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A054208
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Consider all integer triples (i,j,k), j >= k > 0, with i^3 = binomial(j+2,3) + binomial(k+2,3), ordered by increasing i; sequence gives i values.
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3
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2, 11, 45, 65, 109, 280, 644, 1079, 1309, 3180, 3355, 6864, 8284, 9700, 10681, 10856, 16775, 17094, 33506, 35650, 50435, 63196, 84330, 105731, 125220, 145200, 167986, 220545, 503690, 678730, 1046629, 1065426
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OFFSET
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0,1
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COMMENTS
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Another term (not necessarily the next one) is 2630045. - R. J. Mathar, Apr 07 2023
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LINKS
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EXAMPLE
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2^3 = 8 = binomial(2+2,3) + binomial(2+2,3).
11^3 = 1331 = binomial(19+2,3) + binomial(3,3).
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MAPLE
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# is x(x+1)(x+2)/6= A000292(x)=n solvable?
# return true if yes.
isA000292 := proc(n)
local x;
if n = 0 then
return true ;
end if;
x := iroot(6*n, 3) ;
# newton algorithm
while true do
x := x-round((x*(x+1)*(x+2)-6*n)/(3*x^2+6*x+2)) ;
return false ;
return false ;
return true;
end if;
end do:
end proc:
isA054208 := proc(n)
local c, i, ti, tj;
c := n^3 ;
for i from 1 do
if ti > c/2 then
return false ;
end if ;
tj := c-ti ;
if isA000292(tj) then
return true ;
end if;
end do:
end proc:
for n from 1 do
if isA054208(n) then
print(n)
end if;
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MATHEMATICA
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(* Range of j values is merely empirical *) jmin[k_] := Floor[Max[k, 1.86*k - 20000]]; jmax[k_] := Ceiling[1.86*k + 16000]; jmax[3005] = 10^5; ii = Reap[ Do[ Do[i = (Binomial[j+2, 3] + Binomial[k+2, 3])^(1/3); If[IntegerQ[i], Print[{i, j, k}]; Sow[i]; Break[]], {j, jmin[k], jmax[k]}], {k, 1, 40000}] ][[2, 1]]; A054208 = Union[ii] (* Jean-François Alcover, Dec 12 2012 *)
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CROSSREFS
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KEYWORD
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nice,nonn,more
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AUTHOR
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Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 31 2000
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EXTENSIONS
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STATUS
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approved
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