cp's OEIS Frontend

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.

Showing 1-4 of 4 results.

A025397 Numbers that are the sum of 3 positive cubes in exactly 3 ways.

Original entry on oeis.org

5104, 9729, 12104, 12221, 12384, 14175, 17604, 17928, 19034, 20691, 21412, 21888, 24480, 28792, 29457, 30528, 31221, 32850, 34497, 35216, 36288, 38259, 39339, 39376, 40060, 40097, 40832, 40851, 41033, 41040, 41364, 41966, 42056, 42687, 43408, 45144
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Maple
    N:= 10^5: # to get all terms <= N
    Reps:= Matrix(N,3,(i,j) -> {}):
    for i from 1 to floor(N^(1/3)) do
      Reps[i^3,1]:= {[i]}
    od:
    for j from 2 to 3 do
    for i from 1 to floor(N^(1/3)) do
      for x from i^3+1 to N do
        Reps[x,j]:= Reps[x,j] union
          map(t -> if t[-1] <= i then [op(t),i] fi, Reps[x-i^3,j-1]);
      od
    od
    od:
    select(t -> nops(Reps[t,3])=3, [$1..N]); # Robert Israel, Aug 28 2015
  • Mathematica
    Reap[ For[ n = 1, n <= 50000, n++, pr = Select[ PowersRepresentations[n, 3, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 3, Print[n, pr]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 31 2013 *)
  • PARI
    is(n)=k=ceil((n-2)^(1/3)); d=0; for(a=1, k, for(b=a, k, for(c=b, k, if(a^3+b^3+c^3==n, d++)))); d
    n=3; while(n<50000, if(is(n)==3, print1(n, ", ")); n++) \\ Derek Orr, Aug 27 2015

Formula

n such that A025456(n) = 3. - Robert Israel, Aug 28 2015

A018787 Numbers that are the sum of two positive cubes in at least three ways (all solutions).

Original entry on oeis.org

87539319, 119824488, 143604279, 175959000, 327763000, 700314552, 804360375, 958595904, 1148834232, 1407672000, 1840667192, 1915865217, 2363561613, 2622104000, 3080802816, 3235261176, 3499524728, 3623721192, 3877315533, 4750893000, 5544709352, 5602516416
Offset: 1

Views

Author

David W. Wilson, Aug 15 1996

Keywords

References

  • J. Leech, Some solutions of Diophantine equations, Proc. Camb. Phil. Soc., 53 (1957), 778-780.
  • R. K. Guy, Unsolved Problems in Number Theory, D1.

Crossrefs

Programs

  • Mathematica
    a=Sort[Flatten@Table[n^3+m^3,{m,2000},{n,m-1,1,-1}]];f3[l_]:=Module[{t={}},Do[If[l[[n]]==l[[n+2]],AppendTo[t,l[[n]]]],{n,1,Length[l]-2}];t];f3[a] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)

A343708 Numbers that are the sum of two positive cubes in exactly two ways.

Original entry on oeis.org

1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728, 110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656, 314496, 320264, 327763, 373464, 402597, 439101, 443889, 513000, 513856, 515375, 525824, 558441, 593047, 684019, 704977, 805688, 842751, 885248, 886464
Offset: 1

Views

Author

David Consiglio, Jr., Apr 26 2021

Keywords

Comments

This sequence differs from A001235 at term 455 because 87539319 = 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3 = A011541(3). Thus, this term is not in this sequence but is in A001235.

Examples

			13832 is in this sequence because 13832 = 2^3 + 24^3 = 18^3 + 20^3.
		

Crossrefs

Programs

  • Mathematica
    Select[Range@70000,Length@Select[PowersRepresentations[#,2,3],FreeQ[#,0]&]==2&] (* Giorgos Kalogeropoulos, Apr 26 2021 *)
  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 for x in range(1,1000)]#n
    for pos in cwr(power_terms,2):#m
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k,v in keep.items() if v == 2])#s
    for x in range(len(rets)):
        print(rets[x])

A345865 Numbers that are the sum of two cubes in exactly four ways.

Original entry on oeis.org

6963472309248, 12625136269928, 21131226514944, 26059452841000, 55707778473984, 74213505639000, 95773976104625, 101001090159424, 159380205560856, 169049812119552, 174396242861568, 188013752349696, 208475622728000, 300656502205416, 340878679288056
Offset: 1

Views

Author

David Consiglio, Jr., Jul 05 2021

Keywords

Comments

Differs from A023051 at term 143 because 48988659276962496 = 331954^3 + 231518^3 = 336588^3 + 221424^3 = 342952^3 + 205292^3 = 362753^3 + 107839^3 = 365757^3 + 38787^3.

Examples

			12625136269928 is a term because 12625136269928 = 21869^3 + 12939^3 = 22580^3 + 10362^3 = 23066^3 + 7068^3 = 23237^3 + 4275^3.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 for x in range(1, 1000)]
    for pos in cwr(power_terms, 2):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 4])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-4 of 4 results.