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Search: a016754 -id:a016754
Displaying 1-10 of 292 results found. page 1 2 3 4 5 6 7 8 9 10 ... 30
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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). +20
6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
LINKS
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).
FORMULA
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
EXAMPLE
For n=1, b(n)=9, M(n)=17, c(n)=323 (see File Triplets link).
MAPLE
restart: for n from 1 to 50000 do a:= n*(8*n*(n+1)+1): print (a); end do:
MATHEMATICA
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 *)
PROG
(PARI) Vec(x*(x^2+30*x+17)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jan 10 2015
(Magma) [n*(8*n*(n+1)+1): n in [1..40]]; // Vincenzo Librandi, Feb 19 2015
CROSSREFS
Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253680, A253681, A253708, A253709.
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 09 2015
STATUS
approved
A253673 Indices of centered triangular numbers (A005448) that are also centered octagonal numbers (A016754). +20
5
1, 16, 65, 1520, 6321, 148896, 619345, 14590240, 60689441, 1429694576, 5946945825, 140095478160, 582740001361, 13727927165056, 57102573187505, 1345196766697280, 5595469432374081, 131815555209168336, 548298901799472385, 12916579213731799600 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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
LINKS
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).
FORMULA
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)).
EXAMPLE
16 is in the sequence because the 16th centered triangular number is 361, which is also the 10th centered octagonal number.
MATHEMATICA
LinearRecurrence[{1, 98, -98, -1, 1}, {1, 16, 65, 1520, 6321}, 20] (* Harvey P. Dale, Aug 07 2023 *)
PROG
(PARI) 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))
CROSSREFS
Cf. A005448, A016754, A253674, A253675.
Cf. A000567, A001844, A254895, A254896.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Jan 08 2015
STATUS
approved
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). +20
5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
LINKS
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).
FORMULA
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
EXAMPLE
For n=1, b(n)=9, M(n)=17, a(n)=323.
See File Triplets (M,b,c) for a=(2n+1)^2</a>" link.
MAPLE
restart: for n from 1 to 50000 do a:= (n*(n+1)/2)(4*(2*n+1)^4-1): print (a); end do:
MATHEMATICA
f[n_] := (n (n + 1)/2) (4 (2 n + 1)^4 - 1); Array[f, 33] (* Michael De Vlieger, Jan 10 2015 *)
PROG
(PARI) 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
(Magma) [(n*(n+1)/2)*(4*(2*n+1)^4-1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015
CROSSREFS
Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709.
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 09 2015
STATUS
approved
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). +20
5
104329, 56205009, 3319833924, 68869504900, 771665618025, 5755695204609, 32148582480784, 144648440352144, 550265331330225, 1830621686635225, 5457952678249764, 14853496612506084, 37420748658691489, 88243404864147225, 196505988636801600, 416206765369428544, 843426135281228409, 1643334148974958209, 3091319880732100900, 5634162244739340900 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
LINKS
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).
FORMULA
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
EXAMPLE
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).
MAPLE
restart: for n from 1 to 50000 do a:=((n*(n+1)/2)(4*(2*n+1)^4-1))^2: print (a); end do:
MATHEMATICA
f[n_] := ((n (n + 1)/2) (4 (2 n + 1)^4 - 1))^2; Array[f, 20] (* Michael De Vlieger, Jan 10 2015 *)
PROG
(PARI) 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
(Magma) [((n*(n+1)/2)*(4*(2*n+1)^4-1))^2: n in [1..20]]; // Vincenzo Librandi, Jan 15 2015
CROSSREFS
Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253707, A253708.
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 09 2015
STATUS
approved
A253410 Indices of centered pentagonal numbers (A005891) which are also centered octagonal numbers (A016754). +20
4
1, 96, 817, 137712, 1177393, 198579888, 1697799169, 286352060064, 2448225223585, 412919472031680, 3530339074609681, 595429592317621776, 5090746497361935697, 858609059202538568592, 7340852918856836664673, 1238113667940468298287168, 10585504818245061108522049 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
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).
FORMULA
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)).
EXAMPLE
96 is in the sequence because the 96th centered pentagonal number is 22801, which is also the 76th centered octagonal number.
MATHEMATICA
LinearRecurrence[{1, 1442, -1442, -1, 1}, {1, 96, 817, 137712, 1177393}, 20] (* Harvey P. Dale, Jul 12 2021 *)
PROG
(PARI) 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))
CROSSREFS
Cf. A005891, A016754, A253411, A253579.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 31 2014
STATUS
approved
A253826 Indices of centered octagonal numbers (A016754) which are also triangular numbers (A000217). +20
4
1, 18, 595, 20196, 686053, 23305590, 791703991, 26894630088, 913625718985, 31036379815386, 1054323288004123, 35815955412324780, 1216688160731038381, 41331581509442980158, 1404057083160330286975, 47696609245941786776976, 1620280657278860420130193 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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
LINKS
Colin Barker, Table of n, a(n) for n = 1..654
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).
FORMULA
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
EXAMPLE
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
PROG
(PARI) Vec(x*(17*x-1)/((x-1)*(x^2-34*x+1)) + O(x^100))
CROSSREFS
Cf. A000217, A008843, A016754, A046177.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Jan 16 2015
STATUS
approved
A248205 Indices of centered octagonal numbers (A016754) that are also pentagonal numbers (A000326). +20
3
1, 50, 4851, 475300, 46574501, 4563825750, 447208348951, 43821854371400, 4294094520048201, 420777441110352250, 41231895134294472251, 4040304945719747928300, 395908652785401002501101, 38795007668023578497179550, 3801514842813525291721094751 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
Colin Barker, Table of n, a(n) for n = 1..503
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).
FORMULA
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
EXAMPLE
50 is in the sequence because the 50th centered octagonal number is 9801, which is also the 81st pentagonal number.
MATHEMATICA
LinearRecurrence[{99, -99, 1}, {1, 50, 4851}, 20] (* Vincenzo Librandi, Jun 13 2015 *)
PROG
(PARI) Vec(x*(49*x-1)/((x-1)*(x^2-98*x+1)) + O(x^100))
(Magma) 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
CROSSREFS
Cf. A000290, A000326, A046172, A036353.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Jan 11 2015
STATUS
approved
A253411 Indices of centered octagonal numbers (A016754) which are also centered pentagonal numbers (A005891). +20
3
1, 76, 646, 108871, 930811, 156991186, 1342228096, 226381180621, 1935491982901, 326441505463576, 2790978097114426, 470728424497295251, 4024588480547018671, 678790061683594287646, 5803453797970703808436, 978814798219318465489561, 8368576352085274344745321 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
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).
FORMULA
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)).
EXAMPLE
76 is in the sequence because the 76th centered octagonal number is 22801, which is also the 96th centered pentagonal number.
MATHEMATICA
LinearRecurrence[{1, 1442, -1442, -1, 1}, {1, 76, 646, 108871, 930811}, 20] (* Harvey P. Dale, Feb 04 2016 *)
PROG
(PARI) 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))
CROSSREFS
Cf. A005891, A016754, A253410, A253579.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 31 2014
STATUS
approved
A253446 Indices of centered heptagonal numbers (A069099) which are also centered octagonal numbers (A016754). +20
3
1, 16, 465, 13920, 417121, 12499696, 374573745, 11224712640, 336366805441, 10079779450576, 302057016711825, 9051630721904160, 271246864640412961, 8128354308490484656, 243579382390074126705, 7299253117393733316480, 218734014139421925367681 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
Colin Barker, Table of n, a(n) for n = 1..678
Index entries for linear recurrences with constant coefficients, signature (31,-31,1).
FORMULA
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
EXAMPLE
16 is in the sequence because the 16th centered heptagonal number is 841, which is also the 15th centered octagonal number.
MATHEMATICA
LinearRecurrence[{31, -31, 1}, {1, 16, 465}, 20] (* Harvey P. Dale, Oct 04 2023 *)
PROG
(PARI) Vec(x*(15*x-1)/((x-1)*(x^2-30*x+1)) + O(x^100))
CROSSREFS
Cf. A016754, A069099, A253447, A253514.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Jan 01 2015
STATUS
approved
A253447 Indices of centered octagonal numbers (A016754) which are also centered heptagonal numbers (A069099). +20
3
1, 15, 435, 13021, 390181, 11692395, 350381655, 10499757241, 314642335561, 9428770309575, 282548466951675, 8467025238240661, 253728208680268141, 7603379235169803555, 227847648846413838495, 6827826086157245351281, 204606934935870946699921 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
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).
FORMULA
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
EXAMPLE
15 is in the sequence because the 15th centered octagonal number is 841, which is also the 16th centered heptagonal number.
PROG
(PARI) Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-30*x+1)) + O(x^100))
CROSSREFS
Cf. A016754, A069099, A253446, A253514.
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Jan 01 2015
STATUS
approved
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Last modified August 29 12:23 EDT 2024. Contains 375517 sequences. (Running on oeis4.)