/2]}] > 0, n, {}], {n, 1000}] // Flatten
|
|
PROG
|
(Python)
def aupto(lim):
cubes = [i**3 for i in range(1, int(lim**(1/3))+2)]
fours = [i**4 for i in range(1, int(lim**(1/4))+2)]
return sorted(s+t for s in cubes for t in fours if t >= s and s+t <= lim)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|