This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
9 is in the sequence since 2^3 + 1^3 = 9. 35 is in the sequence since 3^3 + 2^3 = 35.
N:= 10000: # to get all terms <= N
S:= select(`<=`,{seq(seq(i^3 + j^3, j = 1 .. i-1), i = 2 .. floor(N^(1/3)))},N);
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(S,list));
# Robert Israel, Oct 07 2014
lst={};Do[Do[x=a^3;Do[y=b^3;If[x+y==n,AppendTo[lst,n]],{b,Floor[(n-x)^(1/3)],a+1,-1}],{a,Floor[n^(1/3)],1,-1}],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Jan 22 2009 *)
Select[Range@ 1700, Total@ Boole@ Map[And[! MemberQ[#, 0], UnsameQ @@ #] &, PowersRepresentations[#, 2, 3]] > 0 &] (* Michael De Vlieger, May 13 2017 *)
isA024670(n)=for( i=ceil(sqrtn( n\2+1,3)),sqrtn(n-.5,3), isA000578(n-i^3) & return(1)) /* One could also use "for( i=2,sqrtn( n\2-1,3),...)" but this is much slower since there are less cubes in [n/2,n] than in [1,n/2]. Replacing the -1 here by +.5 would yield A003325, allowing for a(n)=x^3+x^3. Replacing -1 by 0 may miss some a(n) of this form due to rounding errors. - M. F. Hasler, Apr 12 2008 */
from itertools import count, takewhile
def aupto(limit):
cbs = list(takewhile(lambda x: x <= limit, (i**3 for i in count(1))))
sms = set(c+d for i, c in enumerate(cbs) for d in cbs[i+1:])
return sorted(s for s in sms if s <= limit)
print(aupto(1674)) # Michael S. Branicky, Sep 28 2021
From _Antti Karttunen_, Aug 29 2017: (Start) For n = 36 there is one solution: 36 = 27 + 8 + 1, thus a(36) = 1. For n = 1009 there are two solutions: 1009 = 10^3 + 2^3 + 1^3 = 9^3 + 6^3 + 4^3, thus a(1009) = 2. This is also the first point where sequence attains value greater than one. (End)
A025469 := proc(n) local a, x, y, zcu ; a := 0 ; for x from 1 do if 3*x^3 > n then return a; end if; for y from x+1 do if x^3+2*y^3 > n then break; end if; zcu := n-x^3-y^3 ; if zcu > y^3 and isA000578(zcu) then a := a+1 ; end if; end do: end do: end proc: seq(A025469(n),n=1..80) ; # R. J. Mathar, Jun 15 2018
Table[Count[IntegerPartitions[n, {3}], ?(And[UnsameQ @@ #, AllTrue[#, IntegerQ[#^(1/3)] &]] &)], {n, 105}] (* _Michael De Vlieger, Aug 29 2017 *)
A025469(n) = { my(s=0); for(x=1,n,if(ispower(x,3),for(y=x+1,n-x,if(ispower(y,3),for(z=y+1,n-(x+y),if((ispower(z,3)&&(x+y+z)==n),s++)))))); (s); }; \\ Antti Karttunen, Aug 29 2017
a(1) = 73 = 1^3 + 2^3 + 4^3. a(7) = 521 = 1^3 + 2^3 + 8^3.
lst={};Do[Do[Do[p=n^3+m^3+k^3;If[PrimeQ[p],AppendTo[lst,p]],{n,m+1,4!}],{m,k+1,4!}],{k,4!}];Take[Union[lst],30] (* Vladimir Joseph Stephan Orlovsky, May 23 2009 *)
isA030078 := proc(n) local f ; if n < 8 then false; else f := ifactors(n)[2] ; if nops(f) = 1 and op(2,op(1,f)) = 3 then true; else false; end if; end if; end proc: isA138854 := proc(n) local i,j,p,q,r,rcub ; for i from 1 do p := ithprime(i) ; if p^3+(p+1)^3+(p+2)^3 > n then return false; end if; for j from i+1 do q := ithprime(j) ; rcub := n-q^3-p^3 ; if rcub <= q^3 then break; fi ; if isA030078(rcub) then return true; end if; end do: end do: end proc: for n from 5 do if isA138854(n) then print(n); end if; end do: # R. J. Mathar, Jun 09 2014
f[upto_]:=Module[{maxp=PrimePi[Floor[Power[upto, (3)^-1]]]}, Select[Union[Total/@(Subsets[Prime[Range[maxp]],{3}]^3)],#<=upto&]]; f[9000] (* Harvey P. Dale, Mar 21 2011 *)
isA138854(n)={ if( n%2, isA138853(n), isA120398(n-8)) } for( n=1,10^4, isA138854(n) & print1(n", "))
Comments