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Numbers that are the sum of a cube s and a fourth power t such that 0 < s <= t.
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%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