cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A108369 Coefficients of x/(1+3*x+3*x^2-x^3).

Original entry on oeis.org

0, 1, -3, 6, -8, 3, 21, -80, 180, -279, 217, 366, -2028, 5203, -9159, 9840, 3160, -48159, 144837, -286874, 377952, -128397, -1035539, 3869760, -8631060, 13248361, -9982143, -18429714, 98483932, -250144797, 436552881, -460740320, -177582480, 2351521281
Offset: 0

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Author

Michael Somos, Jun 01 2005

Keywords

References

  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 562.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[x/(1+3x+3x^2-x^3),{x,0,40}],x] (* or *) LinearRecurrence[{-3,-3,1},{0,1,-3},40] (* Harvey P. Dale, Jul 30 2024 *)
  • PARI
    {a(n)=if(n>=0, polcoeff(x/(1+3*x+3*x^2-x^3)+x*O(x^n),n), n=-1-n; polcoeff(x/(1-3*x-3*x^2-x^3)+x*O(x^n),n))}

Formula

x=a(n), z=a(-n-2), y=a(n)+a(n+1), t=a(-1-n)+a(-n-2) is a solution to 2*(x^3+z^3) = y^3+t^3.
G.f.: x/(1+3*x+3*x^2-x^3).
a(n) = -3*a(n-1) - 3*a(n-2) + a(n-3).
a(-1-n) = A108368(n).
a(n+1) = (-1)^n * Sum_{k=0..floor(n/3)} (-2)^k * binomial(n+2,3*k+2). - Seiichi Manyama, Aug 05 2024

A377314 a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

1, 1, 5, 19, 73, 281, 1081, 4159, 16001, 61561, 236845, 911219, 3505753, 13487761, 51891761, 199644319, 768096001, 2955112721, 11369270485, 43741245619, 168286661033, 647452990441, 2490960200041, 9583526232479, 36870912288001, 141854275761481
Offset: 0

Views

Author

Clark Kimberling, Oct 26 2024

Keywords

Comments

See A377109 for a guide to related sequences.

Examples

			((1 + 2^(1/3) + 2^(2/3)))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 19.
		

Crossrefs

Cf. A377109, A377117, A377315, A108368 (coefficients of 2^(2/3)).

Programs

  • Mathematica
    (* Program 1 generates sequences A377314-A377315 and A108368. *)
    tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];
    Take[tbl, 6]
    u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
       Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
    {s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
    s1  (* Peter J. C. Moses, Oct 16 2024 *)
    (* Program 2 generates (a(n)) for n>=1. *)
    LinearRecurrence[{3,3,1}, {1, 1, 5}, 15]

Formula

a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=1, a(1)=1, a(3)=5. [Corrected by Jianing Song, Oct 31 2024]
G.f.: (-1 + 2 x + x^2)/(-1 + 3 x + 3 x^2 + x^3).

A377315 a(n) = coefficient of 2^(1/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.

Original entry on oeis.org

0, 1, 4, 15, 58, 223, 858, 3301, 12700, 48861, 187984, 723235, 2782518, 10705243, 41186518, 158457801, 609638200, 2345474521, 9023795964, 34717449655, 133569211378, 513883779063, 1977076420978, 7606449811501, 29264462476500, 112589813284981, 433169277095944
Offset: 0

Views

Author

Clark Kimberling, Oct 26 2024

Keywords

Comments

See A377109 for a guide to related sequences.

Examples

			(1 + 2^(1/3) + 2^(2/3))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 15.
		

Crossrefs

Cf. A377109, A377117, A377314, A108368 (coefficients of 2^(2/3)).

Programs

  • Mathematica
    (* Program 1 generates sequences A377314-A377315 and A108368. *)
    tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];
    Take[tbl, 6]
    u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
       Map[({#1, #1 /. ^ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
    {s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
    s2  (* Peter J. C. Moses, Oct 16 2024 *)
    (* Program 2 generates (a(n)) for n>=1. *)
    LinearRecurrence[{3,3,1}, {0, 1, 4}, 15]

Formula

a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=0, a(1)=1, a(3)=4.
G.f.: -((x (1 + x))/(-1 + 3 x + 3 x^2 + x^3)).

A374454 Expansion of o.g.f. 1/(1 - 4*x - 6*x^2 - 4*x^3 - x^4).

Original entry on oeis.org

1, 4, 22, 116, 613, 3240, 17124, 90504, 478333, 2528092, 13361506, 70618412, 373233385, 1972618128, 10425707976, 55102092624, 291226324249, 1539193302772, 8134965235054, 42995028146468, 227237903531533, 1201000837247928, 6347545848001836, 33548135057767512
Offset: 0

Views

Author

Enrique Navarrete, Jul 08 2024

Keywords

Comments

a(n) is the number of generalized compositions of n using parts of size at most 4 where there are binomial(4,i) types of i (see example).
The coefficients of 1/(1 - C(k,1)*x - C(k,2)*x^2 - C(k,3)*x^3 - ... - C(k,k)*x^k) give the number of generalized compositions of n using parts of size at most k where there are binomial(k,i) types of i.
Related sequences that count the number of generalized compositions of n using parts of size at most k where there are binomial(k,i) types of i are A108368(n+1), A000129(n+1), and A000012(n) for k = 3, 2, 1, respectively.

Examples

			The following table gives the type of composition, the number of such compositions, and the total number of compositions of n = 6 using parts of size at most 4 where there are binomial(4,i) types of i (ie. 4 types of 1, 6 types of 2, 4 types of 3 and 1 type of 4):
    Type                     Number              Total
    4+2                        2                    12
    3+3                        1                    16
    4+1+1                      3                    48
    3+2+1                      6                   576
    2+2+2                      1                   216
    3+1+1+1                    4                  1024
    2+2+1+1                    6                  3456
    2+1+1+1+1                  5                  7680
    1+1+1+1+1+1                1                  4096,
    adding to a(6) = 17124.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/(1-4*x-6*x^2-4*x^3-x^4),{x,0,23}],x] (* Stefano Spezia, Jul 09 2024 *)

Formula

a(n) = 4*a(n-1) + 6*a(n-2) + 4*a(n-3) + a(n-4), n=>4.
a(n) = Sum_{k>=0} (1/2)^(k+1) * binomial(4*k,n). - Seiichi Manyama, Aug 03 2024

A374455 Expansion of o.g.f. 1/(1 - 5*x - 10*x^2 - 10*x^3 - 5*x^4 - x^5).

Original entry on oeis.org

1, 5, 35, 235, 1580, 10626, 71460, 480570, 3231845, 21734235, 146163251, 982951365, 6610371480, 44454906580, 298960311840, 2010515259876, 13520763292345, 90927457083265, 611489327404315, 4112280377388895, 27655184063541876, 185981775414350150, 1250731895575163300
Offset: 0

Views

Author

Enrique Navarrete, Jul 08 2024

Keywords

Comments

a(n) is the number of generalized compositions of n using parts of size at most 5 where there are binomial(5,i) types of i (see example).
The coefficients of 1/(1 - C(k,1)*x - C(k,2)*x^2 - C(k,3)*x^3 - ... - C(k,k)*x^k) give the number of generalized compositions of n using parts of size at most k where there are binomial(k,i) types of i.

Examples

			The following table gives the type of composition, the number of such compositions, and the total number of compositions of n = 5 using parts of size at most 5 where there are binomial(5,i) types of i (ie. 5 types of 1, 10 types of 2, 10 types of 3, 5 types of 4, and 1 type of 5):
    Type                     Number              Total
    5                          1                     1
    4+1                        2                    50
    3+2                        2                   200
    3+1+1                      3                   750
    2+2+1                      3                  1500
    2+1+1+1                    4                  5000
    1+1+1+1+1+1                1                  3125,
    adding to a(5)=10626.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/(1-5*x-10*x^2-10*x^3-5*x^4-x^5),{x,0,22}],x] (* Stefano Spezia, Jul 09 2024 *)

Formula

a(n) = 5*a(n-1) + 10*a(n-2) + 10*a(n-3) + 5*a(n-4) + a(n-5), n=>5.
a(n) = Sum_{k>=0} (1/2)^(k+1) * binomial(5*k,n). - Seiichi Manyama, Aug 03 2024

Extensions

a(20) corrected by Georg Fischer, Oct 28 2024
Showing 1-5 of 5 results.