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.

Showing 1-4 of 4 results.

A077419 Largest Whitney number of Fibonacci lattices J(Z_n).

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 5, 7, 11, 17, 26, 40, 63, 97, 153, 238, 376, 587, 931, 1458, 2317, 3640, 5794, 9124, 14545, 22951, 36631, 57904, 92512, 146461, 234205, 371281, 594169, 943045, 1510192, 2399460, 3844787, 6114555, 9802895, 15603339, 25027296
Offset: 0

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Author

N. J. A. Sloane, Jan 19 2003

Keywords

Comments

A051286 and A051291, interleaved. a(n) is the maximal element in the n-th row of A079487 or A123245 and in the (n+2)-th row of A078807 or A078808. - Andrey Zabolotskiy, Sep 21 2017

Programs

  • Maple
    with(FormalPowerSeries): with(LREtools): # requires Maple 2022
    gf:= (1 + 2*x + 2*x^4 - x^6 - (1-x^2)*sqrt(1 - 2*x^2 - x^4 - 2*x^6 + x^8))/(2*x*sqrt(1 - 2*x^2 - x^4 - 2*x^6 + x^8));
    re:= FindRE(gf,x,a(n));
    inits:= {seq(a(i-1)=[1,1,1,2,2,3,5,7,11,17,26,40,63,97, 153][i],i=1..14)};
    rm:=  (n+1)*a(n) +(n-2)*a(n-1) +2*(-n+1)*a(n-2) +2*(-n+1)*a(n-3) +(-n-3)*a(n-4) +(-n+8)*a(n-5) +2*(-n+6)*a(n-6) +2*(-n+7)*a(n-7) +(n-9)*a(n-8) +(n-10)*a(n-9)=0;
    minre:= MinimalRecurrence(re, a(n), inits); minrm:= MinimalRecurrence(rm, a(n), inits); # shows that Mathar's recurrence is equivalent
    f:= REtoproc(re,a(n),inits); seq(f(n),n=0..40); # Georg Fischer, Oct 22 2022
  • Mathematica
    gf[x_] = (1 + 2 x + 2 x^4 - x^6 - (1 - x^2) Sqrt[1 - 2 x^2 - x^4 - 2 x^6 + x^8])/(2 x Sqrt[1 - 2 x^2 - x^4 - 2 x^6 + x^8]);
    Table[SeriesCoefficient[gf[x], {x, 0, n}], {n, 0, 40}] (* Hugo Pfoertner, Oct 22 2022 *)

Formula

G.f.: (1 + 2 x + 2 x^4 - x^6 - (1-x^2) sqrt(1 - 2 x^2 - x^4 - 2 x^6 + x^8) )/(2x sqrt(1 - 2 x^2 - x^4 - 2 x^6 + x^8)). - Emanuele Munarini, Mar 05 2007
a(n) ~ phi^(n+2) / (5^(1/4) * sqrt(2*Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 22 2017
D-finite with recurrence: (n+1)*a(n) +(n-2)*a(n-1) +2*(-n+1)*a(n-2) +2*(-n+1)*a(n-3) +(-n-3)*a(n-4) +(-n+8)*a(n-5) +2*(-n+6)*a(n-6) +2*(-n+7)*a(n-7) +(n-9)*a(n-8) +(n-10)*a(n-9)=0. - R. J. Mathar, Nov 19 2019

Extensions

More terms from Emanuele Munarini, Mar 05 2007

A078807 Triangular array T given by T(n,k) = number of 01-words of length n having exactly k 1's, all runlengths odd and first letter 0.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 2, 1, 1, 0, 0, 1, 2, 2, 2, 1, 1, 3, 3, 3, 2, 1, 0, 0, 1, 3, 4, 5, 4, 3, 1, 1, 4, 6, 7, 7, 5, 3, 1, 0, 0, 1, 4, 7, 10, 11, 10, 7, 4, 1, 1, 5, 10, 14, 17, 16, 13, 8, 4, 1, 0, 0, 1, 5, 11, 18, 24, 26, 24, 18, 11, 5, 1, 1, 6, 15, 25, 35, 40, 39, 32, 22, 12, 5, 1, 0, 0
Offset: 1

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Author

Clark Kimberling, Dec 07 2002

Keywords

Comments

Row sums: 1,1,2,3,5,8,13,..., the Fibonacci numbers (A000045).

Examples

			T(6,2) counts the words 010001 and 000101. Top of triangle:
1 = T(1,0)
0 1 = T(2,0) T(2,1)
1 1 0
0 1 1 1
1 2 1 1 0
		

References

  • Clark Kimberling, Binary words with restricted repetitions and associated compositions of integers, in Applications of Fibonacci Numbers, vol.10, Proceedings of the Eleventh International Conference on Fibonacci Numbers and Their Applications, William Webb, editor, Congressus Numerantium, Winnipeg, Manitoba 194 (2009) 141-151.

Crossrefs

Formula

T(n, k)=T(n-1, n-k-1)+T(n-3, n-k-3)+...+T(n-2m-1, n-k-2m-1), where m=[(n-1)/2] and (by definition) T(i, j)=0 if i<0 or j<0 or i=j.

Extensions

Row 0 removed to stick to the triangle format by Andrey Zabolotskiy, Sep 22 2017

A364366 An irregular triangle read by rows, the 3rd row-symmetric Fibonaccian triangle: T(n,k) is the Whitney number of level k of the (3,n)-th symmetric Fibonaccian lattice (0 <= n, 0 <= k <= 2*n).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 4, 5, 4, 3, 1, 1, 4, 7, 10, 11, 10, 7, 4, 1, 1, 5, 11, 18, 24, 26, 24, 18, 11, 5, 1, 1, 6, 16, 30, 46, 58, 63, 58, 46, 30, 16, 6, 1, 1, 7, 22, 47, 81, 116, 143, 158, 143, 116, 81, 47, 22, 7, 1
Offset: 0

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Author

Keywords

Comments

For integers m and n (m >= 2, n > 0), let L be the set of n-tuples S=(S(1),...,S(n)) with each S(j) in {(j-1)*m+1,(j-1)*m+2,...,j*m} and such that S has no consecutive integers. Partially order these '(m,n) Fibonaccian strings' comprising L by the rule R <= S iff R(j) >= S(j) for 1 <= j <= n (so, 'lightest' n-tuples are at the top of the Hasse diagram for L). Then L is a self-dual distributive lattice, the '(m,n)-th symmetric Fibonaccian lattice'. When n=1, L is a chain with m elements. Now allow n=0; in this case, regard L to be a singleton set. Let p(n,x) be the rank generating function of L, so p(n,1)=|L|, p(0,x)=1, and p(1,x)=1+x+...+x^(m-1). For n >= 2, the fact that p(n,x) = p(1,x)*p(n-1,x) - x^(m-1)*p(n-2,x) can be deduced from a recurrence of Whitney numbers of symmetric Fibonaccian lattices proved in Proposition 2.1 of [Donnelly, Dunkum, Lišková, and Nance, 2023].
The (m,n)-th symmetric Fibonaccian lattice realizes a p(n,1)-dimensional representation of the special linear Lie algebra sl(m,C). The representation is reducible exactly when m >= 3 and n >= 3. The polynomial p(n,x) is a natural specialization of the character of this representation, where the latter can be identified as a certain skew Schur function. In [Donnelly, Dunkum, Lišková, and Nance, 2023], these representations are uniformly constructed (as an application of [Donnelly and Dunkum, 2022]) and explicit formulas for p(n,x) are given.
In [Donnelly, Dunkum, Lišková, and Nance, 2023], the (m,n)-th symmetric Fibonaccian lattice L is also described using semistandard tableaux of a specific ribbon shape; the irreducible components of the associated sl(m,C)-representation are in one-to-one correspondence with what are called the 'ballot-admissible' (aka Littlewood-Richardson) tableaux. In terms of Fibonaccian strings, an element S = (S(1),...,S(n)) in L is ballot-admissible iff for any integer q between 1 and n (inclusive) and any integer r between 1 and m-1 (inclusive), the following integer quantity is nonnegative: Sum_{k=n+1-q..n}([n+1-k is odd]*([r+(k-1)*m = S(k)] - [r+(k-1)*m+1 = S(k)]) + [n+1-k is even]*([k*m-r = S(k)]-[k*m+1-r = S(k)])), where '[]' denotes the Iverson bracket. Enumerating the ballot-admissible tableaux or Fibonaccian strings in L seems to be an interesting problem when m >= 3; when m=3, the sizes of the sets of ballot-admissible tableaux conjecturally agree with A004148.
In this OEIS entry, we have m=3. Let L be the (3,n)-th symmetric Fibonaccian lattice. When n=0, we have T(0,0) = |L| = 1. When n=1, we have T(1,0) = T(1,1) = T(1,2) = 1 and p(1,x) = 1+x+x^2, since L is a chain with 3 elements. For n >= 2, we have, by definition, p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k. The Whitney number T(n,k) is the number of (3,n) Fibonaccian strings S=(S(1),...,S(n)) whose coordinate sum S(1)+...+S(n) is equal to 3*(n*(n+1)/2)-k. This irregular triangle of T(n,k)'s is obtained from the regular triangles in entries A079487 and A123245 by removing all even length rows (with the exception of the row of length two, all such even length rows are 'asymmetric').
For m=4, see A364367. For m=5, see A364368. When m=2, the (2,n)-th symmetric Fibonaccian lattice is a chain with n+1 elements and rank generating function 1+x+...+x^(n-1)+x^n. Therefore, the 2nd row-symmetric Fibonaccian triangle is a regular triangle of 1's. The 1st row-symmetric Fibonaccian 'triangle' is regarded to be the signed sequence 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0, ... (A010892). 'Gibonaccian' versions of such triangles are considered in [Donnelly, Dunkum, Huber, and Knupp, 2021].

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k >= 0) starts as follows:
  1;
  1,   1,   1;
  1,   2,   2,   2,   1;
  1,   3,   4,   5,   4,   3,   1;
  1,   4,   7,  10,  11,  10,   7,   4,   1;
  1,   5,  11,  18,  24,  26,  24,  18,  11,   5,   1;
  1,   6,  16,  30,  46,  58,  63,  58,  46,  30,  16,   6,   1;
  1,   7,  22,  47,  81, 116, 143, 158, 143, 116,  81,  47,  22,   7,   1;
...
Below are the 21 (3,3) Fibonaccian strings (organized by rank level) that comprise the (3,3)rd symmetric Fibonaccian lattice:
rank=6: (1,4,7)
rank=5: (1,4,8)  (1,5,7)  (2,4,7)
rank=4: (1,4,9)  (1,5,8)  (2,4,8)  (2,5,7)
rank=3: (1,5,9)  (1,6,8)  (2,4,9)  (2,5,8)  (3,5,7)
rank=2: (1,6,9)  (2,5,9)  (2,6,8)  (3,5,8)
rank=1: (2,6,9)  (3,5,9)  (3,6,8)
rank=0: (3,6,9)
The triples (3,4,7), (3,4,8), (3,4,9), (1,6,7), (2,6,7), and (3,6,7) are disallowed as (3,3) Fibonaccian strings since each contains consecutive integers.
In the (3,5)th symmetric Fibonaccian lattice, rank level 8 consists of exactly the (3,5) Fibonaccian strings whose coordinate sum is 3*(5*(5+1)/2)-8=37: (1,4,7,10,15), (1,4,7,11,14), (1,4,8,10,14), (1,4,8,11,13), (1,5,7,10,14), (1,5,7,11,13), (1,5,8,10,13), (2,4,7,10,14), (2,4,7,11,13), (2,4,8,10,13), and (2,5,7,10,13), confirming that T(5,8)=11.
		

Crossrefs

Sum of row n (n >= 0) is A001906(n+1), cf. row n=3 of the array A316269.

Formula

With T(0,0)=1, then T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k) - T(n-2,k-2) for n >= 1 and 0 <= k <= 2*n, understanding T(i,j) to be zero when j < 0 or j > 2*i. That the preceding recurrence holds is equivalent to the identity p(n,x) = (1+x+x^2)*p(n-1,x) - x^2*p(n-2,x) for n >= 1, where p(0,x)=1 and p(-1,x) is taken to be 0.

A323670 Irregular triangle of the coefficients P_(n,j) of Morier-Genoud and Ovsienko's polynomials P_n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 3, 2, 1, 1, 3, 5, 6, 6, 5, 2, 1, 1, 3, 7, 11, 13, 13, 11, 7, 3, 1, 1, 4, 10, 18, 25, 29, 29, 24, 16, 9, 3, 1, 1, 4, 12, 25, 41, 56, 65, 65, 56, 41, 25, 12, 4, 1, 1, 5, 16, 37, 67, 101, 131, 148, 146, 126, 95, 61, 32, 14, 4, 1, 1
Offset: 0

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Author

Michael De Vlieger, Jan 23 2019

Keywords

Comments

A q-deformation of the Pell numbers.

Examples

			Triangle begins:
1
1   1
1   2   1   1
1   2   3   3   2   1
1   3   5   6   6   5   2   1
1   3   7  11  13  13  11   7   3   1
1   4  10  18  25  29  29  24  16   9   3   1
...
		

Crossrefs

Formula

Coefficients of the polynomials defined by p(n) = p(n-2)*(4,2)_q - q^4*p(n-4) where (4,2)_q = 1+q+2*q^2+q^3+q^4, with p(0)=0, p(1)=1, p(2)=1+q, p(3)=1+q+2*q^2+q^3.
Showing 1-4 of 4 results.