A076116 Start of the smallest string of n consecutive positive numbers with a cube sum, or 0 if no such number exists.

1, 13, 8, 0, 23, 2, 46, 0, 20, 8, 116, 0, 163, 18, 218, 6, 281, 32, 352, 0, 431, 50, 518, 0, 28, 72, 14, 0, 827, 98, 946, 0, 1073, 128, 1208, 0, 1351, 162, 1502, 0, 1661, 200, 1828, 0, 53, 242, 2186, 98, 32, 43, 2576, 0, 2783, 36, 2998, 0, 3221, 392, 3452, 0, 3691, 450

Offset: 1

Author

Amarnath Murthy, Oct 09 2002

Keywords

• nonn
• look

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007814, A019555, A076117, A076114.

Programs

Maple

f:= proc(n) local y,F,t,k,v;
      if n::odd then
         F:= ifactors(n)[2];
         y:= mul(t[1]^ceil(t[2]/3),t=F);
         k:= 1+floor((n*(n-1)/2)^(1/3)/y);
         (k*y)^3/n - (n-1)/2;
      else
         v:= padic:-ordp(n,2);
         if v mod 3 <> 1 then return 0 fi;
         F:= ifactors(n/2^v)[2];
         y:= mul(t[1]^ceil(t[2]/3),t=F)*2^((v-1)/3);
         k:= 1 + floor((n*(n-1)/2)^(1/3)/y);
         if k::even then k:= k+1 fi;
         (k*y)^3/n - (n-1)/2;
      fi
end proc:
map(f, [$1..100]); # Robert Israel, Nov 15 2023

Mathematica

f[n_] := Module[{y, F, t, k, v},
If[OddQ[n],
   F = FactorInteger[n];
   y = Product[t[[1]]^Ceiling[t[[2]]/3], {t, F}];
   k = 1 + Floor[(n*(n-1)/2)^(1/3)/y];
   (k*y)^3/n - (n-1)/2
,
   v = IntegerExponent[n, 2];
   If[Mod[v, 3] != 1, Return[0]];
   F = FactorInteger[n/2^v];
   y = Product[t[[1]]^Ceiling[t[[2]]/3], {t, F}]*2^((v-1)/3);
   k = 1 + Floor[(n*(n-1)/2)^(1/3)/y];
   If[EvenQ[k], k = k+1];
   (k*y)^3/n - (n-1)/2]];
Map[f, Range[100]] (* Jean-François Alcover, Jul 09 2024, after Robert Israel *)

Formula

From Robert Israel, Nov 15 2023: (Start)

If n is odd, then a(n) is the least positive integer of the form (k*A019555(n))^3/n - (n-1)/2 where k is an integer.

If n is even, then let v = A007814(n). If v == 1 (mod 3) then a(n) is the least positive integer of the form (k*A019555(n/2))^3/n - (n-1)/2 where k an odd integer; otherwise, a(n) = 0. (End)

Extensions

More terms from David Wasserman, Apr 02 2005