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.

User: Sarah Nibs

Sarah Nibs's wiki page.

Sarah Nibs has authored 3 sequences.

A237585 Number of structures of size n in class A = o x (o + MSET(A)) where o is a neutral structure of size 1.

Original entry on oeis.org

0, 1, 2, 3, 6, 15, 36, 94, 245, 663, 1815, 5062, 14269, 40706, 117103, 339673, 991834, 2913869, 8605576, 25536300, 76096896, 227634717, 683296679, 2057540487, 6213495745, 18813535942, 57103173296, 173710272584, 529534793886, 1617347972250, 4948744120771
Offset: 0

Author

Sarah Nibs, Feb 09 2014

Keywords

Comments

MSET(A) is the multi-choose function: pick any number of unlabeled structures in A with repetition allowed.
Interpreting the neutral structure of size 1 as a single pointer dereference, A is the class of A-pointers either to null pointers or to a multiset of unlabeled A-pointers, where the size of a pointer is the number of dereferences required to resolve the entire structure, so a null pointer has size 1 and an A-pointer to a null pointer has size 2 and an A-pointer to {A-pointer(null), A-pointer(null), A-pointer({A-pointer(null)})} has size 1+((1+1)+(1+1)+(1+(1+1)))=8.
a(n) is the number of rooted trees of weight n where leaves can have either weight 1 or 2 and non-leaves have weight 1. - Andrew Howroyd, Mar 02 2020

Examples

			For n = 3 the a(3)=3 pointers are the pointer to the multiset of exactly the pointer to the null pointer, the pointer to the multiset of twice the pointer to the empty multiset, and the pointer to the multiset of exactly the pointer to the multiset of the pointer to the empty multiset.
From _Andrew Howroyd_, Mar 02 2020: (Start)
The a(2) = 2 trees are: 2, (1).
The a(3) = 3 trees are: (2), (11), ((1)).
The a(4) = 6 trees are: ((2)), (12), (111), ((11)), (1(1)), (((1))).
(End)
		

Programs

  • PARI
    EulerT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v, n, 1/n))))-1, -#v)}
    seq(n)={my(v=[1]); for(n=2, n, v=concat([1], EulerT(v)); v[2]++); concat([0],v)} \\ Andrew Howroyd, Mar 02 2020

Formula

G.f. A(x) satisfies: A(x) = x * (x + exp(A(x) + A(x^2)/2 + A(x^3)/3 + A(x^4)/4 + ...)). - Ilya Gutkovskiy, Jun 11 2021

Extensions

Terms a(21) and beyond from Andrew Howroyd, Mar 02 2020

A200544 Number of distinct bags of distinct sequences of 1s and 2s such that the sum of all terms is n.

Original entry on oeis.org

1, 1, 3, 6, 14, 28, 61, 122, 253, 505, 1017, 2008, 3976, 7769, 15169, 29379, 56751, 108993, 208725, 397913, 756385, 1432578, 2705744, 5094749, 9568504, 17922756, 33492061, 62438472, 116151352, 215612548, 399451325, 738612472, 1363261171, 2511748010, 4620024202
Offset: 0

Author

Sarah Nibs, Nov 18 2011

Keywords

Comments

This is the number of distinct ways to build minimal Jenga towers out of n blocks. The number of distinct ways to build a single minimal Jenga tower out of n blocks is the Fibonacci number F(n+1) (A000045(n+1)).
To calculate this, first create all partitions of n.
An example partition, for n=4, is
1 1 1 1
1 1 2
1 3
2 2
4
Then each set of towers of the same size gets a configuration. For 2 2 2, for instance, there are two possibilities for each tower (a single level with two blocks or two levels with one block each) but the total possibilities is not 2*2*2=8, since the configuration "1/1,2,2" is the same as "2,1/1,2". Instead we want to choose three towers with repetition from two possibilities which is 3+2-1 choose 3, aka 4C3 = 4.
Multiply all the sets of towers of the same size and sum over partitions for the result.
For n=4, then, 1 1 2 becomes "1 with multiplicity 2" and "2 with multiplicity 1".
There is f(1+1)=1 way to build a tower of size 1, and f(1+1)+2-1 choose 2 = 2C2 = 1 way to build 2 towers of size 1. f(2+1)=2 ways to build a tower of size 2. 1 1 2 has 1*2=2 ways to be built. Sum over each of the 5 partitions of n=4.
This is apparently the limit of the row-reversed rows of the Multiset transform T(n,k) of the Fibonacci sequence in A337009, a(k) = lim_{n->oo} T(n,n-k). - R. J. Mathar, Aug 10 2020

Examples

			For n = 4, a(4)=14 and the bags are: 1/1/1/1; 1/1/1,1; 1/1/2; 1/1,1,1; 1/1,2; 1/2,1; 1,1/1,1; 1,1/2; 2/1,1; 2/2; 1,1,1,1; 1,1,2; 1,2,1; 2,1,1.
		

Crossrefs

Programs

  • Maple
    with(numtheory):with(combinat):
    a:= proc(n) option remember; `if`(n=0, 1, add(add(d*
          fibonacci(d+1), d=divisors(j))*a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Nov 05 2013
  • Mathematica
    CoefficientList[Series[Product[1/(1-x^k)^Fibonacci[k+1], {k, 1, 40}], {x, 0, 40}], x] (* Vaclav Kotesovec, Aug 05 2015 *)
  • SageMath
    # uses[EulerTransform from A166861]
    a = BinaryRecurrenceSequence(1, 1, 1)
    b = EulerTransform(a)
    print([b(n) for n in range(35)]) # Peter Luschny, Nov 11 2020

Formula

sum{m1*a1+m2*a2+...+mk*ak}(prod{k}(binomial(A000045[ak + 1]+mk-1,mk))).
G.f.: Product_{s>=1}(sum{d>=0}(binomial(F(s+1)+d-1,d)*x^(d*s))). - Sarah Nibs, Oct 21 2013
Euler Transform of A000045 starting at index 2, i.e. EULER(1, 2, 3, 5, 8, 13, ...). - Sarah Nibs, Nov 05 2013
a(n) ~ phi^(n+1/4) / (2 * sqrt(Pi) * 5^(1/8) * n^(3/4)) * exp(phi/10 - 3/5 + 2*5^(-1/4)*sqrt(phi*n) + s), where s = Sum_{k>=2} (1+phi^k) / ((phi^(2*k) - phi^k - 1)*k) = 0.7902214013751085262994702391769374769675268259229550490716908... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 06 2015
a(n) = A337009(2*n,n). - Alois P. Heinz, Apr 30 2023

Extensions

Corrected terms from n=8 and onwards by Sarah Nibs, Oct 18 2013
C# program corrected and made much more efficient by Sarah Nibs, Oct 18 2013

A122536 Number of binary sequences of length n with no initial repeats (or, with no final repeats).

Original entry on oeis.org

2, 2, 4, 6, 12, 20, 40, 74, 148, 286, 572, 1124, 2248, 4460, 8920, 17768, 35536, 70930, 141860, 283440, 566880, 1133200, 2266400, 4531686, 9063372, 18124522, 36249044, 72493652, 144987304, 289965744
Offset: 1

Author

Sarah Nibs, Sep 18 2006

Keywords

Comments

An initial repeat of a string S is a number k>=1 such that S(i)=S(i+k) for i=0..k-1. In other words, the first k symbols are the same as the next k symbols, e.g., ABCDABCDZQQ has an initial repeat of size 4.
Equivalently, this is the number of binary sequences of length n with curling number 1. See A216955. - N. J. A. Sloane, Sep 26 2012

Examples

			a(4)=6: 0100, 0110, 0111, 1000, 1001 and 1011. (But not 00**, 11**, 0101, 1010.)
		

Crossrefs

Twice A093371. Leading column of each of the triangles A216955, A217209, A218869, A218870. Different from, but easily confused with, A003000 and A216957. - N. J. A. Sloane, Sep 26 2012
See A121880 for difference from 2^n.

Formula

Conjecture: a_n ~ C * 2^n where C is 0.27004339525895354325... [Chaffin, Linderman, Sloane, Wilks, 2012]
a(2n+1)=2*a(2n) = A211965(n+1), a(2n)=2*a(2n-1)-A216958(n) = A211966(n). - N. J. A. Sloane, Sep 28 2012
a(1) = 2; a(2n) = 2*[a(2n-1) - A216959(n)], n >= 1. - Daniel Forgues, Feb 25 2015

Extensions

a(31)-a(71) computed from recurrence and the first 30 terms of A216958 by N. J. A. Sloane, Sep 28 2012, Oct 25 2012