A010685 -id:A010685 - cp's OEIS frontend

A226044 Period of length 8: 1, 64, 16, 64, 4, 64, 16, 64.

1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64

Offset: 0

Paul Curtz, May 24 2013

A002378(n)/A016754(n) gives 0/1, 2/9, 6/25, 12/49, 20/81, 30/121, 42/169, 56/225,..., where A016754(n) = 4*A002378(n) + 1;
A142705(n)/A154615(n+1) gives 0/1, 3/16,  2/9,  15/64,  6/25,  35/144, 12/49,  63/256,..., where A142705(n) = 4*A154615(n+1) + A010685(n);
A061037(n)/A061038(n) gives 0/1, 5/36,  3/16, 21/100, 2/9,   45/196, 15/64,  77/324,..., where A061038(n) = 4*A061037(n) + A177499(n);
A225948(n)/A226008(n) gives 0/1, 9/100, 5/36, 33/196, 3/16,  65/324, 21/100, 105/484,..., where A226008(n) = 4*A225948(n) + a(n).
See also the triangle in Example lines.

			Triangle in which the terms of each line are repeated:
A000012: 1,   ...
A010685: 1,   4,  ...
A177499: 1,  16,  4,  16,  ...
A226044: 1,  64, 16,  64,  4,  64, 16,  64, ...
         1, 256, 64, 256, 16, 256, 64, 256, 4, 256, 64, 256, 16, 256, 64, 256, ...

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

Cf. A010685, A177499, A205383, A225948.

Table[{1, 64, 16, 64, 4, 64, 16, 64}, {7}] // Flatten (* Jean-François Alcover, May 24 2013 *)

a(n) = A205383(n+7)^2.

G.f.: (1+64*x+16*x^2+64*x^3+4*x^4+64*x^5+16*x^6+64*x^7)/((1-x)*(1+x)*(1+x^2)*(1+x^4)). [Bruno Berselli, May 25 2013]

A136258 a(n) = 2a(n-1) - 2a(n-2), with a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 8, 6, -4, -20, -32, -24, 16, 80, 128, 96, -64, -320, -512, -384, 256, 1280, 2048, 1536, -1024, -5120, -8192, -6144, 4096, 20480, 32768, 24576, -16384, -81920, -131072, -98304, 65536, 327680, 524288, 393216, -262144, -1310720, -2097152, -1572864, 1048576

Offset: 0

Paul Curtz, Mar 18 2008

Sequence opposite in sign to its second differences.
Binomial transform of 1, 4, -1, -4.
A bisection gives A135520.
This sequence with offset 0 is the binomial transform of (-1)^floor(n/2)*A010685(n). - R. J. Mathar, Feb 22 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2).

Cf. A010685, A135520.

[n le 2 select 5^(n-1) else 2*(Self(n-1) - Self(n-2)): n in [1..41]]; // G. C. Greubel, Dec 02 2021

LinearRecurrence[{2,-2},{1,5},50] (* Harvey P. Dale, May 21 2014 *)

vector(100,n,t=if(n<3,[t1=1,5][n],-2*t1+2*t1=t)) \\ M. F. Hasler, May 01 2008

A136258=BinaryRecurrenceSequence(2,-2,1,5)
[A136258(n) for n in (0..40)] # G. C. Greubel, Dec 02 2021

a(4n+1) = 5*(-4)^n, a(4n+3) = 6*(-4)^n. - M. F. Hasler, May 01 2008
G.f.: x*(1+3*x)/(1-2*x+2*x^2). - R. J. Mathar, Feb 22 2009
From Paul Curtz, Apr 27 2011: (Start)
a(n)= -4 * a(n-4).
a(n)= 3*A009545(n) + A009545(n+1). (End)
E.g.f.: exp(x)*( cos(x) + 4*sin(x) ). - G. C. Greubel, Dec 02 2021

Edited and extended by M. F. Hasler, May 01 2008

Offset corrected Paul Curtz, Apr 27 2011

A171475 a(n) = 6a(n-1) - 8a(n-2), for n > 2, with a(0) = 1, a(1) = 6, a(2) = 27.

Original entry on oeis.org

1, 6, 27, 114, 468, 1896, 7632, 30624, 122688, 491136, 1965312, 7862784, 31454208, 125822976, 503304192, 2013241344, 8053014528, 32212156416, 128848822272, 515395682304, 2061583515648, 8246335635456, 32985345687552

Offset: 0

Klaus Brockhaus, Dec 09 2009

Binomial transform of A037480; second binomial transform of A133600.

First differences of A080960.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (6,-8).

Cf. A037480 ((5*3^n +(-1)^n -6)/8), A133600 (row sums of triangle A133599), A080960 (third binomial transform of A010685).

I:=[6,27]; [1] cat [n le 2 select I[n] else 6*Self(n-1) - 8*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 02 2021

Table[If[n==0, 1, 3*(5*4^n - 2*2^n)/8],{n,0,30}] (* G. C. Greubel, Dec 02 2021 *)
LinearRecurrence[{6,-8},{1,6,27},30] (* Harvey P. Dale, Oct 25 2023 *)

{m=21; v=concat([1, 6, 27], vector(m-3)); for(n=4, m, v[n]=6*v[n-1]-8*v[n-2]); v}

[1]+[3*(5*4^n - 2*2^n)/8 for n in (1..30)] # G. C. Greubel, Dec 02 2021

a(n) = 3*(5*4^n - 2*2^n)/8 for n > 0.
G.f.: (1-x)*(1+x)/((1-2*x)*(1-4*x)).
E.g.f.: (1/8)*(-1 - 6*exp(2*x) + 15*exp(4*x)). - G. C. Greubel, Dec 02 2021

A173261 Array T(n,k) read by antidiagonals: T(n,2k)=1, T(n,2k+1)=n, n>=2, k>=0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 1, 2, 1, 5, 1, 3, 1, 1, 6, 1, 4, 1, 2, 1, 7, 1, 5, 1, 3, 1, 1, 8, 1, 6, 1, 4, 1, 2, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2, 1, 13, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 14, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2

Offset: 2

Paul Curtz, Feb 14 2010

One may define another array B(n,0) = -1, B(n,k) = T(n,k-1) + 2*B(n,k-1), n>=2, which also starts in columns k>=0, as follows:
  -1, -1, 0, 1,  4,  9, 20,  41,  84, 169,  340,  681, 1364 ...: A084639;
  -1, -1, 1, 3,  9, 19, 41,  83, 169, 339,  681, 1363, 2729;
  -1, -1, 2, 5, 14, 29, 62, 125, 254, 509, 1022, 2045, 4094;
  -1, -1, 3, 7, 19, 39, 83, 167, 339, 679, 1363, 2727, 5459 ...: -A173114;
B(n,k) = (n-1)*A001045(k) - T(n,k).
First differences are B(n,k+1) - B(n,k) = (n-1)*A001045(k).

			The array T(n,k) starts in row n=2 with columns k>=0 as:
  1,  2, 1,  2, 1,  2, 1,  2, 1,  2, 1,  2 ... A000034;
  1,  3, 1,  3, 1,  3, 1,  3, 1,  3, 1,  3 ... A010684;
  1,  4, 1,  4, 1,  4, 1,  4, 1,  4, 1,  4 ... A010685;
  1,  5, 1,  5, 1,  5, 1,  5, 1,  5, 1,  5 ... A010686;
  1,  6, 1,  6, 1,  6, 1,  6, 1,  6, 1,  6 ... A010687;
  1,  7, 1,  7, 1,  7, 1,  7, 1,  7, 1,  7 ... A010688;
  1,  8, 1,  8, 1,  8, 1,  8, 1,  8, 1,  8 ... A010689;
  1,  9, 1,  9, 1,  9, 1,  9, 1,  9, 1,  9 ... A010690;
  1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10 ... A010691.
Antidiagonal triangle begins as:
  1;
  1,  2;
  1,  3,  1;
  1,  4,  1,  2;
  1,  5,  1,  3,  1;
  1,  6,  1,  4,  1,  2;
  1,  7,  1,  5,  1,  3,  1;
  1,  8,  1,  6,  1,  4,  1,  2;
  1,  9,  1,  7,  1,  5,  1,  3,  1;
  1, 10,  1,  8,  1,  6,  1,  4,  1,  2;
  1, 11,  1,  9,  1,  7,  1,  5,  1,  3,  1;
  1, 12,  1, 10,  1,  8,  1,  6,  1,  4,  1,  2;
  1, 13,  1, 11,  1,  9,  1,  7,  1,  5,  1,  3,  1;
  1, 14,  1, 12,  1, 10,  1,  8,  1,  6,  1,  4,  1,  2;

G. C. Greubel, Antidiagonals n = 0..50 of the array, flattened

Cf. A000034, A010684, A010685, A010686, A010687, A010688, A010689, A010690, A010691.

Cf. A001045, A024206, A084639, A093178, A173114.

T[n_, k_]:= (1/2)*((n+3) - (n+1)*(-1)^k);
Table[T[n-k, k], {n,2,17}, {k,2,n}]//Flatten (* G. C. Greubel, Dec 03 2021 *)

flatten([[(1/2)*((n-k+3) - (n-k+1)*(-1)^k) for k in (2..n)] for n in (2..17)]) # G. C. Greubel, Dec 03 2021

From G. C. Greubel, Dec 03 2021: (Start)
T(n, k) = (1/2)*((n+3) - (n+1)*(-1)^k).
Sum_{k=0..n} T(n-k, k) = A024206(n).
Sum_{k=0..floor((n+2)/2)} T(n-2*k+2, k) = (1/16)*(2*n^2 4*n -5*(1 +(-1)^n) + 4*sin(n*Pi/2)) (diagonal sums).
T(2*n-2, n) = A093178(n). (End)

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

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

Cf. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

A266973 a(n) = 4^n mod 17.

Original entry on oeis.org

1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16, 13, 1, 4, 16

Offset: 0

Vincenzo Librandi, Apr 06 2016

Period 4: repeat [1, 4, 16, 13].

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

Cf. similar sequences of the type 4^n mod p, where p is a prime: A010685 (5), A153727 (7), A168429 (11), A168430 (13), this sequence (17), A187532 (19).

Magma
```
[Modexp(4, n, 17): n in [0..100]];
```

A266973:=n->power(4,n) mod 17: seq(A266973(n), n=0..100); # Wesley Ivan Hurt, Jun 29 2016

Mathematica
```
PowerMod[4, Range[0, 100], 17]
```

G.f.: (1+4*x+16*x^2+13*x^3)/(1-x^4).
a(n) = a(n-4) for n>3.
From Wesley Ivan Hurt, Jun 29 2016: (Start)
a(n) = (34 - 3*(5 + 3*I)*I^(-n) - 3*(5 - 3*I)*I^n)/4 where I=sqrt(-1).
a(n) = (17 - 15*cos(n*Pi/2) - 9*sin(n*Pi/2))/2. (End)

cp's OEIS Frontend

A226044 Period of length 8: 1, 64, 16, 64, 4, 64, 16, 64.

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A136258 a(n) = 2*a(n-1) - 2*a(n-2), with a(0)=1, a(1)=5.

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A171475 a(n) = 6*a(n-1) - 8*a(n-2), for n > 2, with a(0) = 1, a(1) = 6, a(2) = 27.

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A173261 Array T(n,k) read by antidiagonals: T(n,2k)=1, T(n,2k+1)=n, n>=2, k>=0.

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A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

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A266973 a(n) = 4^n mod 17.

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A136258 a(n) = 2a(n-1) - 2a(n-2), with a(0)=1, a(1)=5.

A171475 a(n) = 6a(n-1) - 8a(n-2), for n > 2, with a(0) = 1, a(1) = 6, a(2) = 27.