A016754 -id:A016754 - cp's OEIS frontend

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).

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

A253673 Indices of centered triangular numbers (A005448) that are also centered octagonal numbers (A016754).

Original entry on oeis.org

1, 16, 65, 1520, 6321, 148896, 619345, 14590240, 60689441, 1429694576, 5946945825, 140095478160, 582740001361, 13727927165056, 57102573187505, 1345196766697280, 5595469432374081, 131815555209168336, 548298901799472385, 12916579213731799600

Offset: 1

Colin Barker, Jan 08 2015

Also positive integers x in the solutions to 3*x^2 - 8*y^2 - 3*x + 8*y = 0, the corresponding values of y being A253674.

Also indices of centered square numbers (A001844) that are also octagonal numbers (A000567). - Colin Barker, Feb 10 2015

			16 is in the sequence because the 16th centered triangular number is 361, which is also the 10th centered octagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A005448, A016754, A253674, A253675.

Cf. A000567, A001844, A254895, A254896.

LinearRecurrence[{1,98,-98,-1,1},{1,16,65,1520,6321},20] (* Harvey P. Dale, Aug 07 2023 *)

Vec(x*(3*x-1)*(5*x^2+18*x+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))

a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).

G.f.: x*(3*x-1)*(5*x^2+18*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).

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

A253410 Indices of centered pentagonal numbers (A005891) which are also centered octagonal numbers (A016754).

Original entry on oeis.org

1, 96, 817, 137712, 1177393, 198579888, 1697799169, 286352060064, 2448225223585, 412919472031680, 3530339074609681, 595429592317621776, 5090746497361935697, 858609059202538568592, 7340852918856836664673, 1238113667940468298287168, 10585504818245061108522049

Offset: 1

Colin Barker, Dec 31 2014

Also positive integers x in the solutions to 5*x^2 - 8*y^2 - 5*x + 8*y = 0, the corresponding values of y being A253411.

			96 is in the sequence because the 96th centered pentagonal number is 22801, which is also the 76th centered octagonal number.

Colin Barker, Table of n, a(n) for n = 1..633
Index entries for linear recurrences with constant coefficients, signature (1,1442,-1442,-1,1).

Cf. A005891, A016754, A253411, A253579.

LinearRecurrence[{1,1442,-1442,-1,1},{1,96,817,137712,1177393},20] (* Harvey P. Dale, Jul 12 2021 *)

Vec(x*(95*x^3+721*x^2-95*x-1)/((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))

a(n) = a(n-1) + 1442*a(n-2) - 1442*a(n-3) - a(n-4) + a(n-5).

G.f.: x*(95*x^3 + 721*x^2 - 95*x - 1) / ((x-1)*(x^2 - 38*x + 1)*(x^2 + 38*x + 1)).

A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217).

Original entry on oeis.org

1, 18, 595, 20196, 686053, 23305590, 791703991, 26894630088, 913625718985, 31036379815386, 1054323288004123, 35815955412324780, 1216688160731038381, 41331581509442980158, 1404057083160330286975, 47696609245941786776976, 1620280657278860420130193

Offset: 1

Colin Barker, Jan 16 2015

Also positive integers y in the solutions to x^2 - 8*y^2 + x + 8*y - 2 = 0, the corresponding values of x being A008843.

Also the indices of centered octagonal numbers (A016754) which are also hexagonal numbers (A000384). Also positive numbers y in the solutions to 4x^2-8y^2-2x+8y-2=0. - Colin Barker, Jan 25 2015

			18 is in the sequence because the 18th centered octagonal number is 1225, which is also the 49th triangular number.
18 is in the sequence because the 18th centered octagonal number 1225 is also the 25th hexagonal number. - _Colin Barker_, Jan 25 2015

Colin Barker, Table of n, a(n) for n = 1..654
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A008843, A016754, A046177.

Vec(x*(17*x-1)/((x-1)*(x^2-34*x+1)) + O(x^100))

a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
G.f.: x*(17*x-1) / ((x-1)*(x^2-34*x+1)).
a(n) = sqrt((-2-(17-12*sqrt(2))^n-(17+12*sqrt(2))^n)*(2-(17-12*sqrt(2))^(1+n)-(17+12*sqrt(2))^(1+n)))/(8*sqrt(2)). - Gerry Martens, Jun 04 2015

A248205 Indices of centered octagonal numbers (A016754) that are also pentagonal numbers (A000326).

Original entry on oeis.org

1, 50, 4851, 475300, 46574501, 4563825750, 447208348951, 43821854371400, 4294094520048201, 420777441110352250, 41231895134294472251, 4040304945719747928300, 395908652785401002501101, 38795007668023578497179550, 3801514842813525291721094751

Offset: 1

Colin Barker, Jan 11 2015

Positive integers y in the solutions to 3*x^2 - 8*y^2 - x + 8*y - 2 = 0, the corresponding values of x being A046172.

			50 is in the sequence because the 50th centered octagonal number is 9801, which is also the 81st pentagonal number.

Colin Barker, Table of n, a(n) for n = 1..503
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A000290, A000326, A046172, A036353.

I:=[1,50,4851]; [n le 3 select I[n] else 99*Self(n-1)-99*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jun 13 2015

LinearRecurrence[{99, -99, 1}, {1, 50, 4851}, 20] (* Vincenzo Librandi, Jun 13 2015 *)

Vec(x*(49*x-1)/((x-1)*(x^2-98*x+1)) + O(x^100))

a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
G.f.: x*(49*x-1) / ((x-1)*(x^2 - 98*x + 1)).
a(n) = (1/2+1/48*(49+20*sqrt(6))^(-n)*(-12-5*sqrt(6)+(-12+5*sqrt(6))*(49+20*sqrt(6))^(2*n))). - Colin Barker, Mar 03 2016

A253411 Indices of centered octagonal numbers (A016754) which are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 76, 646, 108871, 930811, 156991186, 1342228096, 226381180621, 1935491982901, 326441505463576, 2790978097114426, 470728424497295251, 4024588480547018671, 678790061683594287646, 5803453797970703808436, 978814798219318465489561, 8368576352085274344745321

Offset: 1

Colin Barker, Dec 31 2014

Also positive integers y in the solutions to 5*x^2 - 8*y^2 - 5*x + 8*y = 0, the corresponding values of x being A253410.

			76 is in the sequence because the 76th centered octagonal number is 22801, which is also the 96th centered pentagonal number.

Colin Barker, Table of n, a(n) for n = 1..633
Index entries for linear recurrences with constant coefficients, signature (1,1442,-1442,-1,1).

Cf. A005891, A016754, A253410, A253579.

LinearRecurrence[{1,1442,-1442,-1,1},{1,76,646,108871,930811},20] (* Harvey P. Dale, Feb 04 2016 *)

Vec(-x*(x^4+75*x^3-872*x^2+75*x+1)/((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))

a(n) = a(n-1) + 1442*a(n-2) - 1442*a(n-3) - a(n-4) + a(n-5).

G.f.: -x*(x^4 + 75*x^3 - 872*x^2 + 75*x + 1) / ((x-1)*(x^2 - 38*x + 1)*(x^2 + 38*x + 1)).

A253446 Indices of centered heptagonal numbers (A069099) which are also centered octagonal numbers (A016754).

Original entry on oeis.org

1, 16, 465, 13920, 417121, 12499696, 374573745, 11224712640, 336366805441, 10079779450576, 302057016711825, 9051630721904160, 271246864640412961, 8128354308490484656, 243579382390074126705, 7299253117393733316480, 218734014139421925367681

Offset: 1

Colin Barker, Jan 01 2015

Also positive integers x in the solutions to 7*x^2 - 8*y^2 - 7*x + 8*y = 0, the corresponding values of y being A253447.

			16 is in the sequence because the 16th centered heptagonal number is 841, which is also the 15th centered octagonal number.

Colin Barker, Table of n, a(n) for n = 1..678
Index entries for linear recurrences with constant coefficients, signature (31,-31,1).

Cf. A016754, A069099, A253447, A253514.

LinearRecurrence[{31,-31,1},{1,16,465},20] (* Harvey P. Dale, Oct 04 2023 *)

Vec(x*(15*x-1)/((x-1)*(x^2-30*x+1)) + O(x^100))

a(n) = 31*a(n-1)-31*a(n-2)+a(n-3).
G.f.: x*(15*x-1) / ((x-1)*(x^2-30*x+1)).
a(n) = sqrt((-2-(15-4*sqrt(14))^n-(15+4*sqrt(14))^n)*(2-(15-4*sqrt(14))^(1+n)-(15+4*sqrt(14))^(1+n)))/(4*sqrt(7)).  - Gerry Martens, Jun 04 2015

A253447 Indices of centered octagonal numbers (A016754) which are also centered heptagonal numbers (A069099).

Original entry on oeis.org

1, 15, 435, 13021, 390181, 11692395, 350381655, 10499757241, 314642335561, 9428770309575, 282548466951675, 8467025238240661, 253728208680268141, 7603379235169803555, 227847648846413838495, 6827826086157245351281, 204606934935870946699921

Offset: 1

Colin Barker, Jan 01 2015

Also positive integers y in the solutions to 7*x^2 - 8*y^2 - 7*x + 8*y = 0, the corresponding values of x being A253446.

			15 is in the sequence because the 15th centered octagonal number is 841, which is also the 16th centered heptagonal number.

Colin Barker, Table of n, a(n) for n = 1..678
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (31,-31,1).

Cf. A016754, A069099, A253446, A253514.

Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-30*x+1)) + O(x^100))

a(n) = 31*a(n-1)-31*a(n-2)+a(n-3).
G.f.: -x*(x^2-16*x+1) / ((x-1)*(x^2-30*x+1)).
a(n) = (8+(4+sqrt(14))*(15+4*sqrt(14))^(-n)-(-4+sqrt(14))*(15+4*sqrt(14))^n)/16. - Colin Barker, Mar 03 2016

cp's OEIS Frontend

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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A253673 Indices of centered triangular numbers (A005448) that are also centered octagonal numbers (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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A253410 Indices of centered pentagonal numbers (A005891) which are also centered octagonal numbers (A016754).

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A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217).

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A248205 Indices of centered octagonal numbers (A016754) that are also pentagonal numbers (A000326).

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