A163392 -id:A163392 - cp's OEIS frontend

A253679 Numbers that begin a run of an odd number of consecutive integers whose cubes sum to a square.

23, 118, 333, 716, 1315, 2178, 3353, 4888, 6831, 9230, 12133, 15588, 19643, 24346, 29745, 35888, 42823, 50598, 59261, 68860, 79443, 91058, 103753, 117576, 132575, 148798, 166293, 185108, 205291, 226890, 249953, 274528, 300663, 328406, 357805, 388908, 421763, 456418, 492921, 531320, 571663, 613998, 658373, 704836, 753435, 804218, 857233, 912528, 970151, 1030150, 1092573, 1157468

Offset: 1

Vladimir Pletser, Jan 08 2015

Numbers k such that k^3 + (k+1)^3 + ... + (k+M-1)^3 = c^2 has nontrivial solutions over the integers where M is an odd positive integer.
To every odd positive integer M corresponds a sum of M consecutive cubes starting at a(n) having at least one nontrivial solution. For n >= 1, M(n) = (2n+1) (A005408), a(n) = M^3 - (3M-1)/2 = (2n+1)^3 - (3n+1) and c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680).
The trivial solutions with M < 1 and k < 2 are not considered here.
Stroeker stated that all odd values of M yield a solution to k^3 + (k+1)^3 + ... + (k+M-1)^3 = c^2. This was further demonstrated by Pletser.

			For n=1, M(n)=3, a(n)=23, c(n)=204.
See "File Triplets (M,a,c) for M=(2n+1)" link.

Vladimir Pletser, Table of n, a(n) for n = 1..50000
Vladimir Pletser, File Triplets (M,a,c) for M=(2n+1)
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253680, A253681.

for n from 1 to 50 do a:=(2*n+1)^3-(3*n+1): print (a); end do:

a253679[n_] := (2 # + 1)^3 - (3 # + 1) & /@ Range@ n; a253679[52] (* Michael De Vlieger, Jan 10 2015 *)

Vec(-x*(x^2-26*x-23)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jan 09 2015

a(n) = (2n+1)^3 - (3n+1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Colin Barker, Jan 09 2015
G.f.: -x*(x^2-26*x-23) / (x-1)^4. - Colin Barker, Jan 09 2015

A253680 Numbers c(n) whose square are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c(n)^2, starting at b(n) (A253679).

Original entry on oeis.org

204, 2940, 16296, 57960, 159060, 368004, 754320, 1412496, 2465820, 4070220, 6418104, 9742200, 14319396, 20474580, 28584480, 39081504, 52457580, 69267996, 90135240, 115752840, 146889204, 184391460, 229189296, 282298800, 344826300, 417972204, 503034840

Offset: 1

Vladimir Pletser, Jan 08 2015

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M being an odd positive integer.
To every odd positive integer M corresponds a sum of M consecutive cubed integers starting at b^3 having at least one nontrivial solution. For n>=1, M(n)=(2n+1) (A005408), b(n) = M^3 -(3M-1)/2 = (2n+1)^3 - (3n+1) (A253679) and c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680).
The trivial solutions with M < 1 and b < 2 are not considered here.
Stroeker stated that all odd values of M yield a solution to  b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2. This was further demonstrated by Pletser.

			For n=1, M(n)=3, b(n)=23, c(n)=204.
See "File Triplets (M,b,c) for M=(2n+1)" link.

Vladimir Pletser, Table of n, a(n) for n = 1..50000
Vladimir Pletser, File Triplets (M,b,c) for M=(2n+1)
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681.

[2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 1 to 50000 do c:=2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): print (c); end do:

f[n_] := 2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1); Array[f, 36] (* Michael De Vlieger, Jan 10 2015 *)

Vec(12*x*(x+1)*(17*x^2+126*x+17)/(x-1)^6 + O(x^100)) \\ Colin Barker, Jan 09 2015

c(n) = 2n(n+1)*(2n+1)*(8n*(n+1)+1).

G.f.: 12*x*(x+1)*(17*x^2+126*x+17) / (x-1)^6. - Colin Barker, Jan 09 2015

A253681 Integer squares c^2 that are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 starting at b(n) (A253679).

Original entry on oeis.org

41616, 8643600, 265559616, 3359361600, 25300083600, 135426944016, 568998662400, 1995144950016, 6080268272400, 16566690848400, 41192058954816, 94910460840000, 205045101804816, 419208426176400, 817072496870400, 1527363954902016, 2751797699456400, 4798055269856016

Offset: 1

Vladimir Pletser, Jan 08 2015

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M being an odd positive integer.
To every odd positive integer M corresponds a sum of M consecutive cubed integers starting at b^3 having at least one nontrivial solution. For n>=1, M(n)=(2n+1) (A005408), b(n) = M^3 - (3M-1)/2 = (2n+1)^3 - (3n+1) (A253679), c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680) and this sequence a(n) = c(n)^2.
The trivial solutions with M < 1 and b < 2 are not considered here.
Stroeker stated that all odd values of M yield a solution to b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2. This was further demonstrated by Pletser.

			For n=1, M(1)=3, b(1)=23, c(1)=204, a(1)=c^2=41616.
See "File Triplets (M,b,c) for M=(2n+1)" 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..50000
Vladimir Pletser, File Triplets (M,b,c) for M=(2n+1)
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680.

[(2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1))^2: n in [1..20]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 1 to 50000 do a:=(2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1))^2: print (a); end do:

f[n_] := (2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1))^2; Array[f, 21] (* Michael De Vlieger, Jan 10 2015 *)

Vec(-144*x*(289*x^8 +56846*x^7 +1199784*x^6 +6296786*x^5 +10697390*x^4 +6296786*x^3 +1199784*x^2 +56846*x +289) / (x -1)^11 + O(x^100)) \\ Colin Barker, Jan 09 2015

a(n) = (2n(n+1)*(2n+1)*(8n*(n+1)+1))^2.

G.f.: -144*x*(289*x^8 + 56846*x^7 + 1199784*x^6 + 6296786*x^5 + 10697390*x^4 + 6296786*x^3 + 1199784*x^2 + 56846*x + 289) / (x -1)^11. - Colin Barker, Jan 09 2015

A253707 Numbers M(n) which are the number of terms in the sums of consecutive cubed integers equaling a squared integer, b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

Original entry on oeis.org

17, 98, 291, 644, 1205, 2022, 3143, 4616, 6489, 8810, 11627, 14988, 18941, 23534, 28815, 34832, 41633, 49266, 57779, 67220, 77637, 89078, 101591, 115224, 130025, 146042, 163323, 181916, 201869, 223230, 246047, 270368, 296241, 323714, 352835, 383652, 416213

Offset: 1

Vladimir Pletser, Jan 09 2015

Numbers M(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for b(n) being an odd squared integer (A016754).
To every odd squared integer b corresponds a sum of a consecutive cubed integers starting at b having at least one nontrivial solution. For n>=1, b(n)= (2n+1)^2 (A016754), M(n) = (sqrt(b)-1) (2b-1)/2 = n(8n(n+1)+1) (this sequence),  and c(n)= (b-1)(4b^2-1)/8 = (n (n+1)/2)(4(2n+1)^4-1) (A253708).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=1, b(n)=9, M(n)=17, c(n)=323 (see File Triplets link).

Vladimir Pletser, Table of n, a(n) for n = 1..50000
Vladimir Pletser, File Triplets (M,b,c) for a=(2n+1)^2
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253680, A253681, A253708, A253709.

[n*(8*n*(n+1)+1): n in [1..40]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 1 to 50000 do a:= n*(8*n*(n+1)+1): print (a); end do:

f[n_] := n*(8 n (n + 1) + 1); Array[f, 52] (* Michael De Vlieger, Jan 10 2015 *)
LinearRecurrence[{4,-6,4,-1},{17,98,291,644},40] (* Harvey P. Dale, Jul 31 2018 *)

Vec(x*(x^2+30*x+17)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jan 10 2015

a(n) = n(8n(n+1)+1).
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Colin Barker, Jan 10 2015
G.f.: x*(x^2+30*x+17) / (x-1)^4. - Colin Barker, Jan 10 2015

A253708 Numbers c(n) whose squares are equal to the sums of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

Original entry on oeis.org

323, 7497, 57618, 262430, 878445, 2399103, 5669972, 12026988, 23457735, 42785765, 73877958, 121874922, 193444433, 297057915, 443289960, 645140888, 918382347, 1281925953, 1758214970, 2373639030, 3158971893, 4149832247, 5387167548, 6917760900, 8794760975

Offset: 1

Vladimir Pletser, Jan 09 2015

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for b being an odd squared integer (A016754).
To every odd squared integer b corresponds a sum of M consecutive cubed integers starting at b^3 equaling a squared integer and having at least one nontrivial solution. For n>=1, b(n) = (2n+1)^2 (A016754), M(n) = (sqrt(b)-1)(2b-1)/2 = n(8n(n+1)+1) (A253707), and c(n)= (b-1)(4b^2-1)/8 = (n(n+1)/2)(4(2n+1)^4-1) (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=1, b(n)=9, M(n)=17, a(n)=323.
See "File Triplets (M,b,c) for a=(2n+1)^2" link.

Vladimir Pletser, Table of n, a(n) for n = 1..50000
Vladimir Pletser, File Triplets (M,b,c) for a=(2n+1)^2
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709.

[(n*(n+1)/2)*(4*(2*n+1)^4-1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 1 to 50000 do a:= (n*(n+1)/2)(4*(2*n+1)^4-1): print (a); end do:

f[n_] := (n (n + 1)/2) (4 (2 n + 1)^4 - 1); Array[f, 33] (* Michael De Vlieger, Jan 10 2015 *)

Vec(-x*(323*x^4+5236*x^3+11922*x^2+5236*x+323)/(x-1)^7 + O(x^100)) \\ Colin Barker, Jan 14 2015

a(n) = (n(n+1)/2)(4(2n+1)^4-1).

G.f.: -x*(323*x^4+5236*x^3+11922*x^2+5236*x+323) / (x-1)^7. - Colin Barker, Jan 14 2015

A253709 Integer squares c^2 that are equal to the sums of M (A253707) consecutive cubed integers equaling a squared integer, b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

Original entry on oeis.org

104329, 56205009, 3319833924, 68869504900, 771665618025, 5755695204609, 32148582480784, 144648440352144, 550265331330225, 1830621686635225, 5457952678249764, 14853496612506084, 37420748658691489, 88243404864147225, 196505988636801600, 416206765369428544, 843426135281228409, 1643334148974958209, 3091319880732100900, 5634162244739340900

Offset: 1

Vladimir Pletser, Jan 09 2015

Numbers a(n)=c^2 such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for b being an odd squared integer (A016754) and M (A253707).
To every odd squared integer b (A016754) corresponds a sum of M (A253707) consecutive cubed integers starting at b^3 having at least one nontrivial solution. For n>=1, b(n)= (2n+1)^2 (A016754), M(n) = (sqrt(b)-1)(2b-1)/2 = n(8n(n+1)+1) (A253707), c(n)= (b-1)(4b^2-1)/8 = (n(n+1)/2)(4(2n+1)^4-1) (A253708) and a(n)=c(n)^2 (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=1, b(1)=9, M(1)=17, c(1)=323, a(1)= 104329 (see File File Triplets (M,b,c) for a=(2n+1)^2 link).

Vladimir Pletser, Table of n, a(n) for n = 1..50000
Vladimir Pletser, File Triplets (M,b,c) for a=(2n+1)^2
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253707, A253708.

[((n*(n+1)/2)*(4*(2*n+1)^4-1))^2: n in [1..20]]; // Vincenzo Librandi, Jan 15 2015

restart: for n from 1 to 50000 do a:=((n*(n+1)/2)(4*(2*n+1)^4-1))^2: print (a); end do:

f[n_] := ((n (n + 1)/2) (4 (2 n + 1)^4 - 1))^2; Array[f, 20] (* Michael De Vlieger, Jan 10 2015 *)

Vec(-x*(104329*x^10 +54848732*x^9 +2597306469*x^8 +30065816496*x^7 +119309063058*x^6 +186443360232*x^5 +119309063058*x^4 +30065816496*x^3 +2597306469*x^2 +54848732*x +104329) / (x -1)^13 + O(x^100)) \\ Colin Barker, Jan 10 2015

a(n) = ((n(n+1)/2)(4(2n+1)^4-1))^2.

G.f.: -x*(104329*x^10 +54848732*x^9 +2597306469*x^8 +30065816496*x^7 +119309063058*x^6 +186443360232*x^5 +119309063058*x^4 +30065816496*x^3 +2597306469*x^2 +54848732*x +104329) / (x -1)^13. - Colin Barker, Jan 10 2015

A253724 Numbers c(n) whose squares are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Original entry on oeis.org

504, 8721, 65472, 312375, 1119528, 3293829, 8388096, 19131147, 39999000, 77947353, 143325504, 250991871, 421651272, 683434125, 1073737728, 1641349779, 2448874296, 3575480097, 5119992000, 7204344903, 9977420904, 13619289621, 18345871872, 24414046875

Offset: 2

Vladimir Pletser, Jan 10 2015

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M(n) being twice a squared integer (A001105) and b(n)=(A002593).
If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting at b^3 and equaling to a squared integer c^2. For n>=1, M(n)= 2n^2 (A001105), b(n) = M(M-1)/2 = n^2(2n^2 - 1) (A002593), and c(n)= sqrt(M/2) (M(M^2-1)/2)= n^3(4n^4 - 1) (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=2, M(n)=8, b(n)=28, c(n)=504.
See "File Triplets (M,b,c) for M=2n^2" link.

Vladimir Pletser, Table of n, a(n) for n = 2..50000
Vladimir Pletser, File Triplets (M,b,c) for M=2n^2
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709, A002593, A253725.

[n^3*(4*n^4 - 1): n in [2..30]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 2 to 50000 do a:= n^3*(4*n^4 - 1): print (a); end do:

f[n_] := n^3 (4 n^4 - 1); Rest@Array[f, 32] (* Michael De Vlieger, Jan 28 2015 *)

Vec(-3*x^2*(x^7-8*x^6+27*x^5-216*x^4-1521*x^3-3272*x^2-1563*x-168)/(x-1)^8 + O(x^100)) \\ Colin Barker, Jan 14 2015

a(n) = n^3(4n^4 - 1).

G.f.: -3*x^2*(x^7-8*x^6+27*x^5-216*x^4-1521*x^3-3272*x^2-1563*x-168) / (x-1)^8. - Colin Barker, Jan 14 2015

A253725 Integer squares c^2 that are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Original entry on oeis.org

254016, 76055841, 4286582784, 97578140625, 1253342942784, 10849309481241, 70360154505216, 366000785535609, 1599920001000000, 6075789839706609, 20542200096854016, 62996919308080641, 177789795179217984, 467082203214515625, 1152912708530601984

Offset: 2

Vladimir Pletser, Jan 10 2015

Numbers a(n)=c^2 such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers where M(n) is twice a squared integer (A001105) and b(n)=(A002593).
If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting at b^3 and equaling a squared integer c^2. For n>=1, M(n)= 2n^2 (A001105), b(n) = M(M-1)/2 = n^2(2n^2 - 1) (A002593), c(n)= sqrt(M/2) (M(M^2-1)/2)= n^3(4n^4 - 1) (A253724) and a(n)=c(n)^2 (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=2, M(n)=8, b(n)=28, c(n)=504, a(n)=c^2=254016.
See "File Triplets (M,b,c) for M=2n^2" link.

Vladimir Pletser, Table of n, a(n) for n = 2..50000
Vladimir Pletser, File Triplets (M,b,c) for M=2n^2
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253707, A253708, A002593, A253724.

[(n^3*(4*n^4-1))^2: n in [2..20]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 2 to 50000 do a:=(n^3*(4*n^4 - 1))^2: print (a); end do:

f[n_] := (n^3 (4 n^4 - 1))^2; Rest[f /@ Range@16] (* Michael De Vlieger, Jan 28 2015 *)
LinearRecurrence[{15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1},{254016,76055841,4286582784,97578140625,1253342942784,10849309481241,70360154505216,366000785535609,1599920001000000,6075789839706609,20542200096854016,62996919308080641,177789795179217984,467082203214515625,1152912708530601984},20] (* Harvey P. Dale, Feb 18 2023 *)

a(n) = (n^3(4n^4 - 1))^2.

G.f.: -9*x^2*(x^14 -15*x^13 +106*x^12 +27754*x^11 +8028759*x^10 +352487303*x^9 +4572193580*x^8 +22833696108*x^7 +49725383807*x^6 +49725372367*x^5 +22833705546*x^4 +4572187210*x^3 +352490761*x^2 +8027289*x +28224) / (x -1)^15. - Colin Barker, Jan 14 2015

A218979 Numbers n such that some sum of n consecutive positive cubes is a square.

Original entry on oeis.org

1, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 37, 39, 40, 41, 42, 43, 45, 47, 48, 49, 50, 51, 53, 54, 55, 57, 59, 60, 61, 63, 64, 65, 67, 69, 71, 72, 73, 75, 76, 77, 79, 81, 82, 83, 85, 87, 89, 91, 92, 93, 94, 95, 97, 98, 99

Offset: 1

Michel Marcus, Nov 08 2012

nonn

The trivial solutions with x = 0 and x = 1 are not considered here.
Numbers n such that x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 has nontrivial solutions over the integers.
The nontrivial solutions are found by solving Y^2 = X^3 + d(n)*X with d(n) = n^2*(n^2-1)/4 (A006011), Y = n*y and X = n*x + (1/2)*n*(n-1). [Corrected by Derek Orr, Aug 30 2014]
x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 can also be written as y^2 = n(x + (n-1)/2)(n(x + (n-1)/2) + x(x-1)). - Vladimir Pletser, Aug 30 2014
There are 892 triples (n,x,y), with n and x less than 10^5 (1 < n,x < 10^5), which are nontrivial solutions of x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 (note that (n,x,y) corresponds to (M,a,c) in A253679, A253680, A253681, A253707, A253708, A253709, A253724, A253725). - Vladimir Pletser, Jan 10 2015

			See "Examples of triples" link.

Michel Marcus, Examples of triples up to n=50
Vladimir Pletser, Triplets (n, x, y) with n,x less than 10^5
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, Fundamental solutions of the Pell equation X^2-(sigma^4-delta^4)Y^2=delta^4 for the first 45 solutions of the sums of consecutive cubed integers equalling integer squares, Research Gate, 2015. See Reference 19.
V. Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253681, A253707, A253708, A253709.

a(n)=for(x=2,10^7, /* note this limit only generates the terms in the data section */ X = n*x + (1/2)*n*(n-1); d=n^2*(n^2-1)/4;if(issquare(X^3+d*X),return(x)))
n=1;while(n<100,if(a(n),print1(n,", "));n++) \\ Derek Orr, Aug 30 2014

Name changed, a(1) = 1 prepended and a(39)-a(68) from Derek Orr, Aug 30 2014

More terms for 50Vladimir Pletser, Jan 10 2015

cp's OEIS Frontend

A253679 Numbers that begin a run of an odd number of consecutive integers whose cubes sum to a square.

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A253680 Numbers c(n) whose square are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c(n)^2, starting at b(n) (A253679).

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A253681 Integer squares c^2 that are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 starting at b(n) (A253679).

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A253707 Numbers M(n) which are the number of terms in the sums of consecutive cubed integers equaling a squared integer, b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

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A253708 Numbers c(n) whose squares are equal to the sums of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

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A253709 Integer squares c^2 that are equal to the sums of M (A253707) consecutive cubed integers equaling a squared integer, b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

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