A116108 -id:A116108 - cp's OEIS frontend

A116145 Squares that are equal to the sum of 5 consecutive cubes.

0, 100, 225, 99225, 4708900, 8643600

Offset: 1

Zak Seidov, Apr 14 2007

m^3+(m+1)^3+(m+2)^3+(m+3)^3+(m+4)^3=5*(2+m)*(10+4*m+m^2). Corresponding values of m are -2,0,1,25,96,118.

Using the theory of elliptic curves one can show that there are no other terms. - Jaap Spies, May 28 2007

Select[Table[n^3 +(n+1)^3 +(n+2)^3 +(n+3)^3 +(n+4)^3, {n, -2, 1000}] ,IntegerQ[Sqrt[#]]&] (* James C. McMahon, Aug 18 2024 *)

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

A091761 a(n) = Pell(4n) / Pell(4).

Original entry on oeis.org

0, 1, 34, 1155, 39236, 1332869, 45278310, 1538129671, 52251130504, 1775000307465, 60297759323306, 2048348816684939, 69583562007964620, 2363792759454112141, 80299370259431848174, 2727814796061228725775, 92665403695822344828176, 3147895910861898495432209

Offset: 0

Paul Barry, Feb 06 2004

A000129(k*n)/A000129(k) = ((sqrt(2)-1)^k(-1)^k-(sqrt(2)+1)^k)((sqrt(2)-1)^(k*n)(-1)^(k*n)-(sqrt(2)+1)^(k*n))/((sqrt(2)-1)^(2k)+(sqrt(2)+1)^(2k)-2(-1)^k).
All squares of the form (3m-1)^3 + (3m)^3 + (3m+1)^3 (cf. A116108) are given for m = 24 b, where b is a square of this sequence. From Ribenboim & McDaniel, it follows there are no squares > 1 in this sequence. - M. F. Hasler, Jun 05 2007
A divisibility sequence, cf. R. K. Guy's post to the SeqFan list. - M. F. Hasler, Feb 05 2013
a(n) gives all nonnegative solutions of the Pell equation b(n)^2 - 32*(3*a(n))^2 = +1, together with b(n) = A056771(n). - Wolfdieter Lang, Mar 09 2019

M. F. Hasler, Table of n, a(n) for n = 0..99
R. K. Guy, A new sequence, post to the SeqFan list, Feb 05 2013.
Tanya Khovanova, Recursive Sequences
Paulo Ribenboim and Wayne L. McDaniel, The Square Terms in Lucas Sequences, Journal of Number Theory 58, 104-123 (1996).
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (34,-1).

A029547 is an essentially identical sequence, cf. formula.

Cf. A000129, A001109, A002965, A029547, A041085, A056771, A116108.

I:=[0,1]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // G. C. Greubel, Mar 11 2019

with (combinat):seq(fibonacci(4*n,2)/12, n=0..17); # Zerinvary Lajos, Apr 21 2008

LinearRecurrence[{34,-1}, {0,1}, 20] (* G. C. Greubel, Mar 11 2019 *)

A091761(n, x=[ -1,17],A=[17,72*4;1,17]) = vector(n,i,(x*=A)[1]) \\ M. F. Hasler, May 26 2007

A091761(n)=([34,1;-1,0]^(n-1))[1,1] \\ M. F. Hasler, Jun 05 2007

[lucas_number1(n,34,1) for n in range(0, 16)]# Zerinvary Lajos, Nov 07 2009

G.f.: x/(1-34*x+x^2).
a(n) = A000129(4n)/A000129(4).
a(n) = ((1+sqrt(2))^(4n) - (1-sqrt(2))^(4n))*sqrt(2)/48.
From M. F. Hasler, Jun 05 2007: (Start)
a(n) = n (mod 2^m) for any m >= 0.
a(n) = sinh(4*n*log(sqrt(2)+1))/(12*sqrt(2)).
a(n) = A[1,1], first element of the 2 X 2 matrix A = (34,1;-1,0)^(n-1). (End)
a(n) = 34*a(n-1) - a(n-2); a(0)=0, a(1)=1. - Philippe Deléham, Nov 03 2008
A029547(n) = a(n+1). - M. F. Hasler, Feb 05 2013
a(n) = sqrt((A056771(n)^2 - 1)/(32*9)), n >= 0. See the Pell remark above. - Wolfdieter Lang, Mar 11 2019
E.g.f.: exp(17*x)*sinh(12*sqrt(2)*x)/(12*sqrt(2)). - Stefano Spezia, Apr 16 2023
a(n) = A002965(8*n)/12. - Gerry Martens, Jul 14 2023

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

A116145 Squares that are equal to the sum of 5 consecutive cubes.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A091761 a(n) = Pell(4n) / Pell(4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Sage

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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).