A005314 -id:A005314 - cp's OEIS frontend

A124279 Riordan array (1/(1-x),x(1-x+x^2)/(1-x)).

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 8, 7, 5, 1, 1, 1, 6, 12, 13, 9, 6, 1, 1, 1, 7, 17, 22, 19, 11, 7, 1, 1, 1, 8, 23, 35, 35, 26, 13, 8, 1, 1, 1, 9, 30, 53

Offset: 0

Paul Barry, Oct 24 2006

Row sums are A005314. Diagonal sums are A124280. T(2n,n) is A002426.

Reversal of A124445. - Paul Barry, Nov 01 2006

			Triangle begins
1,
1, 1,
1, 1, 1,
1, 2, 1, 1,
1, 3, 3, 1, 1,
1, 4, 5, 4, 1, 1,
1, 5, 8, 7, 5, 1, 1

Number triangle T(n,k)=sum{j=0..n-k, C(j,n-k-j)C(k,n-k-j)}

A124445 Expansion of 1/(1-x-xy+x^2y-x^3*y^2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 7, 8, 5, 1, 1, 1, 6, 9, 13, 12, 6, 1, 1, 1, 7, 11, 19, 22, 17, 7, 1, 1, 1, 8, 13, 26, 35, 35, 23, 8, 1, 1, 1, 9, 15, 34, 51, 61, 53, 30, 9, 1

Offset: 0

Paul Barry, Nov 01 2006

Row sums are A107293(n+4). Diagonal sums are A005314(n+1).

Reversal of A124279. T(2n,n) is A002426. - Paul Barry, Nov 01 2006

			Triangle begins:
1,
1, 1,
1, 1, 1,
1, 1, 2, 1,
1, 1, 3, 3, 1,
1, 1, 4, 5, 4, 1,
1, 1, 5, 7, 8, 5, 1,
1, 1, 6, 9, 13, 12, 6, 1,
1, 1, 7, 11, 19, 22, 17, 7, 1

Cf. A180562.

Number triangle T(n,k)=sum{j=0..n, C(j,k-j)*C(n-k,k-j)}*[k<=n];

T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + T(n-3,k-2), T(0,0) = T(1,0) = T(1,1) = T(2,0) = T(2,1) = T(2,2) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2014

A172020 Number of subsets S of {1,2,3,...,n} with the property that if x is a member of S then at least one of x-2 and x+2 is also a member of S.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 28, 49, 84, 144, 252, 441, 777, 1369, 2405, 4225, 7410, 12996, 22800, 40000, 70200, 123201, 216216, 379456, 665896, 1168561, 2050657, 3598609, 6315113, 11082241, 19448018, 34128964, 59892184, 105103504, 184443732, 323676081, 568011852

Offset: 1

John W. Layman, Jan 22 2010

It is interesting that, for k > 0, it appears that a(2k) is the square of A005251(k+2). (This has since been proved by Andrew Weimholt; see link.)

If we denote by d2 the second difference of {a(n)}, it appears that d2(2k) is the square of A005314(k).

Colin Barker, Table of n, a(n) for n = 1..1000
Proof by Andrew Weimholt, SeqFan Jan 2010
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,-1,3,-1,0,1,-1).

Cf. A005251, A005314.

LinearRecurrence[{2, 0, -1, -1, 3, -1, 0, 1, -1}, {1, 1, 2, 4, 8, 16, 28, 49, 84}, 32] (* Jean-François Alcover, Feb 15 2016 *)

Vec(x*(1-x+x^3+2*x^4-x^8)/((1-2*x+x^2-x^3)*(1+x-x^3)*(1-x+x^3)) + O(x^50)) \\ Colin Barker, Feb 15 2016

Andrew Weimholt has shown that a(2*n) = A005251(n+2) ^ 2, and a(2*n+1) = A005251(n+2) * A005251(n+3). (See the link.)

G.f.: x*(1 - x + x^3 + 2*x^4 - x^8) / ((1 - 2*x + x^2 - x^3)*(1 + x - x^3)*(1 - x + x^3)). - Colin Barker, Feb 15 2016

A227300 Rising diagonal sums of triangle of Fibonacci polynomials (rows displayed as centered text).

Original entry on oeis.org

1, 2, 2, 3, 7, 11, 16, 28, 48, 77, 126, 211, 349, 573, 947, 1568, 2588, 4271, 7058, 11661, 19256, 31804, 52538, 86779, 143329, 236744, 391046, 645900, 1066850, 1762163, 2910634, 4807590, 7940870, 13116238, 21664568, 35784145, 59105987, 97627533, 161254953, 266350689

Offset: 1

John Molokach, Jul 09 2013

nonn

Rising diagonal sums of triangle A011973, taken with rows as centered text.

			a(1)  = 1;
a(2)  = 1 +  1;
a(3)  = 1 +  1;
a(4)  = 1 +  1 +  1;
a(5)  = 1 +  1 +  3 +  2;
a(6)  = 1 +  1 +  5 +  4;
a(7)  = 1 +  1 +  7 +  6 +  1;
a(8)  = 1 +  1 +  9 +  8 +  6 +  3;
a(9)  = 1 +  1 + 11 + 10 + 15 + 10;
a(10) = 1 +  1 + 13 + 12 + 28 + 21 +  1.

Index entries for linear recurrences with constant coefficients, signature (1,0,2,0,0,-1).

Cf. A011973 (triangle), A000045 (row sums of triangle), A005314 (falling diagonal sums of triangle). Expansion of terms begin with A055624 at a(1) and adds A016813 at a(4), A016754 at a(7), and A100157 at a(10).

LinearRecurrence[{1, 0, 2, 0, 0, -1}, {1, 2, 2, 3, 7, 11}, 40] (* T. D. Noe, Jul 11 2013 *)

a(n) = if(n<=1, 1, sum(k=0, floor((n-1)/3), binomial(2*n-2-5*k,k)+binomial(2*n-1-5*k,k)) ); \\ Joerg Arndt, Jul 11 2013

a(n) = Sum_{k=0..floor((n-1)/3)} (binomial(2*n-2-5*k,k) + binomial(2*n-3-5*k,k)) for n >= 2; a(1)=1. - John Molokach, Jul 11 2013
a(n) = a(n-1) + 2*a(n-3) - a(n-6), starting with {1, 2, 2, 3, 7, 11}. - T. D. Noe, Jul 11 2013
G.f.: x*(1+x-x^3)/(1-x-2*x^3+x^6) - John Molokach, Jul 15 2013
a(n) = Sum_{k=0..floor((2n-1)/3)} binomial(2n-k-2-3*floor(k/2),floor(k/2)). - John Molokach, Jul 29 2013

A020713 Pisot sequences E(5,9), P(5,9).

Original entry on oeis.org

5, 9, 16, 28, 49, 86, 151, 265, 465, 816, 1432, 2513, 4410, 7739, 13581, 23833, 41824, 73396, 128801, 226030, 396655, 696081, 1221537, 2143648, 3761840, 6601569, 11584946, 20330163, 35676949, 62608681, 109870576, 192809420, 338356945, 593775046, 1042002567

Offset: 0

David W. Wilson

nonn

Colin Barker, Table of n, a(n) for n = 0..1000
Shalosh B. Ekhad, N. J. A. Sloane and Doron Zeilberger, Automated Proof (or Disproof) of Linear Recurrences Satisfied by Pisot Sequences, arXiv:1609.05570 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1).

This is a subsequence of A005314.
See A008776 for definitions of Pisot sequences.
Cf. A099529.

Iv:=[5, 9]; [n le 2 select Iv[n] else Ceiling(Self(n-1)^2/Self(n-2)-1/2): n in [1..40]]; // Bruno Berselli, Feb 04 2016

RecurrenceTable[{a[0] == 5, a[1] == 9, a[n] == Ceiling[a[n - 1]^2/a[n - 2]-1/2]}, a, {n, 0, 40}] (* Bruno Berselli, Feb 04 2016 *)
LinearRecurrence[{2,-1,1},{5,9,16},40] (* Harvey P. Dale, Aug 03 2021 *)

lista(nn) = {print1(x = 5, ", ", y = 9, ", "); for (n=1, nn, z = ceil(y^2/x -1/2); print1(z, ", "); x = y; y = z;);} \\ Michel Marcus, Feb 04 2016

a(n) = 2*a(n-1) - a(n-2) + a(n-3) (holds at least up to n = 1000 but is not known to hold in general).
Empirical g.f.: (5-x+3*x^2) / (1-2*x+x^2-x^3). - Colin Barker, Jun 05 2016
Theorem: E(5,9) satisfies a(n) = 2 a(n - 1) -  a(n - 2) + a(n - 3) for n>=3. Proved using the PtoRv program of Ekhad-Sloane-Zeilberger, and implies the above conjectures. - N. J. A. Sloane, Sep 09 2016
a(n) = (-1)^n * A099529(n+6). - Jinyuan Wang, Mar 10 2020

A232582 Number of (n+1) X (1+1) 0..2 arrays with every element next to itself plus and minus one within the range 0..2 horizontally or antidiagonally, with no adjacent elements equal.

Original entry on oeis.org

0, 2, 4, 6, 10, 18, 32, 56, 98, 172, 302, 530, 930, 1632, 2864, 5026, 8820, 15478, 27162, 47666, 83648, 146792, 257602, 452060, 793310, 1392162, 2443074, 4287296, 7523680, 13203138, 23169892, 40660326, 71353898, 125217362, 219741152, 385618840

Offset: 1

R. H. Hardin, Nov 26 2013

nonn

Column 1 of A232589.

			Some solutions for n=7:
  2 1   0 1   2 1   0 1   0 1   0 1   0 1   2 1   0 1   2 1
  0 1   2 1   0 1   2 0   2 1   2 0   2 1   0 2   2 0   0 1
  2 0   0 1   2 0   1 2   0 2   1 2   0 1   1 0   1 2   2 0
  1 2   2 1   1 2   1 0   1 0   0 1   2 0   1 2   0 1   1 2
  1 0   0 1   0 1   2 1   2 1   2 0   1 2   1 0   2 1   1 0
  1 2   2 0   2 1   0 2   0 1   1 2   0 1   1 2   0 2   2 1
  1 0   1 2   0 2   1 0   2 0   1 0   2 0   1 0   1 0   0 2
  1 2   1 0   1 0   1 2   1 2   1 2   1 2   1 2   1 2   1 0

R. H. Hardin, Table of n, a(n) for n = 1..210

Empirical: a(n) = 2*a(n-1) - a(n-2) + a(n-3) = 2*A005314(n-1).
Empirical: G.f.: -2*x^2 / ( -1+2*x-x^2+x^3 ). - R. J. Mathar, Nov 23 2014
Theorem: a(n) = Sum_{j=1..floor((n-2)/3)} 2* Hypergeometric2F1([2+3*j-n,-(2j+1)], [1], 1). - Richard Turk, Oct 22 2019

A307677 a(0) = a(1) = a(2) = a(3) = 1; thereafter a(n) = a(n-1) + a(n-2) + a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 9, 15, 27, 47, 83, 145, 255, 447, 785, 1377, 2417, 4241, 7443, 13061, 22921, 40223, 70587, 123871, 217379, 381473, 669439, 1174783, 2061601, 3617857, 6348897, 11141537, 19552035, 34311429, 60212361, 105665327, 185429723, 325406479, 571048563, 1002120369

Offset: 0

Joseph Damico, Apr 21 2019

A079398, A103609, A003269, A306276, A126116, and A000288 are the other six sequences which have characteristic equations of the form x^4 = ax^3 + bx^2 + cx + 1 in which a, b, and c are equal to either 0 or 1 -- but not all three of them are equal to zero. (Each of those sequences begins with 1,1,1,1.)
A005251 has the same characteristic equation, and each successive term is determined by the same operation, namely, a(n) = a(n-1) + a(n-2) + a(n-4). However, it has different starting values: (0,1,1,1) instead of (1,1,1,1).
The characteristic equation of this sequence is x^4 = x^3 + x^2 + 1. Lim_{n->infinity} a(n+1)/a(n) = 1.754877666...

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,1).

Cf. A000288, A003269, A005251, A005314, A079398, A103609, A126116, A306276.

[n le 4 select 1 else Self(n-1) +Self(n-2) +Self(n-4): n in [1..51]]; // G. C. Greubel, Oct 23 2024

LinearRecurrence[{1,1,0,1}, {1,1,1,1}, 51] (* G. C. Greubel, Oct 23 2024 *)

Vec((1 - x^2 - x^3) / ((1 + x)*(1 - 2*x + x^2 - x^3)) + O(x^40)) \\ Colin Barker, Apr 25 2020

@CachedFunction # a = A307677
def a(n): return 1 if n<4 else a(n-1) +a(n-2) +a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Oct 23 2024

From Colin Barker, Apr 25 2020: (Start)
G.f.: (1 - x^2 - x^3) / ((1 + x)*(1 - 2*x + x^2 - x^3)).
a(n) = a(n-1) + a(n-2) + a(n-4) for n>3. (End)
a(n) = (1/5)*((-1)^n + 2*(2*A005314(n+1) - A005314(n) - 2*A005314(n-1))). - G. C. Greubel, Oct 23 2024

A347493 a(0) = 1, a(1) = 0, a(2) = a(3) = 1; thereafter, a(n) = a(n-1) + a(n-2) + a(n-4).

Original entry on oeis.org

1, 0, 1, 1, 3, 4, 8, 13, 24, 41, 73, 127, 224, 392, 689, 1208, 2121, 3721, 6531, 11460, 20112, 35293, 61936, 108689, 190737, 334719, 587392, 1030800, 1808929, 3174448, 5570769, 9776017, 17155715, 30106180, 52832664, 92714861, 162703240, 285524281, 501060185, 879299327, 1543062752

Offset: 0

Greg Dresden and Yichen P. Wang, Sep 03 2021

a(n) is also the number of ways to tile a strip of length n with squares, dominoes, and tetrominoes such that the first tile is NOT a square. As such, it completes the set of such tilings with A005251 (first tile is NOT a domino), A005314 (first tile is NOT a tetromino), and A060945 (no restrictions on first tile).

Michael De Vlieger, Table of n, a(n) for n = 0..4096
Jean-Luc Baril, Sergey Kirgizov, and Armen Petrossian, Dyck Paths with catastrophes modulo the positions of a given pattern, Australasian J. Comb. (2022) Vol. 84, No. 2, 398-418.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,1).

Cf. A005251, A005314, A060945.

CoefficientList[Series[(1 - x)/(1 - x - x^2 - x^4), {x, 0, 40}], x] (* Michael De Vlieger, Mar 04 2022 *)
LinearRecurrence[{1,1,0,1},{1,0,1,1},60] (* Harvey P. Dale, Aug 17 2023 *)

a(n) = 2*A060945(n) - A005251(n) - A005314(n).
G.f.: (1 - x)/(1 - x - x^2 - x^4).
Sum_{k=0..n} a(k)*F(n-k) = a(n+3) - F(n+2) for F(n)=A000045(n) the Fibonacci numbers.
5*a(n) = 2*(-1)^n + 3*A005314(n+1) -4*A005314(n) +2*A005314(n-1). - R. J. Mathar, Sep 30 2021

A121230 First Hadamard-Sylvester matrix self -similar matrix based on the Padovan/ Minimal Pisot 3 X 3 matrix as an 9 X 9 matrix: Characteristic Polynomial:1 - x - x^3 - x^4 - x^5 + 3 x^6 + 2 x^7 - x^9.

Original entry on oeis.org

0, 13, 5, 22, 42, 54, 126, 192, 347, 631, 1056, 1914, 3320, 5814, 10276, 17921, 31549, 55338, 97026, 170454, 298914, 524684, 920815, 1615647, 2835660, 4975898, 8732160, 15324202, 26891432, 47191909, 82815621, 145331022, 255039162

Offset: 1

Roger L. Bagula, Aug 13 2006

As far as I can tell by searching the Internet, this matrix and this approach to sequences is entirely new and unique. The second of these matrices at 81 X 81 gives a new fractal that is Cantor dust like. aa = Table[M[[n, m]]*M[[i, j]], {n, 1, 9 }, {m, 1, 9}, {i, 1, 9}, {j, 1, 9}]; M2 = Flatten[Table[{Flatten[Table[aa[[ n, m]][[1, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[n, m]][[2, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[ n, m]][[3, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[ n, m]][[4, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[ n, m]][[5, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[n, m]][[6, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[ n, m]][[7, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[ n, m]][[8, i]], {n, 1, 9}, {i, 1, 9}]], Flatten[Table[aa[[ n, m]][[9, i]], {n, 1, 9}, {i, 1, 9}]]}, {m, 1, 9}], 1]; ListDensityPlot[M2, Mesh -> False]

Index entries for linear recurrences with constant coefficients, signature (1, 1, 1, -1, 1, -1).

Cf. A000931.

Clear[t, M, a, v, a0] t[n_, m_] := {{0, 1, 0}, {0, 0, 1}, {1, 1, 0}}[[n, m]] a0 = Table[t[n, m]*t[i, j], {n, 1, 3}, {m, 1, 3}, {i, 1, 3}, {j, 1, 3}]; M = Flatten[Table[{Flatten[Table[a0[[ n, m]][[1, i]], {n, 1, 3}, {i, 1, 3}]], Flatten[Table[a0[[n, m]][[2, i]], {n, 1, 3}, {i, 1, 3}]], Flatten[Table[a0[[n, m]][[3, i]], {n, 1, 3}, {i, 1, 3}]]}, {m, 1, 3}], 1] v[1] = Table[Fibonacci[n], {n, 0, 8}] v[n_] := v[n] = M.v[n - 1] a = Table[Floor[v[n][[1]]], {n, 1, 50}] Det[M - x*IdentityMatrix[9]] Factor[%] aaa = Table[x /. NSolve[Det[M - x*IdentityMatrix[9]] == 0, x][[n]], {n, 1, 9}] Abs[aaa] a1 = Table[N[a[[n]]/a[[n - 1]]], {n, 7, 50}] ListDensityPlot[M, Mesh -> False]

M={{0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1}, { 0, 0, 0, 0, 0, 0, 1, 1, 0}, {0, 1, 0, 0, 0, 0, 0, 1, 0}, { 0, 0, 1, 0, 0, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1, 0}, { 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0}, { 0, 0, 0, 1, 1, 0, 0, 0, 0}} v[1] = Table[Fibonacci[n], {n, 0, 8}] v[n_] := v[n] = M.v[n - 1] a(n) = v[n][[1]]

G.f.: x^2(13-8x+4x^2+2x^3-2x^4)/((1-2x+x^2-x^3)(1+x-x^3)). a(n) = a(n-1) +a(n-2) +a(n-3) -a(n-4) +a(n-5) -a(n-6). Partial fraction decomposition yield decomposition in terms of A005314 and A050935. [From R. J. Mathar, Nov 26 2008]

A137495 a(n) = A098601(2n) + A098601(2n+1).

Original entry on oeis.org

2, 3, 4, 7, 13, 23, 40, 70, 123, 216, 379, 665, 1167, 2048, 3594, 6307, 11068, 19423, 34085, 59815, 104968, 184206, 323259, 567280, 995507, 1746993, 3065759, 5380032, 9441298, 16568323, 29075380, 51023735, 89540413, 157132471, 275748264, 483904470, 849193147, 1490230088

Offset: 0

Paul Curtz, Apr 27 2008

Nicolas Ollinger and Jeffrey Shallit, The Repetition Threshold for Rote Sequences, arXiv:2406.17867 [math.CO], 2024.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1)

Cf. A005314, A135364, A137584, A137531.

LinearRecurrence[{2, -1, 1}, {2, 3, 4}, 50] (* Paolo Xausa, Jun 29 2024 *)

Vec((x-2)/(-1+2*x-x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Jul 06 2011

a(3n) = A135364(2n+1). a(3n+1) = A137584(2n+1). a(3n+2) = A137531(2n+2).
From R. J. Mathar, Jul 06 2011: (Start)
G.f.: ( -2+x ) / ( -1+2*x-x^2+x^3 ).
a(n) = 2*A005314(n+1) - A005314(n). (End)

cp's OEIS Frontend

A124279 Riordan array (1/(1-x),x(1-x+x^2)/(1-x)).

Views

Author

Keywords

Comments

Examples

Formula

A124445 Expansion of 1/(1-x-x*y+x^2*y-x^3*y^2).

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A172020 Number of subsets S of {1,2,3,...,n} with the property that if x is a member of S then at least one of x-2 and x+2 is also a member of S.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A227300 Rising diagonal sums of triangle of Fibonacci polynomials (rows displayed as centered text).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A020713 Pisot sequences E(5,9), P(5,9).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A232582 Number of (n+1) X (1+1) 0..2 arrays with every element next to itself plus and minus one within the range 0..2 horizontally or antidiagonally, with no adjacent elements equal.

Views

Author

Keywords

Comments

Examples

Links

Formula

A307677 a(0) = a(1) = a(2) = a(3) = 1; thereafter a(n) = a(n-1) + a(n-2) + a(n-4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A347493 a(0) = 1, a(1) = 0, a(2) = a(3) = 1; thereafter, a(n) = a(n-1) + a(n-2) + a(n-4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A124445 Expansion of 1/(1-x-xy+x^2y-x^3*y^2).