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 1-10 of 10 results.

A007297 Number of connected graphs on n labeled nodes on a circle with straight-line edges that don't cross.

Original entry on oeis.org

1, 1, 4, 23, 156, 1162, 9192, 75819, 644908, 5616182, 49826712, 448771622, 4092553752, 37714212564, 350658882768, 3285490743987, 30989950019532, 294031964658430, 2804331954047160, 26870823304476690, 258548658860327880
Offset: 1

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Author

Keywords

Comments

Apart from the initial 1, reversion of g.f. for A162395 (squares with signs): see A263843.

Examples

			G.f. = x*(1 + x + 4*x^2 + 23*x^3 + 156*x^4 + 1162*x^5 + 9192*x^6 + 75819*x^7 + ...).
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A162395, A000290. 4th row of A107111. Row sums of A089434.
See A263843 for a variant.
Cf. A000108 (non-crossing set partitions), A001006, A001187, A054726 (non-crossing graphs), A054921, A099947, A194560, A293510, A323818, A324167, A324169, A324173.

Programs

  • Maple
    A007297:=proc(n) if n = 1 then 1 else add(binomial(3*n - 3, n + j)*binomial(j - 1, j - n + 1), j = n - 1 .. 2*n - 3)/(n - 1); fi; end;
  • Mathematica
    CoefficientList[ InverseSeries[ Series[(x-x^2)/(1+x)^3, {x, 0, 20}], x], x] // Rest (* From Jean-François Alcover, May 19 2011, after PARI prog. *)
    Table[Binomial[3n, 2n+1] Hypergeometric2F1[1-n, n, 2n+2, -1]/n, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 25 2015 *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse((x-x^2)/(1+x)^3+O(x^(n+2))),n+1)) /* Ralf Stephan */

Formula

Apart from initial term, g.f. is the series reversion of (x-x^2)/(1+x)^3 (A162395). See A263843. - Vladimir Kruchinin, Feb 08 2013
G.f.: (g-z)/z, where g=-1/3+(2/3)*sqrt(1+9z)*sin((1/3)*arcsin((2+27z+54z^2)/2/(1+9*z)^(3/2))). - Emeric Deutsch, Dec 02 2002
a(n) = (1/n)*Sum_{k=0..n} binomial(3n, n-k-1)*binomial(n+k-1, k). - Paul Barry, May 11 2005
a(n) = 4^(n-1)*(Gamma(3*n/2-1)/Gamma(n/2+1)/Gamma(n) -Gamma((3*n-1)/2)/ Gamma( (n+1)/2)/Gamma(n+1)). - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
a(n) = 4^n * binomial(3*n/2, n/2) / (9*n-6) - 4^(n-1) * binomial(3*(n-1)/2, (n-1)/2 ) / n. - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
D-finite with recurrence: n*(n-1)*(3*n-4)*a(n) +36*(n-1)*a(n-1) -12*(3*n-8)*(3*n-1)*(3*n-7)*a(n-2)=0. - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
a(n) = (1/n)*Sum_{k=0..n} C(3n, k)*C(2n-k-2, n-1). - Paul Barry, Sep 27 2005
a(n) ~ (2-sqrt(3)) * 6^n * 3^(n/2) / (sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Mar 17 2014
a(n) = binomial(3*n,2*n+1)*hypergeom([1-n,n], [2*n+2], -1)/n. - Vladimir Reshetnikov, Oct 25 2015
a(n) = 2*A078531(n) - A085614(n+1). - Vladimir Reshetnikov, Apr 24 2016

Extensions

Better description from Philippe Flajolet, Apr 20 2000
More terms from James Sellers, Aug 21 2000
Definition revised and initial a(1)=1 added by N. J. A. Sloane, Nov 05 2015 at the suggestion of Axel Boldt. Some of the formulas may now need to be adjusted slightly.

A365754 Expansion of (1/x) * Series_Reversion( x*(1-x)/(1+x)^4 ).

Original entry on oeis.org

1, 5, 36, 305, 2833, 27916, 286632, 3033513, 32858595, 362515725, 4059475368, 46021411644, 527163783916, 6092053249160, 70939443268112, 831558454663449, 9804617762941095, 116201796106426543, 1383557994261012100, 16541672701743657545, 198510770031798279825
Offset: 0

Views

Author

Seiichi Manyama, Sep 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+k, k)*binomial(4*(n+1), n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(4*(n+1),n-k).
a(n) = (1/(n+1)) * [x^n] ( (1+x)^4 / (1-x) )^(n+1). - Seiichi Manyama, Feb 17 2024

A162395 a(n) = -(-1)^n * n^2.

Original entry on oeis.org

1, -4, 9, -16, 25, -36, 49, -64, 81, -100, 121, -144, 169, -196, 225, -256, 289, -324, 361, -400, 441, -484, 529, -576, 625, -676, 729, -784, 841, -900, 961, -1024, 1089, -1156, 1225, -1296, 1369, -1444, 1521, -1600, 1681, -1764, 1849, -1936, 2025, -2116, 2209, -2304, 2401, -2500
Offset: 1

Views

Author

Michael Somos, Jul 02 2009

Keywords

Comments

This sequence is the denominator of (Pi^2)/12 = 1/1-1/4+1/9-1/16+1/25-1/36+... - Mohammad K. Azarian, Dec 29 2011
Also, circulant determinant of [1,2,...,n,n-1,...,1], i.e., determinant of the (2n-1) X (2n-1) matrix which has this as first row (and also first column), where row k+1 is obtained by cyclically shifting row k one place to the left. - M. F. Hasler, Dec 17 2016

Examples

			G.f. = x - 4*x^2 + 9*x^3 - 16*x^4 + 25*x^5 - 36*x^6 + 49*x^7 - 64*x^8 + 81*x^9 + ...
		

Crossrefs

For the reversion of this sequence see A263843 (and also A007297).

Programs

  • Magma
    [(-1)^(n+1) * n^2: n in [1..60]]; // Vincenzo Librandi, Feb 15 2013
  • Mathematica
    Table[(-1)^(n+1) * n^2, {n, 60}] (* Vincenzo Librandi, Feb 15 2013 *)
  • PARI
    {a(n) = -(-1)^n * n^2};
    

Formula

Euler transform of length 2 sequence [-4, 3].
a(n) is multiplicative with a(2^e) = -(4^e) if e>0, a(p^e) = (p^2)^e if p>2.
G.f.: x * (1 - x) / (1 + x)^3.
E.g.f.: exp(-x) * (x - x^2).
a(n) = a(-n) = -(-1)^n * A000290(n) for all n in Z.
Sum_{n>=1} 1/a(n) = Pi^2/12 (A072691). - Amiram Eldar, Dec 10 2022
Dirichlet g.f.: zeta(s-2)*(1-2^(3-s)) = DirichletEta(s-2). - Amiram Eldar, Jan 07 2023

A365755 Expansion of (1/x) * Series_Reversion( x*(1-x)/(1+x)^5 ).

Original entry on oeis.org

1, 6, 52, 530, 5919, 70098, 864784, 10994490, 143042020, 1895316632, 25487708844, 346976558318, 4772478619146, 66222166440780, 925880434336320, 13030945427540170, 184467676431001644, 2624828100099166536, 37521220349342729680, 538573138719587026440
Offset: 0

Views

Author

Seiichi Manyama, Sep 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+k, k)*binomial(5*(n+1), n-k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(5*(n+1),n-k).

A365816 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 3 * A(x)).

Original entry on oeis.org

0, 1, 6, 57, 658, 8442, 115692, 1658505, 24565530, 372999198, 5774883348, 90821581578, 1446901409268, 23301338376916, 378711707274072, 6203898306232233, 102329366764727658, 1698047225583890550, 28327664136201303300, 474821679792884860590, 7992739387298462213340
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Comments

Reversion of g.f. for hexagonal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^3/(1 - 3 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 3 x)/(1 + x)^3, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 n, n - k - 1] 3^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * 3^k for n > 0.
a(n) ~ 6^(3*n + 1/2) / (sqrt((481 + 133*sqrt(13))*Pi) * n^(3/2) * (13*sqrt(13) - 35)^n). - Vaclav Kotesovec, Sep 26 2023

A365817 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 4 * A(x)).

Original entry on oeis.org

0, 1, 7, 80, 1119, 17437, 290532, 5066364, 91311055, 1687341227, 31797227631, 608727899936, 11805599569092, 231454163924700, 4579765707561240, 91340133073920420, 1834295500622405295, 37059418988408887015, 752741444501505866325, 15362331852042084534240, 314860558967057266779495
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Comments

Reversion of g.f. for heptagonal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^3/(1 - 4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 4 x)/(1 + x)^3, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 n, n - k - 1] 4^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * 4^k for n > 0.
a(n) ~ 5 * (81 + 21*sqrt(21))^n / (sqrt((427 + 93*sqrt(21))*Pi) * n^(3/2) * 2^(3*n + 3/2)). - Vaclav Kotesovec, Sep 26 2023

A365818 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 5 * A(x)).

Original entry on oeis.org

0, 1, 8, 107, 1760, 32298, 634128, 13034247, 276943568, 6033834950, 134069957840, 3026476515790, 69213144181888, 1600157697995092, 37337615574348960, 878166685063548639, 20797051344280763184, 495509950454603339310, 11869278747340342255440, 285669061791469915886250, 6904850429493240677872320
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Comments

Reversion of g.f. for octagonal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^3/(1 - 5 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 5 x)/(1 + x)^3, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 n, n - k - 1] 5^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * 5^k for n > 0.
a(n) ~ 3^(3/2) * 2^(n - 1/2) * (154 + 31*sqrt(31))^n / (sqrt((2821 + 506*sqrt(31))*Pi) * n^(3/2) * 5^(2*n)). - Vaclav Kotesovec, Sep 26 2023

A366012 a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(n*(n+1),n-k).

Original entry on oeis.org

1, 2, 13, 156, 2833, 70098, 2214280, 85464984, 3906724321, 206648387550, 12425282899588, 837384222603448, 62539219710804627, 5127758187193514824, 457986530357734020432, 44263628968974498793648, 4602969726808566383149761, 512486177498084438210961270, 60827938291895363867587959628
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[1/(n + 1) Sum[Binomial[n + k, n] Binomial[n (n + 1), n - k], {k, 0, n}], {n, 0, 18}]
    Table[SeriesCoefficient[(1/x) InverseSeries[Series[x (1 - x)/(1 + x)^n, {x, 0, n + 1}], x], {x, 0, n}], {n, 0, 18}]

Formula

a(n) = [x^n] (1/x) * Series_Reversion( x * (1 - x) / (1 + x)^n ).
a(n) ~ exp(n + 3/2) * n^(n - 3/2) / sqrt(2*Pi). - Vaclav Kotesovec, Sep 26 2023

A366203 a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * (n-3)^k.

Original entry on oeis.org

1, 2, 12, 156, 3507, 115692, 5066364, 276943568, 18152243967, 1387267590540, 121106707350928, 11889022355301672, 1296359140925188212, 155440199716271334648, 20327081449263918542412, 2879054747404226046119448, 439060192463001381367975215, 71727764882350305085962745740
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 04 2023

Keywords

Comments

a(n) is the coefficient of x^n in expansion of series reversion of g.f. for n-gonal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    Unprotect[Power]; 0^0 = 1; Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 n, n - k - 1] (n - 3)^k, {k, 0, n - 1}], {n, 1, 18}]
    Table[Binomial[3 n, n - 1] Hypergeometric2F1[1 - n, n, 2 (n + 1), 3 - n]/n, {n, 1, 18}]
    Table[SeriesCoefficient[InverseSeries[Series[x (1 - (n - 3) x)/(1 + x)^3, {x, 0, n}], x], {x, 0, n}], {n, 1, 18}]

Formula

a(n) = [x^n] Series_Reversion( x * (1 - (n - 3) * x) / (1 + x)^3 ).

A291536 Expansion of the series reversion of x*(1 + 4*x + x^2)/(1 - x)^4.

Original entry on oeis.org

1, -8, 101, -1544, 26190, -474144, 8975229, -175492664, 3516970490, -71858843264, 1491301438354, -31349284476496, 666133734882748, -14284509655611840, 308734263333717021, -6718525508918998872, 147085140049822666626, -3237191565662618280384, 71584853778205231503750
Offset: 1

Views

Author

Ilya Gutkovskiy, Aug 25 2017

Keywords

Comments

Reversion of g.f. for the cubes (A000578).

Crossrefs

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 19}], x], x]]

Formula

G.f. A(x) satisfies: A(x)*(1 + 4*A(x) + A(x)^2)/(1 - A(x))^4 = x.
a(n) ~ -(-1)^n * (5*sqrt(6) - 12) * 2^(3*n-2) * 3^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 11 2017
D-finite with recurrence 7*n*(n-1)*(n+1)*a(n) +4*n*(n-1)*(58*n-83)*a(n-1) -36*(n-1)*(16*n^2-80*n+155)*a(n-2) +432*(-96*n^3+720*n^2-1794*n+1495)*a(n-3) +6912*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4)=0. - R. J. Mathar, Jan 25 2023
Showing 1-10 of 10 results.