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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A216349 Triangle T(n,k) in which n-th row lists the values of the n-th derivative at x=1 of all functions that are representable as x^x^...^x with n x's and parentheses inserted in all possible ways; n>=1, 1<=k<=A000081(n).

Original entry on oeis.org

1, 2, 12, 9, 156, 100, 80, 56, 3160, 1880, 1180, 1420, 950, 1360, 890, 660, 480, 87990, 50496, 29682, 35382, 24042, 22008, 14928, 31968, 20268, 14988, 10848, 34974, 21474, 13314, 15114, 10974, 13014, 8874, 6534, 5094, 3218628, 1806476, 1021552, 588756, 1189132
Offset: 1

Views

Author

Alois P. Heinz, Sep 04 2012

Keywords

Comments

The ordering of the functions is the same as in A215703 and is defined by the algorithm below.

Examples

			For n=4 the A000081(4) = 4 functions and their 4th derivatives at x=1 are x^(x^3)->156, x^(x^x*x)->100, x^(x^(x^2))->80, x^(x^(x^x))->56.
Triangle T(n,k) begins:
:     1;
:     2;
:    12,     9;
:   156,   100,    80,    56;
:  3160,  1880,  1180,  1420,   950,  1360,   890,   660,   480;
: 87990, 50496, 29682, 35382, 24042, 22008, 14928, 31968, 20268, ...
		

Crossrefs

First column gives: A216351.
Last elements of rows give: A033917.
A version with sorted row elements is: A216350.
Rows sums give: A216281.

Programs

  • Maple
    with(combinat):
    F:= proc(n) F(n):= `if`(n<2, [x$n], map(h->x^h, g(n-1, n-1))) end:
    g:= proc(n, i) option remember; `if`(n=0 or i=1, [x^n],
         `if`(i<1, [], [seq(seq(seq(mul(F(i)[w[t]-t+1], t=1..j)*v,
          w=choose([$1..nops(F(i))+j-1], j)), v=g(n-i*j, i-1)), j=0..n/i)]))
        end:
    T:= n-> map(f-> n!*coeff(series(subs(x=x+1, f), x, n+1), x, n), F(n))[]:
    seq(T(n), n=1..7);

A216350 Triangle T(n,k) in which n-th row lists in increasing order the values of the n-th derivative at x=1 of all functions that are representable as x^x^...^x with n x's and parentheses inserted in all possible ways; n>=1, 1<=k<=A000081(n).

Original entry on oeis.org

1, 2, 9, 12, 56, 80, 100, 156, 480, 660, 890, 950, 1180, 1360, 1420, 1880, 3160, 5094, 6534, 8874, 10848, 10974, 13014, 13314, 14928, 14988, 15114, 20268, 21474, 22008, 24042, 29682, 31968, 34974, 35382, 50496, 87990, 65534, 78134, 102494, 131684, 141974
Offset: 1

Views

Author

Alois P. Heinz, Sep 04 2012

Keywords

Examples

			For n=4 the A000081(4) = 4 functions and their 4th derivatives at x=1 are x^(x^3)->156, x^(x^x*x)->100, x^(x^(x^2))->80, x^(x^(x^x))->56 => 4th row = [56, 80, 100, 156].
Triangle T(n,k) begins:
:    1;
:    2;
:    9,   12;
:   56,   80,  100,   156;
:  480,  660,  890,   950,  1180,  1360,  1420,  1880,  3160;
: 5094, 6534, 8874, 10848, 10974, 13014, 13314, 14928, 14988, 15114, ...
		

Crossrefs

First column gives: A033917.
Last elements of rows give: A216351.
A version with different ordering of row elements is: A216349.
Rows sums give: A216281.

Programs

  • Maple
    with(combinat):
    F:= proc(n) F(n):= `if`(n<2, [x$n], map(h->x^h, g(n-1, n-1))) end:
    g:= proc(n, i) option remember; `if`(n=0 or i=1, [x^n],
         `if`(i<1, [], [seq(seq(seq(mul(F(i)[w[t]-t+1], t=1..j)*v,
          w=choose([$1..nops(F(i))+j-1], j)), v=g(n-i*j, i-1)), j=0..n/i)]))
        end:
    T:= n-> sort(map(f-> n!*coeff(series(subs(x=x+1, f)
                     , x, n+1), x, n), F(n)))[]:
    seq(T(n), n=1..7);

A199883 Number of distinct values taken by 6th derivative of x^x^...^x (with n x's and parentheses inserted in all possible ways) at x=1.

Original entry on oeis.org

1, 1, 2, 4, 9, 20, 48, 113, 262, 591, 1263, 2505, 4764, 8479, 14285, 22871, 35316, 52755, 76517, 107826, 148914, 202715, 270622
Offset: 1

Views

Author

Alois P. Heinz, Nov 11 2011

Keywords

Examples

			a(4) = 4 because the A000108(3) = 5 possible parenthesizations of x^x^x^x lead to 4 different values of the 6th derivative at x=1: (x^(x^(x^x))) -> 2934; ((x^x)^(x^x)), ((x^(x^x))^x) -> 4908; (x^((x^x)^x)) -> 5034; (((x^x)^x)^x) -> 8322.
		

Crossrefs

Cf. A000081 (distinct functions), A000108 (parenthesizations), A000012 (first derivatives), A028310 (2nd derivatives), A199085 (3rd derivatives), A199205 (4th derivatives), A199296 (5th derivatives), A002845, A003018, A003019, A145545, A145546, A145547, A145548, A145549, A145550, A082499, A196244, A198683, A215703, A215836. Column k=6 of A216368.

Programs

  • Maple
    f:= proc(n) option remember;
          `if`(n=1, {[0, 0, 0, 0, 0]},
                    {seq(seq(seq([2+g[1], 3*(1 +g[1] +h[1]) +g[2],
                     8 +12*g[1] +6*h[1]*(1+g[1]) +4*(g[2]+h[2])+g[3],
                     10+50*h[1]+10*h[2]+5*h[3]+(30+30*h[1]+10*h[2]
                     +15*g[1])*g[1]+(20+10*h[1])*g[2]+5*g[3]+g[4],
                     45*h[1]*g[1]^2+(120+60*h[2]+15*h[3]+60*g[2]+
                     270*h[1])*g[1]+54+15*h[3]+30*g[3]+6*g[4]+
                     60*h[1]*g[2]+15*h[1]*g[3]+30*h[1]+ 20*h[2]*g[2]+
                     100*h[2]+90*h[1]^2+g[5]+60*g[2]+6*h[4]],
                     h=f(n-j)), g=f(j)), j=1..n-1)})
        end:
    a:= n-> nops(map(x-> x[5], f(n))):
    seq(a(n), n=1..15);

Extensions

a(22)-a(23) from Alois P. Heinz, Sep 26 2014

A211192 Consider all distinct functions f representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways; sequence gives difference between numbers of f with f(0)=1 and numbers of f with f(0)=0, with conventions that 0^0=1^0=1^1=1, 0^1=0.

Original entry on oeis.org

0, -1, 1, 0, 2, 1, 8, 10, 39, 72, 225, 506, 1434, 3550, 9767, 25391, 69293, 185061, 505843, 1372744, 3769842, 10339104, 28546539, 78890525, 218945822, 608657861, 1697106780, 4740593393, 13272626627, 37224982494, 104599603493, 294384019508, 829836855332
Offset: 0

Views

Author

Alois P. Heinz, Feb 18 2013

Keywords

Comments

A000081(n) distinct functions are representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways. Some functions are representable in more than one way, the number of valid parenthesizations is A000108(n-1) for n>0.

Examples

			There are A000081(4) = 4 functions f representable as x -> x^x^...^x with 4 x's and parentheses inserted in all possible ways: ((x^x)^x)^x, (x^x)^(x^x) == (x^(x^x))^x, x^((x^x)^x), x^(x^(x^x)).  Only x^((x^x)^x) evaluates to 0 at x=0: 0^((0^0)^0) = 0^(1^0) = 0^1 = 0.  Three functions evaluate to 1 at x=0: ((0^0)^0)^0 = (1^0)^0 = 1^0 = 1, (0^0)^(0^0) = 1^1 = 1, 0^(0^(0^0)) = 0^(0^1) = 0^0 = 1. Thus a(4) = 3-1 = 2.
a(8) = A222380(8) - A222379(8) = 77 - 38 = 39.
		

Crossrefs

Programs

  • Maple
    g:= proc(n, i) option remember; `if`(n=0, [0, 1], `if`(i<1, 0, (v->[v[1]-
          v[2], v[2]])(add(((l, h)-> [binomial(l[2]+l[1]+j-1, j)*(h[1]+h[2]),
          binomial(l[1]+j-1, j)*h[2]])(g(i-1$2), g(n-i*j, i-1)), j=0..n/i))))
        end:
    a:= n-> (f-> f[1]-f[2])(g(n-1$2)):
    seq(a(n), n=0..40);
  • Mathematica
    g[n_, i_] := g[n, i] = If[n==0, {0, 1}, If[i<1, {0, 0}, ({#[[1]]-#[[2]], #[[2]]}&)[Sum[Function[{l, h}, {(h[[1]]+h[[2]])*Binomial[j+l[[1]]+l[[2]] -1, j], h[[2]]*Binomial[j+l[[1]]-1, j]}][g[i-1, i-1]], g[n-i*j, i-1]]], {j, 0, Quotient[n, i]}]];
    a[n_] := (#[[1]]-#[[2]]&)[g[n-1, n-1]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 22 2017, translated from Maple *)

Formula

a(n) = A222380(n) - A222379(n).
From Alois P. Heinz, Mar 01 2019: (Start)
a(n) is even <=> n in { A258592 }.
a(n) is odd <=> n in { A263831 }. (End)

A215522 n-th derivative of (x^x)^(x^x) at x=1.

Original entry on oeis.org

1, 1, 4, 18, 100, 650, 4908, 41090, 382520, 3863808, 42409440, 497972112, 6259762320, 83343114504, 1175904241848, 17442325040520, 272149555445760, 4438451554802880, 75714874759039104, 1343817666163911168, 24837691533530152320, 475811860099666527360
Offset: 0

Views

Author

Alois P. Heinz, Aug 14 2012

Keywords

Comments

Also n-th derivative of (x^(x^x))^x = x^(x^x*x) at x=1.
First term < 0: a(65).

Crossrefs

Column k=6 of A215703.

Programs

  • Maple
    a:= n-> n!*coeff(series(subs(x=x+1, (x^x)^(x^x) ), x, n+1), x, n):
    seq(a(n), n=0..30);
  • Mathematica
    m = 21; CoefficientList[(x+1)^((x+1)^(x+2)) + O[x]^(m+1), x]*Range[0, m]! (* Jean-François Alcover, Feb 07 2021 *)

Formula

E.g.f.: (x+1)^((x+1)^(x+2)).

A222379 Number of distinct functions f representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways giving result f(0)=0, with conventions that 0^0=1^0=1^1=1, 0^1=0.

Original entry on oeis.org

0, 1, 0, 1, 1, 4, 6, 19, 38, 107, 247, 668, 1666, 4468, 11603, 31210, 83044, 224893, 607658, 1657966, 4528193, 12441364, 34254321, 94696165, 262389581, 729258392, 2031264865, 5671570468, 15867219821, 44480785907, 124913622052, 351393746745, 990048748684
Offset: 0

Views

Author

Alois P. Heinz, Feb 17 2013

Keywords

Comments

A000081(n) distinct functions are representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways. Some functions are representable in more than one way, the number of valid parenthesizations is A000108(n-1) for n>0.

Examples

			There are A000081(4) = 4 functions f representable as x -> x^x^...^x with 4 x's and parentheses inserted in all possible ways: ((x^x)^x)^x, (x^x)^(x^x) == (x^(x^x))^x, x^((x^x)^x), x^(x^(x^x)).  Only x^((x^x)^x) evaluates to 0 at x=0: 0^((0^0)^0) = 0^(1^0) = 0^1 = 0. Thus a(4) = 1.
		

Crossrefs

Programs

  • Maple
    g:= proc(n, i) option remember; `if`(n=0, [0, 1], `if`(i<1, 0, (v->[v[1]-
          v[2], v[2]])(add(((l, h)-> [binomial(l[2]+l[1]+j-1, j)*(h[1]+h[2]),
          binomial(l[1]+j-1, j)*h[2]])(g(i-1$2), g(n-i*j, i-1)), j=0..n/i))))
        end:
    a:= n-> g(n-1$2)[2]:
    seq(a(n), n=0..40);
  • Mathematica
    f[l_, h_] := {Binomial[l[[2]] + l[[1]] + j - 1, j]*(h[[1]] + h[[2]]), Binomial[l[[1]] + j - 1, j]*h[[2]]};
    g[n_, i_] := g[n, i] = If[n == 0, {0, 1}, If[i < 1, {0, 0}, Function[v, {v[[1]] - v[[2]], v[[2]]}][Sum[f[g[i - 1, i - 1], g[n - i*j, i - 1]], {j, 0, Quotient[n, i]}]]]];
    a[n_] := g[n - 1, n - 1][[2]];
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 27 2019, after Alois P. Heinz *)

Formula

A222380(n) + a(n) = A000081(n).
A222380(n) - a(n) = A211192(n).
a(n) = Sum_{i=A087803(n-1)+1..A087803(n)} (1-A306710(i)).

A222380 Number of distinct functions f representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways giving result f(0)=1, with conventions that 0^0=1^0=1^1=1, 0^1=0.

Original entry on oeis.org

0, 0, 1, 1, 3, 5, 14, 29, 77, 179, 472, 1174, 3100, 8018, 21370, 56601, 152337, 409954, 1113501, 3030710, 8298035, 22780468, 62800860, 173586690, 481335403, 1337916253, 3728371645, 10412163861, 29139846448, 81705768401, 229513225545, 645777766253
Offset: 0

Views

Author

Alois P. Heinz, Feb 17 2013

Keywords

Comments

A000081(n) distinct functions are representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways. Some functions are representable in more than one way, the number of valid parenthesizations is A000108(n-1) for n>0.

Examples

			There are A000081(4) = 4 functions f representable as x -> x^x^...^x with 4 x's and parentheses inserted in all possible ways: ((x^x)^x)^x, (x^x)^(x^x) == (x^(x^x))^x, x^((x^x)^x), x^(x^(x^x)).  Only x^((x^x)^x) evaluates to 0 at x=0: 0^((0^0)^0) = 0^(1^0) = 0^1 = 0.  Three functions evaluate to 1 at x=0: ((0^0)^0)^0 = (1^0)^0 = 1^0 = 1, (0^0)^(0^0) = 1^1 = 1, 0^(0^(0^0)) = 0^(0^1) = 0^0 = 1. Thus a(4) = 3.
		

Crossrefs

Programs

  • Maple
    g:= proc(n, i) option remember; `if`(n=0, [0, 1], `if`(i<1, 0, (v->[v[1]-
          v[2], v[2]])(add(((l, h)-> [binomial(l[2]+l[1]+j-1, j)*(h[1]+h[2]),
          binomial(l[1]+j-1, j)*h[2]])(g(i-1$2), g(n-i*j, i-1)), j=0..n/i))))
        end:
    a:= n-> g(n-1$2)[1]:
    seq(a(n), n=0..40);
  • Mathematica
    f[l_, h_] := {Binomial[l[[2]] + l[[1]] + j - 1, j]*(h[[1]] + h[[2]]), Binomial[l[[1]] + j - 1, j]*h[[2]]};
    g[n_, i_] := g[n, i] = If[n == 0, {0, 1}, If[i < 1, {0, 0}, Function[v, {v[[1]] - v[[2]], v[[2]]}][Sum[f[g[i - 1, i - 1], g[n - i*j, i - 1]], {j, 0, Quotient[n, i]}]]]];
    a[n_] := g[n - 1, n - 1][[1]];
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 27 2019, after Alois P. Heinz *)

Formula

a(n) + A222379(n) = A000081(n).
a(n) - A222379(n) = A211192(n).
a(n) = Sum_{i=A087803(n-1)+1..A087803(n)} A306710(i).

A179505 n-th derivative of x^(x^(x^(x^x))) at x=1.

Original entry on oeis.org

1, 1, 2, 9, 56, 480, 4374, 47894, 574888, 7829424, 116392080, 1901059512, 33564909432, 639562529424, 13047133134840, 283976169754440, 6563364026374464, 160538113862231808, 4141949353327046592, 112396373034208003008, 3199752121483607518080
Offset: 0

Views

Author

Robert G. Wilson v, Jul 17 2010

Keywords

Crossrefs

Column k=17 of A215703.
Column k=5 of A277537.

Programs

  • Maple
    a:= n-> n!*coeff(series(subs(x=x+1, x^(x^(x^(x^x))) ), x, n+1), x, n):
    seq(a(n), n=0..25);  # Alois P. Heinz, Aug 21 2012
  • Mathematica
    f[n_] := D[x^(x^(x^(x^x))), {x, n}] /. x -> 1; Array[f, 16, 0]
    Range[0, 20]! CoefficientList[ Series[(1 + x)^(1 + x)^(1 + x)^(1 + x)^(1 + x), {x, 0, 20}], x] (* Robert G. Wilson v, Feb 03 2013 *)

Formula

E.g.f.: (x+1)^((x+1)^((x+1)^((x+1)^(x+1)))). - Alois P. Heinz, Aug 21 2012

Extensions

a(16)-a(20) from Alois P. Heinz, Aug 21 2012

A214569 Irregular triangle read by rows: T(n,k) is the number of rooted trees having n vertices and isomorphic (as rooted trees) to k ordered trees (n>=1, k>=1).

Original entry on oeis.org

1, 1, 2, 3, 1, 5, 3, 1, 6, 8, 4, 2, 10, 17, 7, 8, 1, 5, 11, 34, 16, 25, 3, 18, 0, 3, 1, 1, 0, 3, 16, 63, 27, 65, 6, 56, 1, 16, 5, 4, 0, 22, 0, 0, 1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 19, 111, 47, 154, 12, 138, 3, 65, 13, 13, 0, 95, 0, 0, 3, 5, 0, 13, 0, 8, 1, 0, 0, 13, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 26, 186, 73, 348, 18, 319, 6, 208, 35, 32, 0, 308, 0, 2, 13, 34, 0, 58, 0, 29, 1, 0, 0, 88, 0, 0, 1, 1, 0, 16, 0, 0, 0, 0, 1, 18, 0, 0, 0, 8, 0, 2, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6
Offset: 1

Views

Author

Emeric Deutsch, Jul 28 2012

Keywords

Comments

Row n contains A214570(n) entries.
T(n,1) = A003238(n).
Sum(T(n,k), k=1..n) = A000081(n) = number of rooted trees with n vertices.
Sum(k*T(n,k), k=1..n) = A000108(n-1) (the Catalan numbers).
T(n,k) is also the number of size k equivalence classes of function representations as x^x^...^x with n x's and parentheses inserted in all possible ways. T(4,2) = 1: (x^x)^(x^x) == (x^(x^x))^x; T(5,3) = 1: ((x^x)^x)^(x^x) == ((x^x)^(x^x))^x == ((x^(x^x))^x)^x. - Alois P. Heinz, Aug 31 2012

Examples

			Row 4 is 3,1: among the four rooted trees with 4 vertices the path tree P_4, the star tree K_{1,3}, and the tree in the shape of Y are isomorphic only to themselves, while A - B - C - D with root at B is isomorphic to itself and to A - B - C - D with root at C.
Triangle starts:
   1;
   1;
   2;
   3,  1;
   5,  3,  1;
   6,  8,  4,  2;
  10, 17,  7,  8, 1,  5;
  11, 34, 16, 25, 3, 18, 0, 3, 1, 1, 0, 3;
  ...
		

Crossrefs

Programs

  • Maple
    F:= proc(n) option remember; `if`(n=1, [x+1],
          [seq(seq(seq(f^g, g=F(n-i)), f=F(i)), i=1..n-1)])
        end:
    T:= proc(n) option remember; local i, l, p;
          l:= map(f->coeff(series(f, x, n+1), x, n), F(n)):
          p:= proc() 0 end: forget(p);
          for i in l do p(i):= p(i)+1 od:
          l:= map(p, l); forget(p);
          for i in l do p(i):= p(i)+1 od:
          seq(p(i)/i, i=1..max(l[]))
        end:
    seq(T(n), n=1..10);  # Alois P. Heinz, Aug 31 2012
  • Mathematica
    F[n_] := F[n] = If[n == 1, {x+1}, Flatten[Table[Table[Table[f^g, {g, F[n-i]}], {f, F[i]}], {i, 1, n-1}]]]; T[n_] := T[n] = Module[{i, l, p}, l = Map[Function[ {f}, Coefficient[Series[f, {x, 0, n+1}], x, n]], F[n]]; Clear[p]; p[] = 0; Do[ p[i] = p[i]+1 , {i, l}]; l = Map[p, l]; Clear[p]; p[] = 0; Do[p[i] = p[i]+1, {i, l}]; Table[p[i]/i, {i, 1, Max[l]}]]; Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)

Formula

No formula available. Entries have been obtained by counting (using Maple) the rooted trees (identified by their Matula-Goebel numbers) with the required properties (using A061775 and A206487).

A214571 Irregular triangle read by rows: T(n,k) is the number of ordered trees having n vertices and isomorphic (as rooted trees) to k ordered trees (n >= 1, k >= 1).

Original entry on oeis.org

1, 1, 2, 3, 2, 5, 6, 3, 6, 16, 12, 8, 10, 34, 21, 32, 5, 30, 11, 68, 48, 100, 15, 108, 0, 24, 9, 10, 0, 36, 16, 126, 81, 260, 30, 336, 7, 128, 45, 40, 0, 264, 0, 0, 15, 0, 0, 18, 0, 40, 0, 0, 0, 24, 19, 222, 141, 616, 60, 828, 21, 520, 117, 130, 0, 1140, 0, 0, 45, 80, 0, 234, 0, 160, 21, 0, 0, 312, 0, 0, 0, 0, 0, 120, 0, 0, 0, 0, 0, 36, 0, 0, 0, 40, 26, 372, 219, 1392, 90, 1914, 42, 1664, 315, 320, 0, 3696, 0, 28, 195, 544, 0, 1044, 0, 580, 21, 0, 0, 2112, 0, 0, 27, 28, 0, 480, 0, 0, 0, 0, 35, 648, 0, 0, 0, 320, 0, 84, 0, 0, 0, 0, 0, 240, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 360
Offset: 1

Views

Author

Emeric Deutsch, Jul 28 2012

Keywords

Comments

Row n contains A214570(n) entries.
T(n,1) = A003238(n).
Sum_{k=1..n} T(n,k) = A000108(n) (the Catalan numbers).
Sum_{k=1..n} T(n,k)/k = A000081(n) (the number of rooted trees with n vertices).
T(n,k) = k*A214569(n,k).
T(n,k) is also the number of function representations as x^x^...^x with n x's and parentheses inserted in all possible ways that are equivalent to (describe the same function as) k-1 other representations. T(4,2) = 2: (x^x)^(x^x), (x^(x^x))^x; T(5,3) = 3: ((x^x)^x)^(x^x), ((x^x)^(x^x))^x, ((x^(x^x))^x)^x. - Alois P. Heinz, Aug 31 2012

Examples

			Row 4 is 3,2: among the five ordered trees with 4 vertices the path tree P_4, the star tree K_{1,3}, and the tree in the shape of Y are isomorphic only to themselves, while A - B - C - D with root at B and A - B - C - D with root at C are isomorphic among themselves.
Triangle starts:
   1;
   1;
   2;
   3,   2;
   5,   6,   3;
   6,  16,  12,   8;
  10,  34,  21,  32,   5,  30;
  11,  68,  48, 100,  15, 108,   0,  24,   9,  10,   0,  36;
		

Crossrefs

Programs

  • Maple
    F:= proc(n) option remember; `if`(n=1, [x+1],
          [seq(seq(seq(f^g, g=F(n-i)), f=F(i)), i=1..n-1)])
        end:
    T:= proc(n) option remember; local i, l, p;
          l:= map(f->coeff(series(f, x, n+1), x, n), F(n)):
          p:= proc() 0 end: forget(p);
          for i in l do p(i):= p(i)+1 od:
          l:= map(p, l); forget(p);
          for i in l do p(i):= p(i)+1 od:
          seq(p(i), i=1..max(l[]))
        end:
    seq(T(n), n=1..10);  # Alois P. Heinz, Aug 31 2012
  • Mathematica
    F[n_] := F[n] = If[n == 1, {x+1}, Flatten[Table[Table[Table[f^g, {g, F[n-i]}], {f, F[i]}], {i, 1, n-1}]]]; T[n_] := T[n] = Module[{i, l, p}, l = Map[Function[ {f}, Coefficient[Series[f, {x, 0, n+1}], x, n]], F[n]]; Clear[p]; p[] = 0; Do[ p[i] = p[i]+1 , {i, l}]; l = Map[p, l]; Clear[p]; p[] = 0; Do[p[i] = p[i]+1, {i, l}]; Table[p[i], {i, 1, Max[l]}]]; Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)

Formula

No formula available. Entries have been obtained from T(n,k)= k*A214569(n,k).
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