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A253778
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Numbers b that are the first of m^3 consecutive cubes whose sum is a cube c^3, where m^3 is not divisible by 3 (A118719).
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4
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0, -2, 6, 34, 213, 406, 1134, 1735, 3606, 4966, 8790, 11368, 18171, 22534, 33558, 40381, 57084, 67150, 91206, 105406, 138705, 158038, 202686, 228259, 286578, 319606, 394134, 435940, 529431, 581446, 696870, 760633, 901176, 978334, 1147398, 1239706, 1440909, 1550230
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OFFSET
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1,2
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COMMENTS
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Numbers b such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^3 has nontrivial solutions over the integers for M equal to a cube not divisible by 3 (A118719).
If M is a cube not divisible by 3, there always exists at least one nontrivial solution for the sum of M consecutive cubes starting from b^3 and equaling a cube c^3.
There is no nontrivial solution for M=m^3 if m=0(mod 3).
For n>=1, for integers m(n) =A001651(n), all nontrivial solutions for M(n)= m^3=A118719(n+1) are b(n) =(m-1)(m^2 (m-2)-4(m+1))/6 and c(n)= m(m^2-1)(m^2+2)/6.
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LINKS
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FORMULA
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a(n) = (m-1)(m^2 (m-2)-4(m+1))/6 where m = A001651(n).
a(n) = (54*n^4 - 252*n^3 + 312*n^2 - 144*n + 64) / 64 for n even.
a(n) = (54*n^4 - 180*n^3 + 96*n^2 - 12*n + 42) / 64 for n odd.
G.f.: -x^2*(x^7+5*x^6+60*x^5+69*x^4+147*x^3+36*x^2+8*x-2) / ((x-1)^5*(x+1)^4).
(End)
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EXAMPLE
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For n=1, a(1)= 0 and c(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
For n=2, a(2)=-2 and c(2)=6 for M(2)=8= A118719(n+1) = 2^3= (A001651(n))^3 , which is Euler relation: (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 6^3.
For n=3, a(3)=6 and c(3)=180 for M(3)=64= A118719(n+1) = 4^3= (A001651(n))^3.
See "File Triplets (M,a,c) for M=m^3" link.
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MAPLE
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restart: for n from 1 to 15000 do m:=n: if(modp(m, 3)>0) then a:=(m-1)*(m^2*(m-2)-4*(m+1))/6: print (a): fi: od:
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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