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 1-3 of 3 results.

A176085 a(n) = A136431(n,n).

Original entry on oeis.org

0, 1, 3, 11, 41, 155, 591, 2267, 8735, 33775, 130965, 509015, 1982269, 7732659, 30208749, 118167055, 462760369, 1814091011, 7118044023, 27952660883, 109853552255, 432021606103, 1700093447847, 6694137523051, 26372544576331, 103950885100775, 409928481296331
Offset: 0

Views

Author

Paul Curtz, Apr 08 2010

Keywords

Comments

a(n+1) is also the number of sequences of length 2n obeying the regular expression "0^* (1 or 2)^* 3^*" and having sum 3n. For example, a(3)=11 because of the sequences 0033, 0123, 0213, 0222, 1113, 1122, 1212, 1221, 2112, 2121, 2211. - Don Knuth, May 11 2016

Links

Crossrefs

Cf. A000045, A054441, A108617.

Programs

  • GAP
    List([0..30], n-> Sum([0..n], k-> Binomial(2*n-k-1, n-k)*Fibonacci(k) )); # G. C. Greubel, Nov 28 2019
  • Magma
    [(&+[Binomial(2*n-k-1, n-k)*Fibonacci(k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Nov 28 2019
  • Maple
    with(combinat); seq( add(binomial(2*n-k-1, n-k)*fibonacci(k), k=0..n), n=0..30); # G. C. Greubel, Nov 28 2019
1/(sqrt(1 - 4*x) + 1/x - 4): series(%, x, 27):
seq(coeff(%, x, k), k=0..26); # Peter Luschny, May 29 2021
  • Mathematica
    t[n_, k_]:= CoefficientList[ Series[x/(1-x-x^2)/(1-x)^k, {x,0,k}], x][[k+1]]; Array[ t[#, #] &, 20]
Table[Sum[Binomial[2*n-k-1, n-k]*Fibonacci[k], {k,0,n}], {n,0,30}] (* G. C. Greubel, Nov 28 2019 *)
  • Maxima
    a(n):=sum(fib(k)*binomial(2*n-k-1,n-k),k,1,n); /*  Vladimir Kruchinin, Mar 17 2016 */
  • PARI
    a(n) = sum(k=1, n, fibonacci(k)*binomial(2*n-k-1, n-k)) \\ Michel Marcus, Mar 17 2016
  • Sage
    [sum(binomial(2*n-k-1, n-k)*fibonacci(k) for k in (0..n)) for n in (0..30)] # G. C. Greubel, Nov 28 2019

Formula

a(n+1) - 4*a(n) = -A081696(n-1).
From Vaclav Kotesovec, Oct 21 2012: (Start)
G.f.: x*(x-sqrt(1-4*x))/(sqrt(1-4*x)*(x^2+4*x-1)).
Recurrence: (n-2)*a(n) = 2*(4*n-9)*a(n-1) - (15*n-38)*a(n-2) - 2*(2*n-5)* a(n-3).
a(n) ~ 4^n/sqrt(Pi*n). (End)
a(n) = Sum_{k=1..n} (F(k)*binomial(2*n-k-1,n-k)), where F(k) = A000045(k). - Vladimir Kruchinin, Mar 17 2016
Simpler g.f.: x/sqrt(1-4*x)/(x+sqrt(1-4*x)). - Don Knuth, May 11 2016
a(n) = A000045(3*n) - A054441(n). - Hrishikesh Venkataraman, May 27 2021
a(n) = 4*a(n-1) + a(n-2) - binomial(2*n-4,n-2) for n>=2. - Hrishikesh Venkataraman, Jul 02 2021
a(n) = A108617(2n,n)/2. - Alois P. Heinz, Jan 26 2025

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181
Offset: 1

Views

Author

Yuriy Sibirmovsky, Sep 12 2016

Keywords

Comments

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

Examples

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Links

Crossrefs

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Programs

  • Mathematica
    Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid
  • PARI
    T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Formula

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A026677 T(n,0) + T(n,1) + ... + T(n,n), T given by A026670.

Original entry on oeis.org

1, 2, 5, 10, 23, 46, 103, 206, 455, 910, 1993, 3986, 8679, 17358, 37633, 75266, 162643, 325286, 701075, 1402150, 3015563, 6031126, 12948083, 25896166, 55513327, 111026654, 237705547, 475411094, 1016736115, 2033472230
Offset: 0

Views

Author

Clark Kimberling

Keywords

Crossrefs

Partial sums of A026733.

Formula

G.f.: F(x^2) + 2xF(x^2), where xF(x) = g.f. of A054441. - Ralf Stephan, Feb 05 2004
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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