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.

Previous Showing 21-26 of 26 results.

A245230 Triangle T(m,n), 1<=n<=m, read by rows: maximum frustration of complete bipartite graph K(m,n).

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 2, 3, 4, 0, 2, 3, 5, 7, 0, 3, 4, 7, 9, 11, 0, 3, 5, 8, 10, 13, 16, 0, 4, 6, 10, 12, 16, 19, 22
Offset: 1

Views

Author

Robert Israel, Jul 14 2014

Keywords

Comments

The maximum frustration of a graph is the maximum cardinality of a set of edges that contains at most half the edges of any cut-set. Another term that is used is "line index of imbalance". It is also equal to the covering radius of the coset code of the graph.
T(m,n) is symmetric in m and n, so only m>=n is listed here.
T(m,1) = 0.
T(m,2) = floor(m/2) = A004526(m).
T(m,3) = floor(3*m/4) = A057353(m).
T(m,4) = A245231(m).
T(m,5) = A245227(m).
T(m,6) = A245239(m).
T(m,7) = A245314(m).

Examples

			For m=n=3 a set of edges attaining the maximum cardinality T(3,3)=2 is {(1,4),(2,5)}.
Triangle starts
  0;
  0, 1;
  0, 1, 2;
  0, 2, 3, 4;
  0, 2, 3, 5, 7;
  0, 3, 4, 7, 9,  11;
  0, 3, 5, 8, 10, 13, 16;
  0, 4, 6, 10, 12, 16, 19, 22.
		

Crossrefs

Cf. A245231 (row/column 4), A245227 (row/column 5), A245239 (row/column 6), A245314 (row/column 7)

A307707 Lexicographically earliest sequence of nonnegative integers in which, for all k >= 0, there are exactly k pairs of consecutive terms whose sum is k.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8
Offset: 1

Views

Author

Eric Angelini and Jean-Marc Falcoz, Apr 23 2019

Keywords

Comments

The old definition was "Lexicographically earliest sequence starting with a(1) = 0 such that a(n) is the number of pairs of contiguous terms whose sum is a(n)".
From Paul Curtz, Apr 27 2019: This can be written as a triangle:
0
1 1
1 2 1
2 2 2 2
2 3 2 3 2
3 3 3 3 3 3
3 4 3 4 3 4 3
...

Crossrefs

Cf. A002024.
Cf. also A007590, A057353, A106466 and A238410.
For other versions see A307720 and A378117.

Programs

  • Mathematica
    m = 107; a[1]=0;
    a24[n_] := Ceiling[(Sqrt[8n+1]-1)/2];
    Array[a, m] /. Solve[Table[a[n] + a[n+1] == a24[n], {n, 1, m-1}]][[1]] (* Jean-François Alcover, Jun 02 2019, after Rémy Sigrist's formula *)
  • PARI
    v=0; rem=wanted=1; for (n=1, 107, print1 (v", "); v=wanted-v; if (rem--==0, rem=wanted++)) \\ Rémy Sigrist, Apr 23 2019

Formula

a(n) + a(n+1) = A002024(n). - Rémy Sigrist, Apr 24 2019
Let t_m = m*(m+1)/2. Write n = t_m - i with m >= 1 and 0 <= i < m. Then a(n) = m/2 if m is even, or if m is odd, a(n) = (m-1)/2 + (i-1 mod 2). - N. J. A. Sloane, Nov 16 2024

Extensions

Definition clarified by Rémy Sigrist and N. J. A. Sloane, Nov 17 2024

A047379 Numbers that are congruent to {0, 2, 4, 5} mod 7.

Original entry on oeis.org

0, 2, 4, 5, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23, 25, 26, 28, 30, 32, 33, 35, 37, 39, 40, 42, 44, 46, 47, 49, 51, 53, 54, 56, 58, 60, 61, 63, 65, 67, 68, 70, 72, 74, 75, 77, 79, 81, 82, 84, 86, 88, 89, 91, 93, 95, 96
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

Formula

a(n) = floor(floor((7n + 2)/2)/2).
a(n) = floor((7n-5)/4). - Gary Detlefs, Mar 07 2010
G.f.: x^2*(2+2*x+x^2+2*x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
From Wesley Ivan Hurt, Dec 03 2014: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5), n>5;
a(n) = (14*n-13-(-1)^n+2*i^n*(-1)^((3+(-1)^n)/4))/8, where i = sqrt(-1);
a(n) = A047215(n-1) - A057353(n-1). (End)

A090221 Array used for numerators of g.f.s for column sequences of array A090214 ((4,4)-Stirling2).

Original entry on oeis.org

1, 96, 72, 14400, 16, 38400, 3456000, 1, 27000, 22104000, 1270080000, 7200, 34905600, 16111872000, 682795008000, 856, 21154176, 48248363520, 15279164006400, 516193026048000, 48, 6064128, 54644474880, 78083415244800
Offset: 4

Views

Author

Wolfdieter Lang, Dec 01 2003

Keywords

Comments

The row length sequence for this array is A037915(k-4)= floor(3*(k-4)/4)+1, k>=4: [1, 1, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 10, 10, 11, ...].
The g.f. G(k,x) for the k-th column (with leading zeros) of array A090214 is given there. The recurrence is G(k,x) = x*sum(binomial(k-r,4-r)*fallfac(4,4-r)*G(k-r,x),r=1..4))/(1-fallfac(k,4)*x), k>=4, with inputs G(k,x)=0 for k=1,2,3 and G(4,x)=x/(1-4!*x); where fallfac(n,m) := A008279(n,m) (falling factorials with fallfac(n,0) := 1). Computed from the Blasiak et al. reference, eqs. (20) and (21) with r=4: recurrence for S_{4,4}(n,k).

Examples

			[1]; [96]; [72,14400]; [16,38400,3456000]; [1,27000,22104000,1270080000]; ...
G(5,x)/x^2 = 96/((1-4!*x)*(1-5*4*3*2*x)). kmax(5)=0, hence P(5,x)=a(5,0)=96; x^2 from x^ceiling(5/4).
		

Formula

a(k, n) from: sum(a(k, n)*x^n, n=0..kmax(k)) = G(k, x)* product(1-fallfac(p, 4)*x, p=4..k)/x^ceiling(k/4), k>=4, with G(k, x) defined from the recurrence given above and kmax(k) := A057353(k-4)= floor(3*(k-4)/4)= A037915(k-4)-1.

A187228 Rank transform of the sequence floor(3n/4); complement of A187229.

Original entry on oeis.org

1, 2, 4, 6, 7, 8, 10, 11, 12, 14, 16, 18, 19, 20, 22, 24, 25, 27, 28, 30, 31, 32, 34, 35, 36, 38, 40, 42, 43, 44, 46, 47, 48, 50, 52, 53, 54, 56, 58, 59, 60, 62, 64, 66, 67, 68, 70, 72, 73, 75, 76, 78, 79, 80, 82, 83, 84, 86, 88, 90, 91, 92, 94, 96, 97, 99, 100, 102, 103, 104, 106, 108, 109, 111, 112, 114, 115, 116, 118, 120, 121, 123, 124, 126, 127, 128, 130, 131, 132, 134, 136, 138, 139, 140, 142, 143, 144, 146, 148, 149, 150
Offset: 1

Views

Author

Clark Kimberling, Mar 07 2011

Keywords

Comments

See A187224.

Crossrefs

Programs

  • Mathematica
    seqA=Table[Floor[3n/4],{n,1,220}] (*A057353*)
    seqB=Table[n,{n,1,220}];(*A000027*)
    jointRank[{seqA_,seqB_}]:={Flatten@Position[#1,{,1}],Flatten@Position[#1,{,2}]}&[Sort@Flatten[{{#1,1}&/@seqA,{#1,2}&/@seqB},1]];
    limseqU=FixedPoint[jointRank[{seqA,#1[[1]]}]&,jointRank[{seqA,seqB}]][[1]] (*A187228*)
    Complement[Range[Length[seqA]],limseqU] (*A187229*)
    (*by Peter J. C. Moses, Mar 07 2011*)

A187322 a(n) = floor(n/2) + floor(3*n/4).

Original entry on oeis.org

0, 0, 2, 3, 5, 5, 7, 8, 10, 10, 12, 13, 15, 15, 17, 18, 20, 20, 22, 23, 25, 25, 27, 28, 30, 30, 32, 33, 35, 35, 37, 38, 40, 40, 42, 43, 45, 45, 47, 48, 50, 50, 52, 53, 55, 55, 57, 58, 60, 60, 62, 63, 65, 65, 67, 68, 70, 70, 72, 73, 75, 75, 77, 78, 80, 80, 82, 83, 85, 85, 87, 88, 90, 90, 92, 93, 95, 95, 97
Offset: 0

Views

Author

Clark Kimberling, Mar 08 2011

Keywords

Comments

List of quadruples [5*k, 5*k, 5*k+2, 5*k+3]. - Luce ETIENNE, Aug 14 2017

Crossrefs

Programs

  • Mathematica
    Table[Floor[n/2]+Floor[3n/4], {n,0,120}]
    LinearRecurrence[{1,0,0,1,-1},{0,0,2,3,5},80] (* Harvey P. Dale, Dec 05 2018 *)
  • PARI
    a(n) = n\2 + 3*n\4; \\ Altug Alkan, Aug 14 2017
    
  • PARI
    concat(vector(2), Vec(x^2*(2 + x + 2*x^2) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^100))) \\ Colin Barker, Aug 14 2017
    
  • Python
    def A187322(n): return (n>>1)+(3*n>>2) # Chai Wah Wu, Jan 31 2023

Formula

a(n) = A004526(n) + A057353(n). - Michel Marcus, Aug 14 2017
a(n) = (10*n-5+3*cos(n*Pi)+2*(cos(n*Pi/2)-sin(n*Pi/2)))/8. - Luce ETIENNE, Aug 14 2017
From Colin Barker, Aug 14 2017: (Start)
G.f.: x^2*(2 + x + 2*x^2) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
(End)
Previous Showing 21-26 of 26 results.