A000034 -id:A000034 - cp's OEIS frontend

A133087 A133080 * A007318.

1, 2, 1, 1, 2, 1, 2, 5, 4, 1, 1, 4, 6, 4, 1, 2, 9, 16, 14, 6, 1, 1, 6, 15, 20, 15, 6, 1, 2, 13, 36, 55, 50, 27, 8, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1

Offset: 0

Gary W. Adamson, Sep 08 2007

Row sums = A084221: (1, 3, 4, 12, 16, 48, 64, 192, ...).

Subtriangle of (0, 2, -3/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 03 2012

			First few rows of the triangle:
  1;
  2,  1;
  1,  2,  1;
  2,  5,  4,  1;
  1,  4,  6,  4,  1;
  2,  9, 16, 14,  6,  1;
  1,  6, 15, 20, 15,  6,  1;
  2, 13, 36, 55, 50, 27,  8,  1;
  1,  8, 28, 56, 70, 56, 28,  8,  1;
  ...
Triangle (0, 2, -3/2, -1/2, 0, 0, 0, ...) DELTA (1, 0, -1, 0, 0, 0, ...) begins:
  1;
  0,  1;
  0,  2,  1;
  0,  1,  2,  1;
  0,  2,  5,  4,  1;
  0,  1,  4,  6,  4,  1;
  0,  2,  9, 16, 14,  6,  1;
  0,  1,  6, 15, 20, 15,  6,  1;
  0,  2, 13, 36, 55, 50, 27,  8,  1;
  0,  1,  8, 28, 56, 70, 56, 28,  8,  1;
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A133080, A084221.

Columns: A000007, A114752, A133092.

CoefficientList[CoefficientList[Series[(1 + 2*x + y*x)/(1 - (1 + y)^2*x^2), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Oct 21 2017 *)

A133080 * A007318 as infinite lower triangular matrices.
G.f.: (1+2*x+y*x)/(1-(1+y)^2*x^2). - Philippe Deléham, Mar 03 2012
T(n,k) = T(n-2,k) + 2*T(n-2,k-1) + T(n-2,k-1), T(0,0) = 1, T(1,0) = 2, T(1,1) = 1. - Philippe Deléham, Mar 03 2012
Sum_{k=0..n} T(n,k)*x^k = A059841(n), A019590(n+1), A000034(n), A084221(n), A133125(n) for x = -2, -1, 0, 1, 2 respectively. - Philippe Deléham, Mar 03 2012

A178452 Partial sums of floor(2^n/5).

Original entry on oeis.org

0, 0, 1, 4, 10, 22, 47, 98, 200, 404, 813, 1632, 3270, 6546, 13099, 26206, 52420, 104848, 209705, 419420, 838850, 1677710, 3355431, 6710874, 13421760, 26843532, 53687077, 107374168, 214748350, 429496714, 858993443, 1717986902

Offset: 1

Mircea Merca, Dec 22 2010

Partial sums of A077854(n-3).

			a(5) = 0 + 0 + 1 + 3 + 6 = 10.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (4,-6,6,-5,2).

Cf. A077854.

[Floor((4*2^n-5*n-3)/10): n in [1..40]]; // Vincenzo Librandi, Jun 23 2011

Maple
```
seq(round((4*2^n-5*n-4)/10), n=1..50)
```

CoefficientList[Series[x^2 / ((1 - 2 x) (1 + x^2) (1 - x)^2), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)
Accumulate[Floor[2^Range[40]/5]] (* or *) LinearRecurrence[{4,-6,6,-5,2},{0,0,1,4,10},40] (* Harvey P. Dale, Oct 09 2018 *)

a(n) = round((4*2^n - 5*n - 5)/10).
a(n) = floor((4*2^n - 5*n - 3)/10).
a(n) = ceiling((4*2^n - 5*n - 7)/10).
a(n) = round((4*2^n - 5*n - 4)/10).
a(n) = a(n-4) + 3*2^(n-3) - 2, n > 4.
From Bruno Berselli, Jan 18 2011: (Start)
G.f.:  x^3/((1-2*x)*(1+x^2)*(1-x)^2).
a(n) = (4*2^n - 5*n - 5 + A057077(n)*A000034(n))/10.
a(n) = 3*a(n-1) - 2*a(n-2) + a(n-4) - 3*a(n-5) + 2*a(n-6) for n > 6. (End)

A182001 Riordan array ((2*x+1)/(1-x-x^2), x/(1-x-x^2)).

Original entry on oeis.org

1, 3, 1, 4, 4, 1, 7, 9, 5, 1, 11, 20, 15, 6, 1, 18, 40, 40, 22, 7, 1, 29, 78, 95, 68, 30, 8, 1, 47, 147, 213, 185, 105, 39, 9, 1, 76, 272, 455, 466, 320, 152, 49, 10, 1, 123, 495, 940, 1106, 891, 511, 210, 60, 11, 1, 199, 890, 1890, 2512, 2317, 1554, 770, 280, 72, 12, 1

Offset: 0

Philippe Deléham, Apr 05 2012

Subtriangle of the triangle given by (0, 3, -5/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -2/3, 2/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Antidiagonal sums are A001045(n+2).

			Triangle begins :
    1;
    3,   1;
    4,   4,    1;
    7,   9,    5,    1;
   11,  20,   15,    6,    1;
   18,  40,   40,   22,    7,    1;
   29,  78,   95,   68,   30,    8,   1;
   47, 147,  213,  185,  105,   39,   9,   1;
   76, 272,  455,  466,  320,  152,  49,  10, 1;
  123, 495,  940, 1106,  891,  511, 210,  60, 11,  1;
  199, 890, 1890, 2512, 2317, 1554, 770, 280, 72, 12, 1;
(0, 3, -5/3, -1/3, 0, 0, ...) DELTA (1, 0, -2/3, 2/3, 0, 0, ...) begins:
  1;
  0,  1;
  0,  3,  1;
  0,  4,  4,  1;
  0,  7,  9,  5,  1;
  0, 11, 20, 15,  6, 1;
  0, 18, 40, 40, 22, 7, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. Columns : A000032, A023607, A152881

function T(n,k)
  if k lt 0 or k gt n then return 0;
  elif k eq n then return 1;
  elif k eq 0 then return Lucas(n+1);
  else return T(n-1,k) + T(n-1,k-1) + T(n-2,k);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020

with(combinat);
T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=n then 1
    elif k=0 then fibonacci(n+2) + fibonacci(n)
    else T(n-1,k) + T(n-1,k-1) + T(n-2,k)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Feb 18 2020

With[{m = 10}, CoefficientList[CoefficientList[Series[(1+2*x)/(1-x-y*x-x^2), {x, 0, m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 18 2020 *)
T[n_, k_]:= T[n, k]= If[k<0||k>n, 0, If[k==n, 1, If[k==0, LucasL[n+1], T[n-1, k] + T[n-1, k-1] + T[n-2, k] ]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)

G.f.: (1+2*x)/(1-x-y*x-x^2).
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = 3, T(n,k) = 0 if k<0 or if k>n.
Sum_{k=0..nn} T(n,k)*x^k = A000034(n), A000032(n+1), A048654(n), A108300(n), A048875(n) for x = -1, 0, 1, 2, 3 respectively.

a(29) corrected by and a(55)-a(65) from Georg Fischer, Feb 18 2020

A201208 One 1, two 2's, three 1's, four 2's, five 1's, ...

Original entry on oeis.org

1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Paul Curtz, Nov 28 2011

			May be written as a triangle:
  1
  2 2
  1 1 1
  2 2 2 2
  1 1 1 1 1
  2 2 2 2 2 2
  1 1 1 1 1 1 1
Row sums are A022998(n+1).

Cf. A057212, A022998.

Cf. A000034.

a201208 n = a201208_list !! (n-1)
a201208_list = concat $ zipWith ($) (map replicate [1..]) a000034_list
-- Reinhard Zumkeller, Dec 02 2011

ReplaceAll[ColumnForm[Table[Mod[k, 2], {k, 12}, {n, k}], Center], 0 -> 2] (* Alonso del Arte, Nov 28 2011 *)

a(n) = A057212(n) + 1. - T. D. Noe, Nov 28 2011

Edited by N. J. A. Sloane, Dec 02 2011

A207974 Triangle related to A152198.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 5, 2, 4, 1, 1, 1, 6, 3, 6, 3, 2, 1, 1, 7, 3, 9, 3, 5, 1, 1, 1, 8, 4, 12, 6, 8, 4, 2, 1, 1, 9, 4, 16, 6, 14, 4, 6, 1, 1, 1, 10, 5, 20, 10, 20, 10, 10, 5, 2, 1

Offset: 0

Philippe Deléham, Feb 22 2012

Row sums are A027383(n).
Diagonal sums are alternately A014739(n) and A001911(n+1).
The matrix inverse starts
1;
-1,1;
1,-2,1;
1,-1,-1,1;
-1,2,0,-2,1;
-1,1,2,-2,-1,1;
1,-2,-1,4,-1,-2,1;
1,-1,-3,3,3,-3,-1,1;
-1,2,2,-6,0,6,-2,-2,1;
-1,1,4,-4,-6,6,4,-4,-1,1;
1,-2,-3,8,2,-12,2,8,-3,-2,1;
apparently related to A158854. - R. J. Mathar, Apr 08 2013
From Gheorghe Coserea, Jun 11 2016: (Start)
T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
(End)

			Triangle begins :
n\k  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
[0]  1;
[1]  1,  1;
[2]  1,  2,  1;
[3]  1,  3,  1,  1;
[4]  1,  4,  2,  2,  1;
[5]  1,  5,  2,  4,  1,  1;
[6]  1,  6,  3,  6,  3,  2,  1;
[7]  1,  7,  3,  9,  3,  5,  1,  1;
[8]  1,  8,  4,  12, 6,  8,  4,  2,  1;
[9]  1,  9,  4,  16, 6,  14, 4,  6,  1,  1;
[10] ...

Gheorghe Coserea, Rows n = 0..200, flattened

Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
Cf. Diagonals : A000012, A000034, A052938, A097362
Cf. A007318, A110813, A135278, A152201
Related to thickness: A000120, A027383, A057890, A274036.

A207974 := proc(n,k)
    if k = 0 then
        1;
    elif k < 0 or k > n then
        0 ;
    else
        procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 08 2013

seq(N) = {
  my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
  for(n = 2, N+1, for (k = 2, n-1,
      t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
  return(t);
};
concat(seq(10))  \\ Gheorghe Coserea, Jun 09 2016

P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
concat(vector(11, n, Vecrev(P(n-1)))) \\ Gheorghe Coserea, Mar 14 2017

T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
T(2n+1,2k+1) = A110813(n,k).
T(2n+2,2k+1) = 2*A135278(n,k).
T(n,2k) + T(n,2k+1) = A152201(n,k).
T(n,2k) = A152198(n,k).
T(n+1,2k+1) = A152201(n,k).
T(n,k) = T(n-2,k-2) + T(n-2,k).
T(2n,n) = A128014(n+1).
T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - Gheorghe Coserea, Jun 09 2016
P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - Gheorghe Coserea, Mar 14 2017

A215580 Partial sums of A215602.

Original entry on oeis.org

2, 5, 17, 45, 122, 320, 842, 2205, 5777, 15125, 39602, 103680, 271442, 710645, 1860497, 4870845, 12752042, 33385280, 87403802, 228826125, 599074577, 1568397605, 4106118242, 10749957120, 28143753122, 73681302245, 192900153617, 505019158605, 1322157322202, 3461452808000, 9062201101802, 23725150497405, 62113250390417, 162614600673845

Offset: 0

J. M. Bergot, Aug 16 2012

Dividing the terms of this sequence by Fibonacci or Lucas numbers yields symmetric sets of remainders of determinable lengths. For F(n) beginning at n=3: (a) F(2n) will have a set of remainders of length 2n in which the sum of the remainders is 3*(F(2n)-n). Example for F(2*6)=144: the set of remainders is {2,5,17,45,122,32,122,45,17,5,2,0} with 2*6=12 terms and a sum of 3*(144-6)=414. (b) For F(2n+1) there will be 2*(2n+1) terms having a sum equal to (2n+1)*(F(2n+1)-3). Example for F(2*4+1)=34: the remainders are {2,5,7,11,20,14,26,29,31,29,26,14,20,11,17,5,2,0} with 2*9 terms and a sum of 9*(34-1)=279.

Using Lucas numbers starting at n=2: (a) L(2n) has 4n remainders with sum (2n+1)*(L(2n)-6*n). Example for n=4 giving L(2*4)=47, has remainders {2,5,17,45,28,38,43,43,43,38,28,45,17,5,2,0} with a sum of (8+1)*(47)-6*4=399. (B) For L(2n+1) the length of the period is 2*(2n+1) and the sum of the remainders is 4*L(2n+1)-3*(2n+1). Example for n=3 for L(2*3+1)=29 has remainders {2,5,17,16,6,1,11,6,16,17,5,2,0} with length 2*7 and sum of terms 4*29-3*7=95.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A000032, A075269, A064831, A215602.

Cf. A005248, A000034.

LinearRecurrence[{3, 0, -3, 1}, {2, 5, 17, 45}, 35] (* Paolo Xausa, Feb 22 2024 *)

a(2n) = L(4*n)-2, a(2*n+1) = L(4*n+2)-1, where L() are the Lucas numbers A000032.
G.f. ( -2+x-2*x^2 ) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Aug 21 2012
a(n) = A005248(n+1)-A000034(n). - R. J. Mathar, Aug 21 2012

Edited by N. J. A. Sloane, Aug 17 2012

A269266 a(n) = 2^n mod 31.

Original entry on oeis.org

1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 1

Offset: 0

Vincenzo Librandi, Mar 31 2016

Continued fraction expansion of (1651+sqrt(3236405))/2386. - Bruno Berselli, Mar 31 2016

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

Cf. A201912 (11th row of the triangle).

Cf. similar sequences of the type 2^n mod p, where p is a prime: A000034 (p=3), A070402 (p=5), A069705 (p=7), A036117 (p=11), A036118 (p=13), A062116 (p=17), A036120 (p=19), A070335 (p=23), A036122 (p=29), this sequence (p=31), A036124 (p=37), A070348 (p=41), A070349 (p=43), A070351 (p=47), A036128 (p=53), A036129 (p=59), A036130 (p=61), A036131 (p=67).

List([0..70],n->PowerMod(2,n,31)); # Muniru A Asiru, Jan 30 2019

Magma
```
[Modexp(2, n, 31): n in [0..100]];
```

&cat [[1,2,4,8,16]^^20] // Bruno Berselli, Mar 31 2016

Mathematica
```
PowerMod[2, Range[0, 100], 31]
```

a(n)=2^(n%5) \\ Charles R Greathouse IV, Mar 31 2016

x='x+O('x^99); Vec((1+2*x+4*x^2+8*x^3+16*x^4)/(1-x^5)) \\ Altug Alkan, Mar 31 2016

for n in range(0,100):print(2**n%31) # Soumil Mandal, Apr 03 2016

def A269266(n): return pow(2,n,31) # Chai Wah Wu, Jan 03 2022

[2^mod(n,5) for n in (0..100)] # Bruno Berselli, Mar 31 2016

G.f.: (1 + 2*x + 4*x^2 + 8*x^3 + 16*x^4)/(1 - x^5).
a(n) = a(n-5).
a(n) = 2^(n mod 5). - Bruno Berselli, Mar 31 2016

A269837 Irregular triangle read by rows: even terms of A094728(n+1) divided by 4.

Original entry on oeis.org

1, 2, 4, 3, 6, 4, 9, 8, 5, 12, 10, 6, 16, 15, 12, 7, 20, 18, 14, 8, 25, 24, 21, 16, 9, 30, 28, 24, 18, 10, 36, 35, 32, 27, 20, 11, 42, 40, 36, 30, 22, 12, 49, 48, 45, 40, 33, 24, 13, 56, 54, 50, 44, 36, 26, 14, 64, 63, 60, 55, 48, 39, 28, 15

Offset: 0

Paul Curtz, Mar 06 2016

See A264798 and A261046 for the Hydrogen atom and the Janet periodic table.
a(n) odd terms are again A264798.
Decomposition by multiplication i.e. a(n) = b(n)*c(n) by irregular triangle:
1,                   1                 1,
2,                   1                 2,
4,   3,              2, 1,             2, 3,
6,   4,         =    2, 1,        *    3, 4,
9,   8, 5,           3, 2, 1,          3, 4, 5,
12, 10, 6,           3, 2, 1,          4, 5, 6,
16, 15, 12, 7,       4, 3, 2, 1,       4, 5, 6, 7,
etc.                 etc.              etc.
b(n) is duplicated A004736(n) or mirror of A122197(n+1). c(n) = A138099(n+1).
Decomposition by subtraction, a(n) = d(n) - e(n):
1,                    1                    0,
2,                    2,                   0,
4,   3,               4,  3,               0, 0,
6,   4,         =     6,  5,          -    0, 1,
9,   8, 5,            9,  8,  7,           0, 0, 2,
12, 10, 6,           12, 11, 10,           0, 1, 4,
16, 15, 12, 7,       16, 15, 14, 13,       0, 0, 2, 6,
20, 18, 14, 8,       20, 19, 18, 17,       0, 1, 4, 9,
etc.                 etc.                  etc.
d(n) is the natural numbers A000027 inverted by lines.  e(n) will be studied (see A239873).
Sum of a(n) by diagonals: 1, 5, 13, 27, 48, ... . The third differences have the period 2: repeat 2, 1. See A002717.

Cf. A000027, A000034, A002717, A004736, A094728, A122197, A138099, A167430, A239873, A261046, A264798.

Table[(n + 1)^2 - k^2, {n, 15}, {k, 0, n - 1}]/4 /. Rational -> Nothing // Flatten (* _Michael De Vlieger, Mar 07 2016 *)

A281661 The least common multiple of 1 + n^2 and 1 + n^3.

Original entry on oeis.org

1, 2, 45, 140, 1105, 1638, 8029, 8600, 33345, 29930, 101101, 81252, 250705, 186830, 540765, 381488, 1052929, 712530, 1895725, 1241660, 3208401, 2046902, 5164765, 3224520, 7977025, 4890938, 11899629, 7184660, 17233105, 10268190, 24327901, 14329952, 33588225, 19586210

Offset: 0

R. J. Mathar, Jan 26 2017

If d|(1 + n^2) and d|(1 + n^3), then d|((1 + n^2) - (n*(1 + n^2) - (1 + n^3))^2) = 2*n. If k|n and k|(1 + n^2), then k = 1 is only option since k|n^2 and k|(1 + n^2). So d must be 1 or 2, exactly. Obviously if n is odd, then the greatest d must be 2 since 1 + n^2 and 1 + n^3 are even. If n is even, then d must be 1 since 1 + n^2 and 1 + n^3 are odd.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-15,0,20,0,-15,0,6,0,-1).

Cf. A001093, A002522, A281660.

A281661 := proc(n)
        ilcm(1+n^2,1+n^3);
end proc:

Table[LCM[n^2+1,n^3+1],{n,0,50}] (* Harvey P. Dale, Jun 10 2023 *)

a(n) = lcm(n^2+1, n^3+1); \\ Michel Marcus, Jan 29 2017

a(n) = (n^2 + 1)*(n^3 + 1)/(1 + n%2); \\ Altug Alkan, Jan 29 2017

a(n) = lcm(1+n^2, 1+n^3) = (1+n^2)*(1+n^3)/gcd(1+n^2, 1+n^3).
a(n) = (1+n^2)*(1+n^3)/ A000034(n) with g.f. ( 1 +2*x +39*x^2 +128*x^3 +850*x^4 +828*x^5 +2054*x^6 +832*x^7 +861*x^8 +130*x^9 +35*x^10 ) / ( (x-1)^6 *(1+x)^6 ).
A006530(a(n)) = max( A081256(n), A014442(n)). - R. J. Mathar, Jan 28 2017
a(n) = (3 + (-1)^n)*(1 + n^2 + n^3 + n^5)/4. - Colin Barker, Feb 07 2017

A289203 Number of maximum independent vertex sets in the n X n knight graph.

Original entry on oeis.org

1, 1, 2, 6, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2

Offset: 1

Eric W. Weisstein, Jun 28 2017

Eric Weisstein's World of Mathematics, Independent Vertex Set
Eric Weisstein's World of Mathematics, Knight Graph
Eric Weisstein's World of Mathematics, Maximum Independent Vertex Set
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A000034.

Table[Length[With[{g = KnightTourGraph[n, n]}, FindIndependentVertexSet[g, Length /@ FindIndependentVertexSet[g], All]]], {n, 8}]
Table[Piecewise[{{1, n == 2}, {2, n == 3}, {6, n == 4}, {2, Mod[n, 2] == 0}, {1, Mod[n, 2] == 1}}], {n, 100}]
Table[Piecewise[{{1, n == 2}, {2, n == 3}, {6, n == 4}}, ((-1)^n + 3)/2], {n, 100}]
CoefficientList[Series[(-1 - x - x^2 - 5 x^3 + x^4 + 4 x^5)/(-1 + x^2), {x, 0, 20}],x]

def A289203(n): return (1,1,2,6)[n-1] if n<5 else 2-(n&1) # Chai Wah Wu, Feb 12 2024

For n > 4, a(n) = ((-1)^n + 3)/2.

G.f.: (x*(-1 - x - x^2 - 5*x^3 + x^4 + 4*x^5))/(-1 + x^2).

cp's OEIS Frontend

A133087 A133080 * A007318.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A178452 Partial sums of floor(2^n/5).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A182001 Riordan array ((2*x+1)/(1-x-x^2), x/(1-x-x^2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

Extensions

A201208 One 1, two 2's, three 1's, four 2's, five 1's, ...

Views

Author

Keywords

Examples

Crossrefs

Programs

Haskell

Mathematica

Formula

Extensions

A207974 Triangle related to A152198.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

PARI

Formula

A215580 Partial sums of A215602.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A269266 a(n) = 2^n mod 31.

Views

Author

Keywords

References

Links