A240970 -id:A240970 - cp's OEIS frontend

A253778 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).

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

Offset: 1

Vladimir Pletser, Jan 12 2015

A253778 is a subset of A240970.
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.

			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.

Vladimir Pletser, Table of n, a(n) for n = 1..10000
K. S. Brown's Mathpages, Sum of Consecutive Nth Powers Equals an Nth Power
Vladimir Pletser, File Triplets (M,b,c) for M=m^3
Ben Vitale, Sum of Cubes Equals a Cube

Cf. A240970, A253779, A253780.

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:

a(n) = (m-1)(m^2 (m-2)-4(m+1))/6 where m = A001651(n).
Conjectures from Colin Barker, Jan 13 2015: (Start)
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)

A253779 Numbers c whose cubes are equal to the sum of m^3 consecutive cubes for m^3 not divisible by 3 (A118719).

Original entry on oeis.org

0, 6, 180, 540, 2856, 5544, 16830, 27060, 62244, 90090, 175440, 237456, 413820, 534660, 860706, 1074744, 1630200, 1983150, 2872044, 3422580, 4776480, 5597856, 7579110, 8760780, 11565756, 13214994, 17077320, 19320840, 24514644, 27500220, 34343370, 38241456, 47098800

Offset: 1

Vladimir Pletser, Jan 12 2015

Numbers c 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 are no nontrivial solutions 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.

			For n=1, b(1)= 0 and a(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
For n=2, b(2)=-2 and a(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, b(3)=6 and a(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, [where in this File, M is the number of term, a the first term and c the square root of the sum].

Vladimir Pletser, Table of n, a(n) for n = 1..10000
K. S. Brown's Mathpages, Sum of Consecutive Nth Powers Equals an Nth Power
Vladimir Pletser, File Triplets (M,b,c) for M=m^3
Ben Vitale, Sum of Cubes Equals a Cube

Cf. A240970, A253778, A253780.

restart: for n from 1 to 15000 do m:=n: if(modp(m,3)>0) then c:=m*(m^2-1)*(m^2+2))/6: print (c): fi: od:

a(n) = m(m^2-1)(m^2+2)/6 where m = A001651(n).
Conjectures from Colin Barker, Jan 13 2015: (Start)
a(n) = (81*n^5 - 270*n^4 + 396*n^3 - 312*n^2 + 96*n) / 64 for n even.
a(n) = (81*n^5 - 135*n^4 + 126*n^3 - 66*n^2 - 15*n + 9) / 64 for n odd.
G.f.: 6*x^2*(x^8+29*x^7+55*x^6+241*x^5+158*x^4+241*x^3+55*x^2+29*x+1) / ((x-1)^6*(x+1)^5).
(End)

A253780 Cubes c^3 that are equal to the sum of m^3 consecutive cubes starting at b^3 with b (A253778) for m^3 not divisible by 3 (A118719).

Original entry on oeis.org

0, 216, 5832000, 157464000, 23295638016, 170400029184, 4767078987000, 19814511816000, 241152896222784, 731189187729000, 5399901725184000, 13389040129314816, 70865430394968000, 152838610998696000, 637623759116775816, 1241409566336822784, 4332341335608000000

Offset: 1

Vladimir Pletser, Jan 12 2015

Consider the set of all nontrivial solutions of the equation b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^3 for integers b, M and c, with M equal to a cube not divisible by 3 (A118719). This sequence gives the values of c^3.
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 a=c^3.
There are no nontrivial solutions 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, c(n) = m(m^2-1)(m^2+2)/6.

			For n=1, b(1)=0, c(1)=0 and a(1)=C^3=0 for M(1)=1=A118719(n+1) = 1^3 = (A001651(n))^3.
For n=2, b(2)=-2, c(2)=6 and a(2)=c^3=216 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, b(3)=6, c(3)=180 and a(3)=c^3=5832000 for M(3)=64= A118719(n+1) = 4^3 = (A001651(n))^3.
See "File Triplets (M,b,c) for M=m^3" link.

Vladimir Pletser, Table of n, a(n) for n = 1..10000
K. S. Brown's Mathpages, Sum of Consecutive Nth Powers Equals an Nth Power
Vladimir Pletser, File Triplets (M,b,c) for M=m^3
Ben Vitale, Sum of Cubes Equals a Cube

Cf. A240970, A253778, A253779.

restart: for n from 1 to 15000 do m:=n: if(modp(m,3)>0) then a:=(m*(m^2-1)*(m^2+2))/6)^3: print (a): fi: od:

a253780[n_] := (# (#^2 - 1) (#^2 + 2)/6)^3 & /@ Select[Range@ n, Mod[#, 3] != 0 &]; a253780[25] (* Michael De Vlieger, Jan 19 2015 *)

a(n) = m=floor((3*n-1)/2); (m*(m^2-1)*(m^2+2)/6)^3 \\ Colin Barker, Jan 14 2015

a(n) = (m(m^2-1)(m^2+2)/6)^3 where m = A001651(n).

Empirical g.f.: 216*x^2*(x^28 +26999*x^27 +701985*x^26 +106716191*x^25 +670508953*x^24 +19676938356*x^23 +59522167700*x^22 +716736365044*x^21 +1294915592031*x^20 +8429482190425*x^19 +9962263692743*x^18 +39936619145457*x^17 +32146645170615*x^16 +84954433749528*x^15 +47129019463944*x^14 +84954433749528*x^13 +32146645170615*x^12 +39936619145457*x^11 +9962263692743*x^10 +8429482190425*x^9 +1294915592031*x^8 +716736365044*x^7 +59522167700*x^6 +19676938356*x^5 +670508953*x^4 +106716191*x^3 +701985*x^2 +26999*x +1) / ((x-1)^16*(x +1)^15). - Colin Barker, Jan 14 2015

A250304 Four-column array read by rows: T(n,k) = the coefficient of x^k in the expanded polynomial x^3 + (x+1)^3 + ... + (x+n-1)^3, for 0 <= k <= 3.

Original entry on oeis.org

0, 0, 0, 1, 1, 3, 3, 2, 9, 15, 9, 3, 36, 42, 18, 4, 100, 90, 30, 5, 225, 165, 45, 6, 441, 273, 63, 7, 784, 420, 84, 8, 1296, 612, 108, 9, 2025, 855, 135, 10, 3025, 1155, 165, 11, 4356, 1518, 198, 12, 6084, 1950, 234, 13, 8281, 2457, 273, 14, 11025, 3045, 315, 15, 14400, 3720, 360, 16

Offset: 1

Derek Orr, Jan 15 2015

A240970 solves the Diophantine equation: k^3 + (k+1)^3 + ... + (k+n-1)^3 = y^3. This array gives the coefficients of the left hand side for specified n.

			Array starts:
n = 1:   0,   0,  0, 1;
n = 2:   1,   3,  3, 2;
n = 3:   9,  15,  9, 3;
n = 4:  36,  42, 18, 4;
n = 5: 100,  90, 30, 5;
n = 6: 225, 165, 45, 6;
n = 7: 441, 273, 63, 7;
n = 8: 784, 420, 84, 8;
...

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,5,0,0,0,-10,0,0,0,10,0,0,0,-5,0,0,0,1).

Cf. A000537, A059270, A045943, A240970.

for(n=1,50,for(k=0,3,print1(polcoeff(sum(i=1,n,(x+i-1)^3),k),", ")))

concat([0,0,0], Vec(x^4*(x^12-3*x^11+3*x^10-x^9-3*x^8+6*x^7-4*x^5+3*x^4-3*x^3-3*x^2-x-1) / ((x-1)^5*(x+1)^5*(x^2+1)^5) + O(x^100))) \\ Colin Barker, Jun 02 2015

a(4*k+1) = A000537(k), for k >= 0.
a(4*k+2) = A059270(k), for k >= 0.
a(4*k+3) = A045943(k), for k >= 0.
a(4*k) = k, for k >= 1.
a(n) = ((2*n^4+40*n^3+188*n^2-24*n-558-(2*n^4-24*n^3+188*n^2-792*n-558)*(-1)^n+(2*n^4-20*n^3-130*n^2+772*n+377)*(-1)^((2*n-1+(-1)^n)/4)-(2*n^4+40*n^3-196*n^2-280*n+594)*(-1)^((6*n-1+(-1)^n)/4)-(4*n^3+66*n^2-228*n-217)*(-1)^((10*n-1+(-1)^n)/4)))/8192. - Luce ETIENNE, May 22 2015
G.f.: x^4*(x^12-3*x^11+3*x^10-x^9-3*x^8+6*x^7-4*x^5+3*x^4-3*x^3-3*x^2-x-1) / ((x-1)^5*(x+1)^5*(x^2+1)^5). - Colin Barker, Jun 02 2015

cp's OEIS Frontend

A253778 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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A253779 Numbers c whose cubes are equal to the sum of m^3 consecutive cubes for m^3 not divisible by 3 (A118719).

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A253780 Cubes c^3 that are equal to the sum of m^3 consecutive cubes starting at b^3 with b (A253778) for m^3 not divisible by 3 (A118719).

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A250304 Four-column array read by rows: T(n,k) = the coefficient of x^k in the expanded polynomial x^3 + (x+1)^3 + ... + (x+n-1)^3, for 0 <= k <= 3.

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