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.

Previous Showing 31-40 of 49 results. Next

A386380 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} a(6*k) * a(n-1-6*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 15, 24, 34, 45, 57, 70, 154, 253, 368, 500, 650, 819, 1827, 3045, 4495, 6200, 8184, 10472, 23562, 39627, 59052, 82251, 109668, 141778, 320866, 543004, 814506, 1142295, 1533939, 1997688, 4540200, 7718340, 11633440, 16398200, 22137570
Offset: 0

Views

Author

Seiichi Manyama, Jul 20 2025

Keywords

Crossrefs

Programs

  • Maple
    A386380 := proc(n)
        option remember ;
        if n = 0 then
            1;
        else
            add(procname(6*k)*procname(n-1-6*k),k=0..floor((n-1)/6)) ;
        end if;
    end proc:
    seq(A386380(n),n=0..80) ; # R. J. Mathar, Jul 30 2025
  • PARI
    apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
    a(n) = apr(n\6, 7, n%6+1);

Formula

For k=0..5, a(6*n+k) = (k+1) * binomial(7*n+k+1,n)/(7*n+k+1).
G.f. A(x) satisfies A(x) = 1/(1 - x * Product_{k=0..5} A(w^k*x)), where w = exp(Pi*i/3).

A047755 a(n) = A047752(12n+5).

Original entry on oeis.org

0, 0, 1, 5, 26, 133, 708, 3861, 21604, 123266, 715221, 4206956, 25032840, 150413348, 911379384, 5562367173, 34164355848, 211015212580, 1309815397995, 8166460799805, 51120054233490, 321156223592865, 2024257417812240, 12797201858645100
Offset: 0

Views

Author

Keywords

Comments

Also A047754(4n+1). The proof is elementary and involves reducing both expressions via the various linear expansions with A001764, A047753, A047751 and A047749 which are found in these sequences. - R. J. Mathar, Oct 18 2008

Extensions

Edited by N. J. A. Sloane, Jan 17 2009 at the suggestion of R. J. Mathar

A082938 Number of solid 2-trees with 2n+1 edges.

Original entry on oeis.org

1, 1, 1, 2, 5, 13, 49, 201, 940, 4643, 24037, 127859, 696365, 3858759, 21704863, 123619126, 711787259, 4137614454, 24256010068, 143271593982, 852001881614, 5097719884665, 30670572676389, 185466705697057
Offset: 0

Views

Author

N. J. A. Sloane, May 26 2003

Keywords

Comments

Also, the number of noncrossing partitions up to rotation and reflection composed of n blocks of size 3. - Andrew Howroyd, May 03 2018

Crossrefs

Column k=3 of A303929.

Programs

  • Mathematica
    u[n_, k_, r_] := (r*Binomial[k*n + r, n]/(k*n + r));
    e[n_, k_] := Sum[ u[j, k, 1 + (n - 2*j)*k/2], {j, 0, n/2}]
    c[n_, k_] := If[n == 0, 1, (DivisorSum[n, EulerPhi[n/#]*Binomial[k*#, #] &] + DivisorSum[GCD[n-1, k], EulerPhi[#]*Binomial[n*k/#, (n-1)/#] &])/(k*n) - Binomial[k*n, n]/(n*(k - 1) + 1)];
    T[n_, k_] := (1/2)*(c[n, k] + If[n == 0, 1, If[OddQ[k], If[OddQ[n], 2*u[ Quotient[n, 2], k, (k + 1)/2], u[n/2, k, 1] + u[n/2 - 1, k, k]], e[n, k] + If[OddQ[n], u[Quotient[n, 2], k, k/2]]]/2]) /. Null -> 0;
    a[n_] := T[n, 3];
    Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 14 2018, after Andrew Howroyd and A303929 *)

Formula

a(n) = (A047749(n)+A054423(n))/2. - Vladeta Jovovic, Sep 11 2004
a(n) ~ 3^(3*n - 1/2) / (sqrt(Pi) * n^(5/2) * 2^(2*n + 3)). - Vaclav Kotesovec, Jun 01 2022

Extensions

More terms from Vladeta Jovovic, Sep 11 2004

A084081 Sum of lists created by n substitutions k -> Range[k+1,0,-2] starting with {0}, counting down from k+1 to 0 step -2.

Original entry on oeis.org

0, 1, 2, 5, 10, 24, 50, 121, 260, 637, 1400, 3468, 7752, 19380, 43890, 110561, 253000, 641355, 1480050, 3771885, 8765250, 22439040, 52451256, 134796060, 316663760, 816540124, 1926501200, 4982228488, 11798983280, 30593078076, 72690164850
Offset: 0

Views

Author

Wouter Meeussen, May 11 2003

Keywords

Comments

Lengths of lists is A047749.

Examples

			Lists {0}, {1}, {2, 0}, {3, 1, 1}, {4, 2, 0, 2, 0, 2, 0} sum to 0, 1, 2, 5, 10.
		

Crossrefs

Programs

  • Magma
    F:=Floor; B:=Binomial;
    function A084081(n)
      if (n mod 2) eq 0 then return 10*B(F((3*n+2)/2), F((n-2)/2))/(n+3);
      else return 2*(3*n+1)*B(F((3*n+5)/2), F((n+1)/2))/((n+3)*(3*n+5));
      end if; return A084081;
    end function;
    [A084081(n): n in [0..40]]; // G. C. Greubel, Oct 17 2022
    
  • Mathematica
    Plus@@@Flatten/@NestList[ # /. k_Integer :> Range[k+1, 0, -2]&, {0}, 8]
    A084081[n_]:= If[EvenQ[n], 10*Binomial[(3*n+2)/2, (n-2)/2]/(n+3), 2*(3*n + 1)*Binomial[(3*n+5)/2, (n+1)/2]/((n+3)*(3*n+5))];
    Table[A084081[n], {n, 40}] (* G. C. Greubel, Oct 17 2022 *)
  • SageMath
    def A084081(n):
        if (n%2==0): return 10*binomial(int((3*n+2)/2), int((n-2)/2))/(n+3)
        else: return 2*(3*n+1)*binomial(int((3*n+5)/2), int((n+1)/2))/((n+3)*(3*n+5))
    [A084081(n) for n in range(40)] # G. C. Greubel, Oct 17 2022

Formula

Equals A093951(n) - A047749(n).
From G. C. Greubel, Oct 17 2022: (Start)
a(2*n+1) = (3*n-1)*binomial[3*n+1, n]/((n+1)*(3*n+1)).
a(2*n) = 10*binomial(3*n+1, n-1)/(2*n+3). (End)

A093951 Sum of integers generated by n-1 substitutions, starting with 1, k -> k+1, k-1, .., 1.

Original entry on oeis.org

1, 2, 4, 8, 17, 36, 80, 176, 403, 910, 2128, 4896, 11628, 27132, 65208, 153824, 373175, 888030, 2170740, 5202600, 12797265, 30853680, 76292736, 184863168, 459162452, 1117370696, 2786017120, 6804995008, 17024247304, 41717833740, 104673837384
Offset: 1

Views

Author

Wouter Meeussen, Apr 18 2004

Keywords

Comments

Substitutions 1 -> {2}, 2 -> {3,1}, 3 -> {4,2}, 4 -> {5,3,1}, 5 -> {6,4,2}, 6 -> {7,5,3,1}, 7 -> {8,6,4,2}, etc. The function f(n) gives determinant of (I_{n} - x * A(n)) where I_{n} is the identity matrix and A(n) = 0 if j > i + 1 otherwise (i+j) mod 2, for i = 1, ..., n and j = 1, ..., n, and can be written in terms of Dickson polynomials as g(w) = x*D_(w-1)(1+x, x*(1+x)) + (1-2*x)*E_(w-1)(1+x, x*(1+x)). - Francisco Salinas (franciscodesalinas(AT)hotmail.com), Apr 13 2004
Count of integers is A047749.
Sum of integers with substitution starting from 0 is A084081.

Examples

			GF(12) = (1 + 2*x - 7*x^2 - 14*x^3 + 9*x^4 + 20*x^5 + 2*x^6 - 2*x^7 + 2*x^11)/(1 - 11*x^2 + 36*x^4 - 35*x^6 + 5*x^8) produces a(1) to a(12).
a(4)=8 since 4-1 = 3 substitutions on 1 produce 1 -> 2 -> 3+1 -> 4 + 2 + 2 = 8.
		

Crossrefs

Programs

  • Magma
    function A093951(n)
      if (n mod 2) eq 0 then return 8*Binomial(Floor(3*n/2), Floor((n-2)/2))/(n+2);
      else return 6*Binomial(Floor((3*n+1)/2), Floor((n-1)/2))/(n+2) - 2*Binomial(Floor((3*n-1)/2), Floor((n-1)/2))/(n+1);
      end if; return A093951;
    end function;
    [A093951(n): n in [1..40]]; // G. C. Greubel, Oct 17 2022
    
  • Mathematica
    Plus@@@Flatten/@NestList[ #/.k_Integer:>Range[k+1, 1, -2]&, {1}, 8];(*or for n>16 *); f[1]=1; f[2]=1-x^2; f[3]=1-2x^2; f[n_]:=f[n]=Expand[f[n-1]-x^2 f[n-3]]; g[1]=1; g[2]=1+2x; g[3]=1+2x+2x^2; g[n_]:=g[n]=Expand[g[n-1] -x^2 g[n-3]+2 x^(n-1)]; GF[n_]:=g[n]/f[n]; CoefficientList[Series[GF[36], {x, 0, 36}], x]
  • PARI
    {a(n)=if(n%2==0,4*binomial(3*n/2,n/2-1)/(n/2+1), 6*binomial(3*(n\2)+2, n\2)/(2*(n\2)+3) - binomial(3*(n\2)+1,n\2)/(n\2+1))} \\ Paul D. Hanna, Apr 24 2006
    
  • SageMath
    def A093951(n):
        if (n%2==0): return 8*binomial(3*n/2, (n-2)/2)/(n+2)
        else: return 6*binomial((3*n+1)/2, (n-1)/2)/(n+2) - 2*binomial((3*n-1)/2, (n-1)/2)/(n+1)
    [A093951(n) for n in range(1,40)] # G. C. Greubel, Oct 17 2022

Formula

a(n) = [x^n] GF(n) with GF(n) = g(n)/f(n) and f(1)=1, f(2)=1-x^2, f(3)=1-2*x^2, f(n) = f(n-1) - x^2*f(n-3) and g(1)=1, g(2)=1+2*x, g(3)=1+2*x+2*x^2, g(n) = g(n-1) - x^2*g(n-3) + 2*x^(n-1).
From Paul D. Hanna, Apr 24 2006: (Start)
a(2*n) = 4*binomial(3*n, n-1)/(n+1) = 2*A006629(n-1).
a(2*n+1) = 6*binomial(3*n+2, n)/(2*n+3) - binomial(3*n+1, n)/(n+1) = A056096(n+3). (End)

A219795 Sum of the absolute values of the antidiagonals of the triangle A135929(n) companion. See the comment.

Original entry on oeis.org

2, 2, 2, 2, 3, 3, 5, 7, 10, 11, 16, 23, 33, 44, 58, 81, 114, 158, 212, 293, 407, 565, 777, 1064, 1471, 2036, 2813, 3863, 5334, 7370, 10183, 14046, 19356, 26726, 36909, 50955, 70251, 96977, 133886, 184841, 255092
Offset: 0

Views

Author

Paul Curtz, Nov 28 2012

Keywords

Comments

The companion to A135929(n) is the triangle
2;
2, 0;
2, 0, 1;
2, 0, -1, 0;
2, 0, -3, 0, -1;
2, 0, -5, 0, 0, 0;
2, 0, -7, 0, 3, 0, 1;
2, 0, -9, 0, 8, 0, 1, 0;
(A192011(n) beginning with 2 instead of -1).
Consider a(1),a(5),a(10),a(14), that is, a(A193910(n) -1).
a(1)+a(4)-a(5) = 2, a(5)+a(8)-a(9) = 2, a(10)+a(13)-a(14) = 2, a(14)+a(17)-a(18) = 4, a(19)+a(22)-a(23) = 6, a(23)+a(26)-a(27) = 14, yields 2,2,2,4,6,14,24,60,... = 2*A047749(n) or 2, followed with A116637(n+1).

Examples

			a(0)=2, a(1)=2, a(2)=2+0, a(3)=2+0, a(4)=2+0+1, a(5)=2+0+1.
		

Programs

  • Maple
    A219795 := proc(n)
        if n=0 then
            2;
        else
            add(abs(A192011(n-k,k)),k=0..floor(n/2)) ;
        end if;
    end proc: # R. J. Mathar, Jan 06 2013

Formula

a(n) = sum abs ( [k=0..floor(n/2)] A192011(n-k,k) ), a(0)=2.

Extensions

a(24)-a(40) from Jean-Francois Alcover, Nov 28 2012

A124747 Inverse of number triangle A124744.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 2, 2, 2, 1, 0, 0, 3, 3, 3, 2, 1, 0, 0, 7, 7, 7, 5, 3, 1, 0, 0, 12, 12, 12, 9, 6, 3, 1, 0, 0, 30, 30, 30, 23, 16, 9, 4, 1, 0, 0, 55, 55, 55, 43, 31, 19, 10, 4, 1
Offset: 0

Views

Author

Paul Barry, Nov 06 2006

Keywords

Comments

Third column and row sums give A047749.

Examples

			Triangle begins
1,
0, 1,
0, 0, 1,
0, 0, 1, 1,
0, 0, 1, 1, 1,
0, 0, 2, 2, 2, 1,
0, 0, 3, 3, 3, 2, 1,
0, 0, 7, 7, 7, 5, 3, 1,
0, 0, 12, 12, 12, 9, 6, 3, 1
		

A124816 Product of Riordan array (1,x(1-x^2))^(-1) and number triangle T(n,k)=C(floor(k/2),n-k).

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 2, 1, 1, 0, 3, 3, 3, 2, 1, 0, 0, 7, 4, 5, 2, 1, 0, 12, 12, 12, 10, 6, 3, 1, 0, 0, 30, 18, 24, 12, 9, 3, 1, 0, 55, 55, 55, 50, 32, 22, 10, 4, 1, 0, 0, 143, 88, 121, 66, 57, 25, 14, 4, 1, 0, 273, 273, 273
Offset: 0

Views

Author

Paul Barry, Nov 08 2006

Keywords

Comments

Product of A124305 and A124788. Columns include aerated A001764,A047749(n+1),A124817,A084081. Row sums are A124818.

Examples

			Triangle begins
1,
0, 1,
0, 0, 1,
0, 1, 1, 1,
0, 0, 2, 1, 1,
0, 3, 3, 3, 2, 1,
0, 0, 7, 4, 5, 2, 1,
0, 12, 12, 12, 10, 6, 3, 1,
0, 0, 30, 18, 24, 12, 9, 3, 1,
0, 55, 55, 55, 50, 32, 22, 10, 4, 1
		

Crossrefs

Cf. A124790.

A192894 Number of symmetric 13-ary factorizations of the n-cycle (1,2...n).

Original entry on oeis.org

1, 1, 1, 7, 13, 112, 247, 2310, 5525, 53998, 135408, 1360289, 3518515, 36017352, 95223414, 988172368, 2655417765, 27844071255, 75769712590, 801012669457, 2201663313200, 23428926096576, 64924369564353, 694644371065372, 1938034271677595, 20829931845958872, 58448142042957576
Offset: 0

Views

Author

N. J. A. Sloane, Jul 12 2011

Keywords

Comments

The six sequences displayed in Table 1 of the Bousquet-Lamathe reference are A047749, A143546, A143547, A143554, A192893, A192894. From this one should be able to guess a g.f.

Crossrefs

Column k=13 of A369929 and k=14 of A370062.
Cf. A143049.

Formula

From Seiichi Manyama, Jul 07 2025: (Start)
G.f. A(x) satisfies A(x) = 1/( 1 - x*(A(x)*A(-x))^6 ).
G.f. A(x) satisfies A(x)*A(-x) = (A(x) + A(-x))/2 = G(x^2), where G(x) = 1 + x*G(x)^13.
a(0) = 1; a(n) = Sum_{x_1, x_2, ..., x_7>=0 and x_1+2*(x_2+x_3+...+x_7)=n-1} a(x_1) * Product_{k=2..7} a(2*x_k). (End)
a(0) = 1; a(n) = Sum_{x_1, x_2, ..., x_13>=0 and x_1+x_2+...+x_13=n-1} (-1)^(x_1+x_2+x_3+x_4+x_5+x_6) * Product_{k=1..13} a(x_k). - Seiichi Manyama, Jul 09 2025

Extensions

a(11) onwards from Andrew Howroyd, Jan 26 2024
a(0)=1 prepended by Seiichi Manyama, Jul 07 2025

A217138 G.f. A(x) satisfies A(x) = 1 + x*A(x)^2*(A(x) + A(-x))/2.

Original entry on oeis.org

1, 1, 2, 7, 22, 94, 340, 1579, 6118, 29746, 120060, 600934, 2492028, 12725756, 53798888, 278786739, 1195684230, 6265816042, 27175425004, 143671870034, 628705751828, 3347680236132, 14756641134872, 79039468217086, 350529497005532, 1886818634445044, 8410852483002200
Offset: 0

Views

Author

Paul D. Hanna, Sep 27 2012

Keywords

Comments

Compare to: G(x) = 1 + x*G(x)*(G(x) + G(-x))/2, which is the g.f. of A047749.
The radius of convergence r of g.f. A(x) is
r = 0.192450089729875254836382926833985818549200... with
A(r) = (3 - sqrt(9-6*sqrt(2)))/sqrt(2) = 1.614014407382354328773...
A(-r) = (sqrt(9+6*sqrt(2)) - 3)/sqrt(2) = 0.835475335400823769423...
where y = A(r) and y = A(-r) solves y^4 = 18*(1-y)^2.
In closed form, r = 1/(3*sqrt(3)). - Vaclav Kotesovec, Feb 17 2014

Examples

			A(x) = 1 + x + 2*x^2 + 7*x^3 + 22*x^4 + 94*x^5 + 340*x^6 + 1579*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 5*x^2 + 18*x^3 + 62*x^4 + 260*x^5 + 1005*x^6 + 4522*x^7 +...
(A(x) + A(-x))/2 = 1 + 2*x^2 + 22*x^4 + 340*x^6 + 6118*x^8 +...
		

Crossrefs

Cf. A047749.

Programs

  • PARI
    {a(n)=local(A=1+O(x^(n+1))); for(i=0, n, A=1+x*A^2*(A+subst(A, x, -x))/2); polcoeff(A, n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies: 2*(A(x)-1)^3 + x*(A(x)-1)*(2-A(x))*A(x)^3 - x^2*A(x)^6 = 0.
The formal inverse of g.f. A(x) is (x-1)*(2-x + sqrt(x^2 + 4*x - 4))/(2*x^3).
Recurrence: (n-1)*n*(n+1)*(2*n-1)*(2*n+1)*(198*n^5 - 1890*n^4 + 6915*n^3 - 12015*n^2 + 9832*n - 3060)*a(n) = 30*(n-1)*n*(2*n - 1)*(3*n - 7)*(36*n^4 - 246*n^3 + 567*n^2 - 468*n + 55)*a(n-1) + 3*(n-1)*(14256*n^9 - 193104*n^8 + 1111626*n^7 - 3526794*n^6 + 6652659*n^5 - 7420161*n^4 + 4436059*n^3 - 971921*n^2 - 90480*n - 23100)*a(n-2) - 90*(2*n - 1)*(3*n - 8)*(3*n - 7)*(3*n - 4)*(36*n^4 - 246*n^3 + 567*n^2 - 468*n + 55)*a(n-3) - 36*(n-3)*(3*n - 11)*(3*n - 10)*(3*n - 8)*(3*n - 7)*(198*n^5 - 900*n^4 + 1335*n^3 - 630*n^2 - 23*n - 20)*a(n-4). - Vaclav Kotesovec, Feb 17 2014
a(n) ~ (1+2*sqrt(2)-sqrt(3) + (-1)^n*(1-2*sqrt(2)+sqrt(3))) * 3^(3*n/2) / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 17 2014
a(0) = 1; a(n) = Sum_{i, j, k>=0 and i+j+2*k=n-1} a(i) * a(j) * a(2*k). - Seiichi Manyama, Jul 07 2025
Previous Showing 31-40 of 49 results. Next