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.

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A038044 Shifts left under transform T where Ta is a DCONV a.

Original entry on oeis.org

1, 1, 2, 4, 9, 18, 40, 80, 168, 340, 698, 1396, 2844, 5688, 11456, 22948, 46072, 92144, 184696, 369392, 739536, 1479232, 2959860, 5919720, 11842696, 23685473, 47376634, 94753940, 189519576, 379039152, 758102900, 1516205800
Offset: 1

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Author

Keywords

Crossrefs

Positions of odd terms are given by A003095. Other self-convolved sequences: A000108, A007460 - A007464, A025192, A061922, A062177.
Column k=1 of A144324 and A144823. - Alois P. Heinz, Nov 04 2012
Cf. A038040.
Cf. A000010.

Programs

  • Haskell
    import Data.Function (on)
    a038044 n = a038044_list !! (n-1)
    a038044_list = 1 : f 1 [1] where
       f x ys = y : f (x + 1) (y:ys) where
         y = sum $ zipWith ((*) `on` a038044) divs $ reverse divs
             where divs = a027750_row x
    -- Reinhard Zumkeller, Jan 21 2014
  • Maple
    with(numtheory); EIGENbyDIRCONV := proc(upto_n) local n,a,j,i,s,m; a := [1]; for i from 1 to upto_n do s := 0; m := convert(divisors(i),set); n := nops(m); for j from 1 to n do s := s+(a[m[j]]*a[m[(n-j)+1]]); od; a := [op(a),s]; od; RETURN(a); end;
  • Mathematica
    dc[b_, c_] := Module[{p}, p[n_] := p[n] = Sum[b[d]*c[n/d], {d, If[n<0, {}, Divisors[n]]}]; p]; A[n_, k_] := Module[{f, b, t}, b[1] = dc[f, f]; For[t = 2, t <= k, t++, b[t] = dc[b[t-1], b[t-1]]]; f = Function[m, If[m == 1, 1, b[k][m-1]]]; f[n]]; a[n_] := A[n, 1]; Array[a, 40] (* Jean-François Alcover, Mar 20 2017, after A144324 *)

Formula

From Benoit Cloitre, Aug 29 2004: (Start)
a(n+1) = Sum_{d|n} a(d)*a(n/d), a(1) = 1.
a(prime(k)+1) = 2*a(prime(k));
a(n) is asymptotic to c*2^n where c=0.353030198... (End)
G.f.: A(x) = Sum_{n>=1} a(n)*x^n = x * (1 + Sum_{i>=1} Sum_{j>=1} a(i)*a(j)*x^(i*j)). - Ilya Gutkovskiy, May 01 2019 [modified by Ilya Gutkovskiy, May 09 2019]
a(n+1) = Sum_{k=1..n} a(gcd(n,k))*a(n/gcd(n,k))/phi(n/gcd(n,k)) where phi = A000010. - Richard L. Ollerton, May 19 2021

A050383 Permutation rooted trees with n nodes.

Original entry on oeis.org

1, 1, 3, 8, 25, 77, 262, 897, 3208, 11658, 43243, 162477, 618219, 2374699, 9200541, 35903017, 140997527, 556798525, 2209685939, 8807924914, 35248187347, 141564134395, 570402287162, 2305138038036, 9340981510156, 37946616550787
Offset: 1

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Author

Christian G. Bower, Nov 15 1999

Keywords

Crossrefs

Programs

  • Mathematica
    m = 26; A[_] = 0;
    Do[A[x_] = 1/Product[1 - x^n A[x^n], {n, 1, m}] + O[x]^m // Normal, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Oct 02 2019 *)
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1/prod(k=1, n, (1-x^k*subst(A,x,x^k+x*O(x^n))))); polcoeff(A, n)} /* Paul D. Hanna */

Formula

G.f. (with offset 0) satisfies: A(x) = 1/Product_{n>=1} (1 - x^n*A(x^n)). - Paul D. Hanna, Sep 28 2011
Shifts left under transform T where Ta is EULER(CIK(a)).
a(n) ~ c * d^n / n^(3/2), where d = 4.313133937842504228... and c = 0.153549235191409889... - Vaclav Kotesovec, Nov 05 2021

A107595 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n^2).

Original entry on oeis.org

1, 1, 2, 7, 31, 158, 884, 5292, 33385, 219797, 1500449, 10573815, 76688602, 571232869, 4363912280, 34161879247, 273906591562, 2248935278231, 18909284838057, 162842178607893, 1436660527685476, 12988076148036405, 120345643023918566, 1143054910071718088, 11129160383826078389
Offset: 0

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Author

Paul D. Hanna, May 17 2005

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 + 5292*x^7 +...
Let A = g.f. A(x) then
A = 1 + x*A^1 + x^2*A^4 + x^3*A^9 + x^4*A^16 + x^5*A^25 ...
= 1 + x*(1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 +...)
+ x^2*(1 + 4*x + 14*x^2 + 56*x^3 + 257*x^4 + 1312*x^5 +...)
+ x^3*(1 + 9*x + 54*x^2 + 291*x^3 + 1557*x^4 + 8568*x^5 +..)
+ x^4*(1 + 16*x + 152*x^2 + 1152*x^3 + 7836*x^4 +...)
+ x^5*(1 + 25*x + 350*x^2 + 3675*x^3 + 32625*x^4 +...)
+ x^6*(1 + 36*x + 702*x^2 + 9912*x^3 + 114201*x^4 +...) +...
= 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 +...
		

Crossrefs

Programs

  • Mathematica
    m = 25; A[_] = 0;
    Do[A[x_] = 1 + Sum[x^k A[x]^(k^2) + O[x]^j, {k, 1, j}], {j, m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Nov 05 2019 *)
  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(k=1,n,A=1+sum(j=1,n,x^j*A^(j^2)+x*O(x^n)));polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = (1/x)*Series_Reversion(x/F(x)) and thus A(x) = F(x*A(x)) where F(x) is the g.f. of A107594.
G.f. A(x) = x/Series_Reversion(x*G(x)) and thus A(x) = G(x/A(x)) where G(x) is the g.f. of A107596.
From Paul D. Hanna, Apr 23 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
A = 1/(1 - A*x/(1 - (A^3-A)*x/(1 - A^5*x/(1 - (A^7-A^3)*x/(1 - A^9*x/(1- (A^11-A^5)*x/(1 - A^13*x/(1 - (A^15-A^7)*x/(1 - ...)))))))))
due to an identity of a partial elliptic theta function. (End)
From Paul D. Hanna, May 04 2010: (Start)
Let A = g.f. A(x), then A satisfies:
A = Sum_{n>=0} x^n*A^n * Product_{k=1..n} (1 - x*A^(4k-3)) / (1 - x*A^(4k-1))
due to a q-series identity. (End)

A179420 E.g.f. A(x) satisfies: A(A(x)) = x*A'(x) with A(0)=0, A'(0)=1.

Original entry on oeis.org

0, 1, 2, 12, 132, 2200, 50280, 1482768, 54171376, 2381590944, 123292821600, 7390709937600, 506182300962624, 39180896544097152, 3396777800819754624, 327323946734658720000, 34831825328790915321600
Offset: 0

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Author

Paul D. Hanna, Jul 13 2010

Keywords

Examples

			E.g.f.: A(x) = x + 2*x^2/2! + 12*x^3/3! + 132*x^4/4! + 2200*x^5/5! +...
E.g.f. satisfies: A(A(x)) = x*A'(x) where:
A'(x) = 1 + 2*x + 12*x^2/2! + 132*x^3/3! + 2200*x^4/4! +...
A(A(x)) = x + 4*x^2/2! + 36*x^3/3! + 528*x^4/4! + 11000*x^5/5! +...
Related expansions begin:
A*Dx(A)/2! = 2*x^2/2! + 15*x^3/3! + 180*x^4/4! + 3150*x^5/5! +...
A*Dx(A*Dx(A))/3! = 6*x^3/3! + 104*x^4/4! + 2140*x^5/5! +...
A*Dx(A*Dx(A*Dx(A)))/4! = 24*x^4/4! + 770*x^5/5! + 24600*x^6/6! +...
A*Dx(A*Dx(A*Dx(A*Dx(A))))/5! = 120*x^5/5! + 6264*x^6/6! +...
which generate iterations of A=A(x) as illustrated by:
A(A(x))/x = 1 + 2*A + 2^2*A*Dx(A)/2! + 2^3*A*Dx(A*Dx(A))/3! +...
A(A(A(x)))/x = 1 + 3*A + 3^2*A*Dx(A)/2! + 3^3*A*Dx(A*Dx(A))/3! +...
A_{-1}(x)/x = 1 - A + A*Dx(A)/2! - A*Dx(A*Dx(A))/3! +-...(inverse).
Illustrate a main property of the iterations A_n(x) of A(x) by:
A(x) = A(A(x)) * A(x)/[x*d/dx A(x)];
A(x) = A_3(x) * A_2(x)/[x*d/dx A_2(x)];
A(x) = A_4(x) * A_3(x)/[x*d/dx A_3(x)]; ...
which can be shown consistent by the chain rule of differentiation.
...
The RIORDAN ARRAY (A(x)/x, A(x)) begins:
. 1;
. 1, 1;
. 4/2!, 2, 1;
. 33/3!, 10/2!, 3, 1;
. 440/4!, 90/3!, 18/2!, 4, 1;
. 8380/5!, 1240/4!, 177/3!, 28/2!, 5, 1;
. 211824/6!, 23800/5!, 2544/4!, 300/3!, 40/2!, 6, 1;
. 6771422/7!, 598788/6!, 49680/5!, 4520/4!, 465/3!, 54/2!, 7, 1; ...
where the e.g.f. of column k = A(x)^(k+1)/x for k>=0.
...
The MATRIX LOG of the above Riordan array (A(x)/x, A(x)) begins:
. 0;
. 1, 0;
. 2/2!, 2, 0;
. 12/3!, 4/2!, 3, 0;
. 132/4!, 24/3!, 6/2!, 4, 0;
. 2200/5!, 264/4!, 36/3!, 8/2!, 5, 0;
. 50280/6!, 4400/5!, 396/4!, 48/3!, 10/2!, 6, 0;
. 1482768/7!, 100560/6!, 6600/5!, 528/4!, 60/3!, 12/2!, 7, 0; ...
where the e.g.f. of column k = (k+1)*A(x) for k>=0.
		

Crossrefs

a(n)/n! = A221019(n)/A221020(n).

Programs

  • Mathematica
    a[n_] := a[n] = Module[{A}, A[x_] = x+x^2+Sum[a[m]*x^m/m!, {m, 3, n-1}]; If[n<3, n!*Coefficient[A[x], x, n], n!*Coefficient[A[A[x]], x, n]/(n-2)] ]; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Jan 15 2018, translated from PARI *)
  • Maxima
    Co(n, k, F):=if k=1  then F(n) else sum(F(i+1)*Co(n-i-1, k-1, F), i, 0, n-k);
    a(n):=if n=0 then 0 else if n<3 then 1 else sum(Co(n,k,a)*a(k),k,2,n-1)/(n-2); /* Vladimir Kruchinin, Jun 29 2011 */
  • PARI
    {a(n)=local(A=x+x^2+sum(m=3,n-1,a(m)*x^m/m!)+x*O(x^n));if(n<3,n!*polcoeff(A,n),n!*polcoeff(subst(A,x,A),n)/(n-2))}
    

Formula

E.g.f. A(x) equals the e.g.f. of column 0 in the matrix log of the Riordan array (A(x)/x, A(x)).
Let A_n(x) denote the n-th iteration of e.g.f. A(x) with A_0(x)=x,
then A=A(x) satisfies:
A(x)/x = 1 + A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! +...
A_{-1}(x)/x = 1 - A + A*Dx(A)/2! - A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! -+...
A_n(x)/x = 1 + n*A + n^2*A*Dx(A)/2! + n^3*A*Dx(A*Dx(A))/3! + n^4*A*Dx(A*Dx(A*Dx(A)))/4! +...
where Dx(F) = d/dx(x*F).
Further, we have: A(x) = A_{n+1}(x) * A_n(x)/[x*d/dx A_n(x)] which holds for all n.
a(n)=sum(k=2..n-1, R(n-1,k-1)*a(k))/(n-2), n>2, a(1)=1, a(2)=1, where R is the Riordan array (A(x)/x, A(x)). [Vladimir Kruchinin, Jun 29 2011]
E.g.f. satisfies: A(x) = Series_Reversion(-G(-x)) where G(x) is the e.g.f. of A193202 and satisfies: G(G(x)) = x*G'(G(x)). [Paul D. Hanna, Jul 22 2011]

A007439 Number of planted trees: all sub-rooted trees from any node are identical; non-root, non-leaf nodes an even distance from the root are of degree 2.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 3, 7, 4, 11, 6, 15, 7, 24, 8, 29, 12, 40, 13, 51, 14, 68, 19, 76, 20, 107, 23, 116, 29, 147, 30, 175, 31, 215, 39, 229, 45, 297, 46, 312, 55, 387, 56, 435, 57, 513, 73, 534, 74, 670, 78, 705, 92, 823, 93, 897, 102, 1051, 117, 1082
Offset: 1

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Author

Keywords

References

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

Crossrefs

Programs

  • Haskell
    a007439 n = a007439_list !! (n-1)
    a007439_list = 1 : 1 : f 2 where
       f x = (sum $ map a007439 $ a027750_row (x - 1)) : f (x + 1)
    -- Reinhard Zumkeller, Dec 20 2014
  • Mathematica
    a[n_] := a[n] = Sum[a[k], {k, Divisors[n-2]}]; a[1] = a[2] = 1; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, May 15 2013 *)

Formula

a(n+2) = Sum a(k), k|n. Shifts left two places under inverse Moebius transformation.
G.f. A(x) satisfies: A(x) = x + x^2 * (1 + A(x) + A(x^2) + A(x^3) + ...). - Ilya Gutkovskiy, May 09 2019

Extensions

New description from Christian G. Bower, Oct 15 1998

A007560 Number of planted identity trees where non-root, non-leaf nodes an even distance from root are of degree 2.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 4, 6, 10, 17, 29, 51, 89, 159, 284, 512, 930, 1692, 3101, 5698, 10515, 19464, 36143, 67296, 125622, 235050, 440756, 828142, 1558955, 2939761, 5552744, 10504222, 19899760, 37750091, 71704061, 136361602, 259618770, 494821629, 944074665
Offset: 1

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Author

Keywords

References

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

Crossrefs

Cf. A007562.
Column k=2 of A316074.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(binomial(a(i), j)*b(n-i*j, i-1), j=0..n/i)))
        end:
    a:= n-> `if`(n<2, n, b(n-2, n-2)):
    seq(a(n), n=1..50);  # Alois P. Heinz, May 19 2013
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Binomial[a[i], j]*b[n-i*j, i-1], {j, 0, n/i}]]]; a[n_] := If[n<2, n, b[n-2, n-2]]; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Jan 27 2014, after Alois P. Heinz *)

Formula

Shifts 2 places left under weigh transform.
a(n) ~ c * d^n / n^(3/2), d = 1.983229991815043367273184141..., c = 0.5857451140002020594085469... . - Vaclav Kotesovec, Aug 25 2014
G.f.: x + x^2 * Product_{n>=1} (1 + x^n)^a(n). - Ilya Gutkovskiy, May 09 2019

Extensions

Better description from Christian G. Bower, May 15 1998

A032027 Number of planted planar trees (n+1 nodes) where any 2 subtrees extending from the same node are different.

Original entry on oeis.org

1, 1, 1, 3, 5, 13, 35, 95, 255, 715, 2081, 6003, 17645, 52127, 155863, 468129, 1415521, 4301055, 13134789, 40275109, 123970669, 382919917, 1186475687, 3686899725, 11487023793, 35876838669, 112304155021, 352276801491
Offset: 1

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Keywords

Examples

			From _Gus Wiseman_, Nov 15 2022: (Start)
The a(1) = 1 through a(6) = 13 ordered rooted identity trees (ranked by A358374):
  o  (o)  ((o))  ((o)o)   (((o))o)   (((o)o)o)
                 (o(o))   (((o)o))   ((o(o))o)
                 (((o)))  ((o(o)))   (o((o)o))
                          (o((o)))   (o(o(o)))
                          ((((o))))  ((((o)))o)
                                     ((((o))o))
                                     ((((o)o)))
                                     (((o))(o))
                                     (((o(o))))
                                     ((o)((o)))
                                     ((o((o))))
                                     (o(((o))))
                                     (((((o)))))
(End)
		

Crossrefs

The unordered version is A004111, ranked by A276625.
These trees (ordered rooted identity) are ranked by A358374.

Programs

  • Mathematica
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],FreeQ[#,[_]?(!UnsameQ@@#&)]&]],{n,1,10}] (* Gus Wiseman, Nov 15 2022 *)
  • PARI
    AGK(v)={apply(p->subst(serlaplace(y^0*p),y,1), Vec(prod(k=1, #v, (1 + x^k*y + O(x*x^#v))^v[k])-1, -#v))}
    seq(n)={my(v=[1]); for(i=2, n, v=concat([1], AGK(v))); v} \\ Andrew Howroyd, Sep 20 2018

Formula

Shifts left under "AGK" (ordered, elements, unlabeled) transform.

A032188 Number of labeled series-reduced mobiles (circular rooted trees) with n leaves (root has degree 0 or >= 2).

Original entry on oeis.org

1, 1, 5, 41, 469, 6889, 123605, 2620169, 64074901, 1775623081, 54989743445, 1882140936521, 70552399533589, 2874543652787689, 126484802362553045, 5977683917752887689, 301983995802099667861, 16239818347465293071401, 926248570498763547197525, 55847464116157184894240201
Offset: 1

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Keywords

Comments

With offset 0, a(n) = number of partitions of the multiset {1,1,2,2,...,n,n} into lists of strictly decreasing lists, called blocks, such that the concatenation of all blocks in a list has the Stirling property: all entries between the two occurrences of i exceed i, 1<=i<=n. For example, with slashes separating blocks, a(2) = 5 counts 1/1/2/2; 1/2/2/1; 2/2/1/1; 1/2/2 1; 2/2 1/1, but not, for instance, 2 1/2/1 because it fails the Stirling test for i=2. - David Callan, Nov 21 2011

Examples

			D^3(1) = (24*x^2-64*x+41)/(2*x-1)^6. Evaluated at x = 0 this gives a(4) = 41.
a(3) = 5: Denote the colors of the vertices by the letters a,b,c, .... The 5 possible increasing plane trees on 3 vertices with vertices of outdegree k coming in 2^(k-1) colors are
.
   1a       1a        1b        1a        1b
   |       /  \      /  \      /  \      /  \
   2a     2    3    2    3    3    2    3    2
   |
   3
		

Crossrefs

Programs

  • Maple
    Order := 20; t1 := solve(series((ln(1-A)+2*A),A)=x,A); A000311 := n->n!*coeff(t1,x,n);
    # With offset 0:
    a := n -> add(combinat:-eulerian2(n,k)*2^k,k=0..n):
    seq(a(n),n=0..19); # Peter Luschny, Jul 09 2015
  • Mathematica
    For[y=x+O[x]^21; oldy=0, y=!=oldy, oldy=y; y=((1-y)Log[1-y]+x*y+y-x)/(2y-1), Null]; Table[n!Coefficient[y, x, n], {n, 1, 20}]
    Rest[CoefficientList[InverseSeries[Series[2*x + Log[1-x],{x,0,20}],x],x] * Range[0,20]!] (* Vaclav Kotesovec, Jan 08 2014 *)
  • Maxima
    a(n):=sum((n+k-1)!*sum(1/(k-j)!*sum((2^l*(-1)^(n+l+1)*stirling1(n-l+j-1,j-l))/(l!*(n-l+j-1)!),l,0,j),j,0,k),k,0,n-1); /* Vladimir Kruchinin, Feb 06 2012 */
    
  • PARI
    N = 66; x = 'x + O('x^N);
    Q(k) = if(k>N, 1,  1 + (k+1)*x - 2*x*(k+1)/Q(k+1) );
    gf = 1/Q(0); Vec(gf) \\ Joerg Arndt, May 01 2013
    
  • PARI
    {my(n=20); Vec(serlaplace(serreverse(2*x+log(1-x + O(x*x^n)))))} \\ Andrew Howroyd, Jan 16 2018

Formula

Doubles (index 2+) under "CIJ" (necklace, indistinct, labeled) transform.
E.g.f. A(x) satisfies log(1-A(x))+2*A(x)-x = 0. - Vladeta Jovovic, Dec 06 2002
With offset 0, second Eulerian transform of the powers of 2 [A000079]. See A001147 for definition of SET. - Ross La Haye, Feb 14 2005
From Peter Bala, Sep 05 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A'(x) = (1-A)/(1-2*A) with A(0) = 0. Hence the inverse function A^-1(x) = int {t = 0..x} (1-2*t)/(1-t) = 2*x+log(1-x). The expansion of A(x) can be found by inverting the above integral using the method of [Dominici, Theorem 4.1] to arrive at the result a(n) = D^(n-1)(1) evaluated at x = 0, where D denotes the operator g(x) -> d/dx((1-x)/(1-2*x)*g(x)). Compare with A006351.
Applying [Bergeron et al., Theorem 1] to the result x = int {t = 0..A(x)} 1/phi(t), where phi(t) = (1-t)/(1-2*t) = 1+t+2*t^2+4*t^3+8*t^4+... leads to the following combinatorial interpretation for this sequence: a(n) gives the number of plane increasing trees on n vertices where each vertex of outdegree k >= 1 can be colored in 2^(k-1) ways. An example is given below. (End)
The integral from 0 to infinity w.r.t. w of exp(-2w)(1-z*w)^(-1/z) gives an o.g.f. for the series with offset 0. Consequently, a(n)= sum(j=1 to infinity): St1d(n,j)/(2^(n+j-1)) where St1d(n,j) is the j-th element of the n-th diagonal of A132393 with offset=1; e.g., a(3)= 5 = 0/2^3 + 2/2^4 + 11/2^5 + 35/2^6 + 85/2^7 + ... . - Tom Copeland, Sep 15 2011
A signed o.g.f., with Γ(v,x) the incomplete gamma function (see A111999 with u=1), is g(z) = (2/z)^(-(1/z)-1) exp(2/z) Γ((1/z)+1,2/z)/z. - Tom Copeland, Sep 16 2011
With offset 0, a(n) = Sum[T(n+k,k), k=1..n] where T(n,k) are the associated Stirling numbers of the first kind (A008306). For example, a(3) = 41 = 6 + 20 + 15. - David Callan, Nov 21 2011
a(n) = sum(k=0..n-1, (n+k-1)!*sum(j=0..k, 1/(k-j)!*sum(l=0..j, (2^l*(-1)^(n+l+1)*stirling1(n-l+j-1,j-l))/(l!*(n-l+j-1)!)))), n>0. - Vladimir Kruchinin, Feb 06 2012
G.f.: 1/Q(0), where Q(k)= 1 + (k+1)*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) ~ n^(n-1) / (2 * exp(n) * (1-log(2))^(n-1/2)). - Vaclav Kotesovec, Jan 08 2014
a(n) = A032034(n)/2. - Alois P. Heinz, Jul 04 2018
E.g.f: series reversion of 2*x + log(1-x). - Andrew Howroyd, Sep 19 2018

A007563 Number of rooted connected graphs where every block is a complete graph.

Original entry on oeis.org

0, 1, 1, 3, 8, 25, 77, 258, 871, 3049, 10834, 39207, 143609, 532193, 1990163, 7503471, 28486071, 108809503, 417862340, 1612440612, 6248778642, 24309992576, 94905791606, 371691137827, 1459935388202, 5749666477454
Offset: 0

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Author

Keywords

References

  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 71, (3.4.13).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=2 of A144042.
Cf. A245566.

Programs

  • Maple
    with(numtheory): etr:= proc(p) local b; b:= proc(n) option remember; if n=0 then 1 else (add(d*p(d), d=divisors(n)) +add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n-1))/n fi end end: b:= etr(a): c:= etr(b): a:= n-> if n=0 then 0 else c(n-1) fi: seq(a(n), n=0..25); # Alois P. Heinz, Sep 06 2008
  • Mathematica
    etr[p_] := etr[p] = Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[ Sum[ d*p[d], {d, Divisors[j]}]*b[n-j], {j, 1, n}]/n]; b]; a[0] = 0; a[n_] := etr[etr[a]][n-1]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, May 28 2013, after Alois P. Heinz *)
  • PARI
    EulerT(v)={Vec(exp(x*Ser(dirmul(v,vector(#v,n,1/n))))-1, -#v)}
    seq(n)={my(v=[1]); for(i=2, n, v=concat([1], EulerT(EulerT(v)))); concat([0], v)} \\ Andrew Howroyd, May 20 2018

Formula

Shifts left when Euler transform is applied twice.
a(n) ~ c * d^n / n^(3/2), where d = 4.189610958393826965527036454524044275... (see A245566), c = 0.1977574301782950818433893126632477845870281049591883888... . - Vaclav Kotesovec, Jul 26 2014

Extensions

New description from Christian G. Bower, Oct 15 1998

A029768 Number of increasing mobiles with n elements.

Original entry on oeis.org

0, 1, 1, 2, 7, 36, 245, 2076, 21059, 248836, 3356609, 50896380, 856958911, 15864014388, 320245960333, 7001257954796, 164792092647355, 4154906594518116, 111719929072986521, 3191216673497748444
Offset: 0

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Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

A labeled tree of size n is a rooted tree on n nodes that are labeled by distinct integers from the set {1,...,n}. An increasing tree is a labeled tree such that the sequence of labels along any branch starting at the root is increasing.
a(n) counts increasing trees with cyclically ordered branches.
a(n+1) counts the non-plane (where the subtrees stemming from a node are not ordered between themselves) increasing trees on n nodes where the nodes of outdegree k come in k+1 colors. An example is given below. The number of plane increasing trees on n nodes where the nodes of outdegree k come in k+1 colors is given by the triple factorial numbers A008544. - Peter Bala, Aug 30 2011
a(n+1)/a(n)/n tends to 1/A073003 = 1.676875... . - Vaclav Kotesovec, Mar 11 2014

Examples

			a(4) = 7: D^2[(1+x)*exp(x)] = exp(2*x)*(2*x^2+8*x+7). Evaluated at x = 0 this gives a(4) = 7. Denote the colors of the nodes by the letters a,b,c,.... The 7 possible trees on 3 nodes with nodes of outdegree k coming in k+1 colors are:
........................................................
...1a....1b....1a....1b........1a.......1b........1c....
...|.....|.....|.....|......../.\....../..\....../..\...
...2a....2b....2b....2a......2...3....2....3....2....3..
...|.....|.....|.....|..................................
...3.....3.....3.....3..................................
G.f. = x + x^2 + 2*x^3 + 7*x^4 + 36*x^5 + 245*x^6 + 2076*x^7 + 21059*x^8 + ...
		

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 392.

Crossrefs

Programs

  • Maple
    S:= rhs(dsolve({diff(a(x),x) = log(1/(1-a(x)))+1,a(0)=0},a(x),series,order=101)):
    seq(coeff(S,x,j)*j!,j=0..100); # Robert Israel, Apr 17 2015
  • Mathematica
    Multinomial1[list_] := Apply[Plus, list]!/Apply[Times, (#1! & ) /@ list]; a[1]=1; a[n_]/;n>=2 := a[n] = Sum[Map[Multinomial1[ # ]Product[Map[a,# ]]/Length[ # ]&,Compositions[n-1]]]; Table[a[n],{n,8}] (* David Callan, Nov 29 2007 *)
    nmax=20; b = ConstantArray[0,nmax]; b[[1]]=0; b[[2]]=1; Do[b[[n+1]] = b[[n]] + Sum[Binomial[n-2,i]*b[[i+1]]*b[[n-i+1]],{i,1,n-2}],{n,2,nmax-1}]; b (* Vaclav Kotesovec after Vladimir Kruchinin, Mar 11 2014 *)
    terms = 20; A[x_] := x; Do[A[x_] = Integrate[(1 + A[x])*Exp[A[x] + O[x]^j], x] + O[x]^j // Normal // Simplify, {j, 1, terms - 1}]; Join[{0, 1}, CoefficientList[A[x], x]*Range[0, terms - 2]! // Rest] (* Jean-François Alcover, May 22 2014, updated Jan 12 2018 (after PARI script by Michael Somos) *)
  • PARI
    {a(n) = my(A = x + O(x^2)); if( n<2, n==1, n--; for(k=1, n-1, A = intformal( (1 + A) * exp(A)));  n! * polcoeff(A, n))}; /* Michael Somos, Apr 17 2015 */
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    seq(N) = {
      my(a = vector(N)); a[1] = 1;
      for (n = 2, N, a[n] = a[n-1] + sum(k=1, n-2, binomial(n-2, k)*a[k]*a[n-k]));
      concat(0, a);
    };
    seq(19)
    \\ test: N=200; y=serconvol(Ser(seq(N),'x), exp('x+O('x^N))); y' == y''*(1-y)
    \\ Gheorghe Coserea, Jun 26 2018

Formula

Bergeron et al. give several formulas. Shifts left under "CIJ" (necklace, indistinct, labeled) transform.
E.g.f.: A(x) =
x + (1/2)*x^2 + (1/3)*x^3 + (7/24)*x^4 + (3/10)*x^5 + (49/144)*x^6 + (173/420)*x^7 + (21059/40320)*x^8 + (8887/12960)*x^9 + ...
and satisfies the differential equation A'(x)=log(1/(1-A(x)))+1. - Vladimir Kruchinin, Jan 22 2011
E.g.f. A(x) satisfies: A''(x) = A'(x) * exp(A'(x)-1). - Paul D. Hanna, Apr 17 2015
From Robert Israel, Apr 17 2015 (Start):
E.g.f. A(x) satisfies e*(Ei(1,A'(x)) - Ei(1,1)) = integral(s = 1 .. A'(x), exp(1-s)/s ds) = -x.
a(n) = e^(1-n)*limit(w -> 1, (d^(n-2)/dw^(n-2))(((w-1)/(Ei(1,1)-Ei(1,w)))^(n-1))) for n >= 2. (End)
a(n) = sum(i=1..n-2,binomial(n-2,i)*a(i)*a(n-i))+a(n-1), a(0)=0, a(1)=1. - Vladimir Kruchinin, Jan 24 2011
The following remarks refer to the interpretation of this sequence as counting increasing trees where the nodes of outdegree k come in k+1 colors. Thus we work with the generating function B(x) = A'(x)-1 = x + 2*x^2/2!+7*x^3/3!+36*x^4/4!+.... The degree function phi(x) (see [Bergeron et al.] for definition) for this variety of trees is phi(x) = 1+2*x+3*x^2/2!+4*x^3/3!+5*x^4/4!+... = (1+x)*exp(x). The generating function B(x) satisfies the autonomous differential equation B' = phi(B(x)) with initial condition B(0) = 0. It follows that the inverse function B(x)^(-1) may be expressed as an integral B(x)^(-1) = int {t = 0..x} 1/phi(t) dt = int {t = 0..x} exp(-t)/(1+t) dt. Applying [Dominici, Theorem 4.1] to invert the integral produces the result B(x) = sum {n>=1} D^(n-1)[(1+x)*exp(x)](0)*x^n/n!, where the nested derivative D^n[f](x) of a function f(x) is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. Thus a(n+1) = D^(n-1)[(1+x)*exp(x)](0). - Peter Bala, Aug 30 2011

Extensions

More terms from Christian G. Bower
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