A032128 -id:A032128 - cp's OEIS frontend

A319436 Number of palindromic plane trees with n nodes.

1, 1, 2, 3, 6, 10, 20, 35, 68, 122, 234, 426, 808, 1484, 2798, 5167, 9700, 17974, 33656, 62498, 116826, 217236, 405646, 754938, 1408736, 2623188, 4892848, 9114036, 16995110, 31664136, 59034488, 110004243, 205068892, 382156686, 712363344, 1327600346, 2474618434

Offset: 1

Gus Wiseman, Sep 18 2018

nonn

A rooted plane tree is palindromic if the sequence of branches directly under any given node is a palindrome.

			The a(7) = 20 palindromic plane trees:
  ((((((o))))))  (((((oo)))))  ((((ooo))))  (((oooo)))  ((ooooo))  (oooooo)
                 ((((o)(o))))  (((o(o)o)))  ((o(oo)o))  (o(ooo)o)
                 (((o))((o)))  ((o((o))o))  (o((oo))o)  (oo(o)oo)
                               (((o)o(o)))  ((oo)(oo))
                               (o(((o)))o)  ((o)oo(o))
                               ((o)(o)(o))  (o(o)(o)o)

Andrew Howroyd, Table of n, a(n) for n = 1..500
Gus Wiseman, The a(8) = 35 palindromic plane trees.
Gus Wiseman, The a(11) = 234 palindromic plane trees.

Cf. A000108, A000670, A001003, A005043, A008965, A025065, A032128, A118376, A242414, A317085, A317086, A317087, A319122, A319437.

panplane[n_]:=If[n==1,{{}},Join@@Table[Select[Tuples[panplane/@c],#==Reverse[#]&],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
Table[Length[panplane[n]],{n,10}]

PAL(p)={(1+p)/subst(1-p, x, x^2)}
seq(n)={my(p=O(1));for(i=1, n, p=PAL(x*p)); Vec(p)} \\ Andrew Howroyd, Sep 19 2018

a(n) ~ c * d^n, where d = 1.86383559155190653688720443906758855085492625375... and c = 0.24457511051198663873739022949952908293770055... - Vaclav Kotesovec, Nov 16 2021

A032130 Shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.

Original entry on oeis.org

2, 2, 5, 15, 52, 193, 765, 3143, 13323, 57670, 254040, 1134249, 5122124, 23349966, 107310784, 496633774, 2312539465, 10826481544, 50929829953, 240616214596, 1141195080020, 5431477088428, 25933525825389, 124185539096075, 596268057962349, 2869992942831031, 13845453533124431, 66934180769445444, 324218809545624984

Offset: 1

Christian G. Bower

nonn

From Petros Hadjicostas, Jan 14 2018: (Start)
For this sequence, if (b(n): n>=1) = BIK((a(n): n>=1)), then b(n) = a(n+1) for n>=1.
Let A(x) = Sum_{n>=1} a(n)*x^n be the g.f. for this sequence. For an explanation on how to derive the formula BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1 - A(x^2))) from Bower's formulae in the link below about transforms, see the comments for sequence A001224. (For that sequence, the roles of sequences (a(n): n>=1) and (b(n): n>=1) are reversed.)
(End)

			From _Petros Hadjicostas_, Jan 14 2018: (Start)
According to Bower's theory in the link above, we have boxes of different sizes and colors. The size of a box is determined by the number of balls it can hold. Two boxes of the same size and color are considered identical (indistinct and unlabeled). We place the boxes on a line that can be read in either direction; i.e., we have a reversible line.
Here, a(n) = number of colors a box holding n balls can be, while b(n) = number of ways of placing boxes in a line that can be read in either direction when the total number of balls is n.
Since a(1) = 2, a(2) = 2, a(3) = 5, a(4) = 15, etc.,  a box with 1 ball can be of 2 colors only, a box with 2 balls can be of 2 colors only, a box with 3 balls can be of 5 colors only, a box with 4 balls can be of 15 colors, and so on.
When we have n=3 balls, we have b(3) = a(4) = 15. Here, we consider three cases. In the first case, we have one box holding 3 balls and we have 5 possibilities. In the second case, we have a box with 2 balls and a box with 1 ball, and we have 2 x 2 = 4 possibilities here because the line is reversible (i.e., 21 is considered the same as 12). In the third case, we have three identical boxes each holding 1 ball and we have 6 possibilities (if the colors are a and b, we have the possibilities aaa, aab, aba, bba, bab, and bbb). Thus, b(3) = 5 + 4 + 6 = 15 = a(4).
When we have n=4 balls, we have b(4) = a(5) = 52. Here we consider 5 cases: a single box with 4 balls (a(4) = 15 possibilities); a box with 3 balls and a box with 1 ball (a(3) x a(1) = 5 x 2 = 10 possibilities); two identical boxes each with 2 balls (3 possibilities, aa, ab, and bb); a box with 2 balls and two identical boxes each with 1 ball (14 possibilities, see below); and 4 identical boxes each with 1 ball (10 possibilities, aaaa, aaab, aaba, aabb, abba, baab, abab, abbb, babb, bbbb). Hence, b(4) = 15 + 10 + 3 + 14 + 10 = 52 = a(5).
We explain the fourth case above in more detail. Here, we have a box with 2 balls and two identical boxes each with 1 ball. Let a and b be the two colors for the 1-ball boxes and A and B be the colors for the 2-ball boxes. Then we have the following 14 cases: Aaa, Aab, Abb, Baa, Bab, Bbb, aAa, aAb, bAb, aBa, aBb, bBb, abA, and abB. Note that Aab = baA <> abA = Aba and abB = Bba <> Bab = baB.
(End)

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)

Cf. A001224, A032131.

When a(1) = 1, we get sequence A032128.

m = 30; a[1] = 2; A[_] = 0;
Do[A[x_] = x(a[1]+(1/2)(A[x]/(1-A[x])+(A[x]+A[x^2])/(1-A[x^2]))) + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] // Rest (* Jean-François Alcover, Sep 17 2019 *)

BIK(p)={(1/(1-p) + (1+p)/subst(1-p, x, x^2))/2}
seq(n)={my(p=O(1));for(i=1, n, p=1+BIK(x*p)); Vec(p)} \\ Andrew Howroyd, Aug 30 2018

From Petros Hadjicostas, Jan 14 2018: (Start)
The sequence (a(n): n>=1) shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.
G.f.: If A(x) = Sum_{n>=1} a(n)*x^n, then -a(1) + A(x)/x = BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1-A(x^2))). Here a(1) = 2.
In general, if we let a(1) = c, we get:
a(2) = c,
a(3) = (1/2)*(c + 3)*c,
a(4) = (1/2)*(c + 3)*(c + 1)*c,
a(5) = (1/2)*(c^3 + 5*c^2 + 10*c + 4)*c,
a(6) = (1/4)*(2*c^4 + 15*c^3 + 38*c^2 + 37*c + 8)*c,
a(7) = (1/8)*(4*c^4 + 38*c^3 + 103*c^2 + 109*c + 22)*(c + 1)*c,
a(8) = (1/8)*(4*c^6 + 56*c^5 + 251*c^4 + 511*c^3 + 499*c^2 + 201*c + 22)*c,
and so on. No pattern is apparent in these formulae.
(End)

Name edited by Petros Hadjicostas, Jan 14 2018

a(24)-a(29) from Petros Hadjicostas, Jan 14 2018

A032131 Shifts left 2 places under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 7, 13, 31, 66, 160, 369, 907, 2191, 5461, 13558, 34209, 86426, 220359, 563475, 1449282, 3739365, 9688104, 25173917, 65621067, 171498288, 449361649, 1180078602, 3105740797, 8189749105, 21636207962, 57257857968, 151771200002, 402899862910, 1071076810324, 2851165864937

Offset: 1

Christian G. Bower

nonn

From Petros Hadjicostas, Jan 14 2018: (Start)
For this sequence, if (b(n): n>=1) = BIK((a(n): n>=1)), then b(n) = a(n+2) for n>=1.
Let A(x) = Sum_{n>=1} a(n)*x^n be the g.f. for this sequence. For an explanation on how to derive the formula BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1 - A(x^2))) from Bower's formulae in the link below about transforms, see the comments for sequence A001224. (For that sequence, the roles of sequences (a(n): n>=1) and (b(n): n>=1) are reversed.)
(End)

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)

Cf. A001224, A032128, A032130.

m = 36; a[1] = a[2] = 1; A[_] = 0;
Do[A[x_] = x^2 (a[1]/x + a[2] + (1/2)(A[x]/(1 - A[x]) + (A[x] + A[x^2])/(1 - A[x^2]))) + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] // Rest (* Jean-François Alcover, Sep 17 2019 *)

BIK(p)={(1/(1-p) + (1+p)/subst(1-p, x, x^2))/2}
seq(n)={my(p=1+O(x^(n%2)));for(i=1, n\2, p=1+x*BIK(x*p)); Vec(p)} \\ Andrew Howroyd, Aug 30 2018

From Petros Hadjicostas, Jan 14 2018: (Start)
G.f.: If A(x) = Sum_{n>=1} a(n)*x^n, then (A(x) - a(1)*x - a(2)*x^2)/x^2 = BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1-A(x^2))). Here, a(1) = a(2) = 1.
In general, we have:
a(3) = a(1),
a(4) = (1/2)*(a(1)^2 + a(1) + 2*a(2)),
a(5) = (1/2)*(a(1)^2 + a(1) + 2*a(2) + 2)*a(1),
a(6) = (1/2)*(a(1)^4 + 4*a(1)^2 + (3*a(1)^2 + a(1) + 3)*a(2) + a(2)^2 + a(1)),
a(7) = (1/2)*(a(1)^4 + 4*a(1)^2*a(2) + 6*a(1)^2 + 3*a(2)^2 + 3*a(1) + 7*a(2) + 2)*a(1),
and so on. No pattern is apparent here.
(End)

Name edited by Petros Hadjicostas, Jan 14 2018

a(31)-a(35) from Petros Hadjicostas, Jan 14 2018

A038035 Number of labeled dyslexic planted planar trees with n+1 nodes.

Original entry on oeis.org

1, 2, 9, 72, 840, 12780, 238770, 5281920, 134946000, 3909578400, 126638542800, 4535037460800, 177904622095200, 7586967310322400, 349479111223242000, 17292052928037888000, 914673660594613920000, 51506610632458293312000, 3076341001739003430432000

Offset: 1

Christian G. Bower, Sep 15 1998

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)
Index entries for sequences related to rooted trees

Cf. A005470, A032128.

m = 20;
CoefficientList[InverseSeries[2*x*(1 - x)/(2 - x^2) + O[x]^m], x]*Range[0, m - 1]! // Rest (* Jean-François Alcover, Sep 08 2019 *)

Vec(serlaplace(serreverse(2*x*(1 - x)/(2 - x^2) + O(x^20)))) \\ Andrew Howroyd, Sep 19 2018

Divides by n and shifts left under "BIJ" (reversible, indistinct, labeled) transform.

E.g.f.: series reversion of 2*x*(1 - x)/(2 - x^2). - Andrew Howroyd, Sep 19 2018

Terms a(16) and beyond from Andrew Howroyd, Sep 19 2018

A032129 Number of dyslexic rooted planar trees with n nodes.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 55, 146, 413, 1194, 3553, 10756, 33134, 103273, 325484, 1034734, 3314870, 10688513, 34662777, 112976023, 369876832, 1215811262, 4010932603, 13275356936, 44070010202, 146698487202, 489550622528, 1637472527602, 5488829461525, 18435194140301

Offset: 1

Christian G. Bower

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)
Index entries for sequences related to rooted trees

Cf. A032128.

BIK(p)={(1/(1-p) + (1+p)/subst(1-p, x, x^2))/2}
DIK(p,n)={(sum(d=1, n, eulerphi(d)/d*log(subst(1/(1+O(x*x^(n\d))-p), x, x^d))) + ((1+p)^2/(1-subst(p, x, x^2))-1)/2)/2}
seq(n)={my(p=O(1));for(i=1, n-1, p=BIK(x*p)); Vec(1+DIK(x*p, n))} \\ Andrew Howroyd, Aug 30 2018

"DIK" (bracelet, indistinct, unlabeled) transform of A032128 (shifted right one place).

Terms a(28) and beyond from Andrew Howroyd, Aug 30 2018

cp's OEIS Frontend

A319436 Number of palindromic plane trees with n nodes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A032130 Shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A032131 Shifts left 2 places under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = a(2) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A038035 Number of labeled dyslexic planted planar trees with n+1 nodes.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A032129 Number of dyslexic rooted planar trees with n nodes.

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

Extensions