%I #19 Feb 17 2021 11:47:18
%S 2,17,24,82,89,108,145,257,264,283,320,381,472,626,633,652,689,750,
%T 841,968,1137,1297,1304,1323,1360,1421,1512,1639,1808,2025,2296,2402,
%U 2409,2428,2465,2526,2617,2744,2913,3130,3401,3732,4097,4104,4123,4129,4160,4221,4312
%N Numbers that are the sum of a cube s and a fourth power t such that 0 < s <= t.
%C Contains the entries of A340050 and numbers like 2, 8192, 1062882,.. which are 2 times 12th powers (A008456). - _R. J. Mathar_, Jan 05 2021
%e 24 is in the sequence since 2^3 + 2^4 = 8 + 16 = 24, where 0 < 8 <= 16.
%p isA340048 := proc(n)
%p local t,s3 ;
%p for t from 0 do
%p s3 := n-t^4 ;
%p if s3 <= 0 then
%p return false ;
%p elif s3 <= t^4 and isA000578(s3) then
%p return true;
%p end if;
%p end do:
%p end proc:
%p for n from 1 do
%p if isA340048(n) then
%p printf("%d,\n",n) ;
%p end if;
%p end do: # _R. J. Mathar_, Jan 05 2021
%t 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
%o (Python)
%o def aupto(lim):
%o cubes = [i**3 for i in range(1, int(lim**(1/3))+2)]
%o fours = [i**4 for i in range(1, int(lim**(1/4))+2)]
%o return sorted(s+t for s in cubes for t in fours if t >= s and s+t <= lim)
%o print(aupto(4312)) # _Michael S. Branicky_, Feb 17 2021
%Y Cf. A010057.
%Y Cf. A008456, A340050.
%K nonn
%O 1,1
%A _Wesley Ivan Hurt_, Dec 26 2020