A126203 -id:A126203 - cp's OEIS frontend

A128931 Erroneous version of A126203.

2, 3, 98, 120

Offset: 1

dead

L. E. Dickson, History of the Theory of Numbers, Volume 2, Chapter 21, page 587.
E. Lucas, Recherches sur l'analyse indeterminee, Moulins, 1873, 92. Extract from Bull. Soc. d'Emulation du Departement de l'Allier, 12, 1873, 532.

A126200 Numbers n such that n^2 is a sum of consecutive cubes larger than 1.

8, 27, 64, 125, 204, 216, 312, 315, 323, 343, 504, 512, 588, 720, 729, 1000, 1331, 1728, 2079, 2170, 2197, 2744, 2940, 3375, 4096, 4472, 4913, 4914, 5187, 5832, 5880, 5984, 6630, 6859, 7497, 8000, 8721, 8778, 9261, 9360, 10296, 10648, 10695, 11024, 12167, 13104

Offset: 1

Zak Seidov, Mar 11 2007

nonn

Note that all triangular numbers A000217(i) have squares A000217(i)^2=A000537(i), which are sums of consecutive cubes starting with 1. But such decompositions are not counted here. - R. J. Mathar, Nov 02 2007
Also, the positive integers n such that n^2 is the difference of squares of two positive triangular numbers. - Max Alekseyev, Jul 27 2014
Included all cubes > 1.

			204^2=23^3+24^3+25^3, 312^2=14^3+15^3+...24^3+25^3;
n^2=sum[i^3, (i=i1...i2)]; {n, i1=initial index of cube, i2=final index of cube}: {8, 4, 4}, {27, 9, 9}, {64, 16, 16}, {125, 25, 25}, {204, 23, 25}, {216, 36, 36}, {312, 14, 25}, {315, 25, 29}, {323, 9, 25}, {343, 49, 49}, {504, 28, 35}, {512, 64, 64}, {588, 14, 34}, {720, 25, 39}, {729, 81, 81}, {1000, 100, 100}, {1331, 121, 121}, {1728, 144, 144}, {2079, 33, 65}, {2170, 96, 100}, {2197, 169, 169}, {2744, 196, 196}.

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A126203.

mc=335241; cb=vector(mc); for(i=2, mc, cb[i]=i^3); v=vector(1000); mx=194104539^2; n=0; for(i=2, mc, s=0; for(j=i, mc, s=s+cb[j]; if(s>mx, next(2)); if(issquare(s,&sr), n++; v[n]=sr))); v=vecsort(v); for(i=1, 1000, write("b126200.txt", i " " v[i])) /* Donovan Johnson, Feb 02 2013 */

Better definition from Dean Hickerson, Dec 02 2007

Many terms were missing - thanks to Donovan Johnson for catching this. (Feb 02 2013)

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

Original entry on oeis.org

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

cp's OEIS Frontend

A128931 Erroneous version of A126203.

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A126200 Numbers n such that n^2 is a sum of consecutive cubes larger than 1.

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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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