A010684 -id:A010684 - cp's OEIS frontend

A206787 Sum of the odd squarefree divisors of n.

1, 1, 4, 1, 6, 4, 8, 1, 4, 6, 12, 4, 14, 8, 24, 1, 18, 4, 20, 6, 32, 12, 24, 4, 6, 14, 4, 8, 30, 24, 32, 1, 48, 18, 48, 4, 38, 20, 56, 6, 42, 32, 44, 12, 24, 24, 48, 4, 8, 6, 72, 14, 54, 4, 72, 8, 80, 30, 60, 24, 62, 32, 32, 1, 84, 48, 68, 18, 96, 48, 72, 4

Offset: 1

Reinhard Zumkeller, Feb 12 2012

a(A000079(n)) = 1; a(A057716(n)) > 1; a(A065119(n)) = 4; a(A086761(n)) = 6.

Inverse Mobius transform of 1, 0, 3, 0, 5, 0, 7, 0, 0, 0, 11, 0, 13, 0, 15, 0, 17, 0, 19, 0, 21, 0, 23, 0, 0, 0, 0, 0, 29... - R. J. Mathar, Jul 12 2012

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Jon Maiga, Computer-generated formulas for A206787, Sequence Machine.

Cf. A000203, A000593, A007947, A008683, A056911, A048250, A010684, A192066, A204455.
Cf. A000079, A057716, A065119, A086761.
Inverse Möbius transform of the absolute values of A349343.

a206787 = sum . filter odd . a206778_row

[&+[d:d in Divisors(m)|IsOdd(d) and IsSquarefree(d)]:m in [1..72]]; // Marius A. Burtea, Aug 14 2019

seq(add(d*mobius(2*d)^2, d in divisors(n)), n=1 .. 80); # Ridouane Oudra, Aug 14 2019

a[n_] := DivisorSum[n, #*Boole[OddQ[#] && SquareFreeQ[#]]&]; Array[a, 80] (* Jean-François Alcover, Dec 05 2015 *)
f[2, e_] := 1; f[p_, e_] := p + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = sumdiv(n, d, d*(d % 2)*issquarefree(d)); \\ Michel Marcus, Sep 21 2014

from math import prod
from sympy import primefactors
def A206787(n): return prod(1+(p if p>2 else 0) for p in primefactors(n)) # Chai Wah Wu, Oct 10 2024

a(n) = Sum_{k = 1..A034444(n)} (A206778(n,k) mod 2) * A206778(n,k).
a(n) = Sum_{d|n} d*mu(2*d)^2, where mu is the Möbius function (A008683). - Ridouane Oudra, Aug 14 2019
Multiplicative with a(2^e) = 1, and a(p^e) = p + 1 for p > 2. - Amiram Eldar, Sep 18 2020
Sum_{k=1..n} a(k) ~ (1/3) * n^2. - Amiram Eldar, Nov 17 2022
Dirichlet g.f.: (zeta(s)*zeta(s-1)/zeta(2*s-2))*(2^s/(2^s+2)). - Amiram Eldar, Jan 03 2023
From Antti Karttunen, Nov 22 2023: (Start)
a(n) = A000203(A204455(n)) = A000593(A007947(n)) = A048250(n)/A010684(n-1). [From Sequence Machine]
a(n) = Sum_{d|n} abs(A349343(d)). [See R. J. Mathar's Jul 12 2012 comment above]
(End)
a(n) = Sum_{d divides n, d odd} d * mu(d)^2. - Peter Bala, Feb 01 2024

A112030 a(n) = (2 + (-1)^n) * (-1)^floor(n/2).

Original entry on oeis.org

3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1

Offset: 0

Reinhard Zumkeller, Aug 27 2005

The fractions A112031(n)/A112032(n) give the partial sums of a(n)/floor((n+4)/2).

Sum of the two Cartesian coordinates from the image of the point (2,1) after n 90-degree counterclockwise rotations about the origin. - Wesley Ivan Hurt, Jul 06 2013

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

Cf. A010684, A016116, A112031, A112032, A112033.

A112030 := proc(n)
        (2 + (-1)^n) * (-1)^floor(n/2) ;
end proc: # R. J. Mathar, Jul 09 2013

LinearRecurrence[{0, -1}, {3, 1}, 100] (* Jean-François Alcover, Nov 24 2020 *)

a(n)=[3,1,-3,-1][n%4+1] \\ Charles R Greathouse IV, Aug 21 2011

def A112030(n): return (3, 1, -3, -1)[n&3] # Chai Wah Wu, Jan 31 2023

a(n) = A010684(n+1) * (-1)^floor(n/2).

O.g.f.: (3+x)/(1+x^2). - R. J. Mathar, Jan 09 2008

A100320 A Catalan transform of (1 + 2x)/(1 - 2x).

Original entry on oeis.org

1, 4, 12, 40, 140, 504, 1848, 6864, 25740, 97240, 369512, 1410864, 5408312, 20801200, 80233200, 310235040, 1202160780, 4667212440, 18150270600, 70690527600, 275693057640, 1076515748880, 4208197927440, 16466861455200, 64495207366200, 252821212875504, 991837065896208

Offset: 0

Paul Barry, Nov 14 2004

A Catalan transform of (1 + 2*x)/(1 - 2*x) under the mapping g(x) -> g(x*c(x)). (Here c(x) is the g.f. of A000108.) The original sequence can be retrieved by g(x) -> g(x*(1-x)).
Hankel transform is A144704. - Paul Barry, Sep 19 2008
Central terms of the triangle in A124927. - Reinhard Zumkeller, Mar 04 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.

Cf. A000108, A010684, A028329, A039599, A095660, A124927, A132046, A144704.

a100320 n = a124927 (2 * n) n  -- Reinhard Zumkeller, Mar 04 2012

[4*Binomial(2*n-1, n) - 3*0^n: n in [0..40]]; // G. C. Greubel, Feb 01 2023

a[0]= 1; a[n_]:= 2 Binomial[2 n, n];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 31 2018 *)

def A100320(n): return 4*binomial(2*n-1, n) - 3*0^n
[A100320(n) for n in range(41)] # G. C. Greubel, Feb 01 2023

G.f.: (1 + 2*x*c(x))/(1 - 2*x*c(x)), where c(x) is the g.f. of A000108.
a(n) = 4*binomial(2*n-1, n) - 3*0^n.
a(n) = binomial(2*n, n)*(4*2^(n-1) - 0^n)/2^n.
a(n) = Sum_{j=0..n} Sum_{k=0..n} C(2*n, n-k)*((2*k + 1)/(n + k + 1))*C(k, j)*(-1)^(j-k)*(4*2^(j-1) - 0^j).
a(n) = A028329(n), n > 0. - R. J. Mathar, Sep 02 2008
a(n) = T(2*n,n), where T(n,k) = A132046(n,k). - Paul Barry, Sep 19 2008
a(n) = Sum_{k=0..n} A039599(n,k)*A010684(k). - Philippe Deléham, Oct 29 2008
a(n) = A095660(2*n,n) for n > 0. - Reinhard Zumkeller, Apr 08 2012
G.f.: G(0) - 1, where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) + (k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
a(n) = [x^n] (1 + 2*x)/(1 - x)^(n+1). - Ilya Gutkovskiy, Oct 12 2017
a(n) = 2*(2*n-1)*a(n-1)/n. - G. C. Greubel, Feb 01 2023
E.g.f.: 2*exp(2*x)*BesselI(0, 2*x) - 1. - Stefano Spezia, May 11 2024

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A010696 Periodic sequence: Repeat 2,6.

Original entry on oeis.org

2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2, 6, 2

Offset: 0

N. J. A. Sloane

Original name: Period 2.

Also continued fraction expansion of 1+(2/3)*sqrt(3). - Bruno Berselli, Sep 22 2011

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

Cf. A174114. [From Reinhard Zumkeller, Mar 08 2010]

PadRight[{}, 100, {2, 6}] (* Paolo Xausa, Feb 22 2024 *)

G.f.: ( -2-6*x ) / ( (x-1)*(1+x) ). - R. J. Mathar, Jul 07 2011

a(n) = a(-n) = 2*A010684(n) = A131800(2n+1) = A010123(2n+2). - Bruno Berselli, Sep 22 2011

Definition rewritten by Bruno Berselli, Sep 22 2011

A084101 Expansion of (1+x)^2/((1-x)*(1+x^2)).

Original entry on oeis.org

1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1

Offset: 0

Paul Barry, May 15 2003

Partial sums of A084099. Inverse binomial transform of A000749 (without leading zeros).
From Klaus Brockhaus, May 31 2010: (Start)
Periodic sequence: Repeat 1, 3, 3, 1.
Interleaving of A010684 and A176040.
Continued fraction expansion of (7 + 5*sqrt(29))/26.
Decimal expansion of 121/909.
a(n) = A143432(n+3) + 1 = 2*A021913(n+1) + 1 = 2*A133872(n+3) + 1.
a(n) = A165207(n+1) - 1.
First differences of A047538.
Binomial transform of A084102. (End)
From Wolfdieter Lang, Feb 09 2012: (Start)
a(n) = A045572(n+1) (Modd 5) := A203571(A045572(n+1)), n >= 0.
For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the five residue classes Modd 5, called [m] for m=0,1,...,4, are shown in the array A090298 if there the last row is taken as class [0] after inclusion of 0.
(End)

			From _Wolfdieter Lang_, Feb 09 2012: (Start)
Modd 5 of nonnegative odd numbers restricted mod 5:
A045572: 1, 3, 7, 9, 11, 13, 17, 19, 21, 23, ...
Modd 5:  1, 3, 3, 1,  1,  3,  3,  1,  1,  3, ...
(End)

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

Cf. A084102.

Cf. A010684 (repeat 1, 3), A176040 (repeat 3, 1), A178593 (decimal expansion of (7+5*sqrt(29))/26), A143432 (expansion of (1+x^4)/((1-x)*(1+x^2))), A021913 (repeat 0, 0, 1, 1), A133872 (repeat 1, 1, 0, 0), A165207 (repeat 2, 2, 4, 4), A047538 (congruent to 0, 1, 4 or 7 mod 8), A084099 (expansion of (1+x)^2/(1+x^2)), A000749 (expansion of x^3/((1-x)^4-x^4)). - Klaus Brockhaus, May 31 2010

R:=PowerSeriesRing(Integers(), 100); Coefficients(R!( (1+x)^2/((1-x)*(1+x^2)) )); // G. C. Greubel, Feb 28 2019

CoefficientList[Series[(1+x)^2/((1-x)(1+x^2)),{x,0,110}],x] (* or *) PadRight[{},110,{1,3,3,1}] (* Harvey P. Dale, Nov 21 2012 *)

x='x+O('x^100); Vec((1+x)^2/((1-x)*(1+x^2))) \\ Altug Alkan, Dec 24 2015

((1+x)^2/((1-x)*(1+x^2))).series(x, 100).coefficients(x, sparse=False) # G. C. Greubel, Feb 28 2019

a(n) = binomial(3, n mod 4). - Paul Barry, May 25 2003
From Klaus Brockhaus, May 31 2010: (Start)
a(n) = a(n-4) for n > 3; a(0) = a(3) = 1, a(1) = a(2) = 3.
a(n) = (4 - (1+i)*i^n - (1-i)*(-i)^n)/2 where i = sqrt(-1). (End)
E.g.f.: 2*exp(x) + sin(x) - cos(x). - Arkadiusz Wesolowski, Nov 04 2017
a(n) = 2 - (-1)^(n*(n+1)/2). - Guenther Schrack, Feb 26 2019

A169609 Period 3: repeat [1, 3, 3].

Original entry on oeis.org

1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3

Offset: 0

Klaus Brockhaus, Dec 03 2009

Interleaving of A000012, A010701 and A010701.
Also continued fraction expansion of (5+sqrt(65))/10 = 1.3062257748....
Also decimal expansion of 133/999.
a(n) = A144437(n) for n > 0.
Unsigned version of A154595.
Binomial transform of A168615.
Inverse binomial transform of A168673.
Essentially first differences of A047347.

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

Cf. A000012 (all 1's sequence), A010701 (all 3's sequence), A144437 (repeat 3, 3, 1), A154595 (repeat 1, 3, 3, -1, -3, -3), A168615, A168673, A047347 (congruent to {0, 1, 4} mod 7), A010684 (repeat 1, 3).
Cf. A171419 (decimal expansion of (5+sqrt(65))/10).
Cf. A146094.

[ n mod 3 eq 0 select 1 else 3: n in [0..104] ];

&cat [[1, 3, 3]^^30]; // Wesley Ivan Hurt, Jul 02 2016

seq(op([1, 3, 3]), n=0..50); # Wesley Ivan Hurt, Jul 02 2016

PadRight[{},120,{1,3,3}] (* or *) LinearRecurrence[{0,0,1},{1,3,3},120] (* Harvey P. Dale, Apr 29 2015 *)

a(n) = a(n-3) for n > 2, with a(0) = 1, a(1) = 3, a(2) = 3.
G.f.: (1+3*x+3*x^2)/(1-x^3).
a(n) = (7/3)+(2/3)*cos((2*Pi/3)*(n+1))-(2*sqrt(3)/3)*sin((2*Pi/3)*(n+1)). [Richard Choulet, Mar 15 2010]
a(n) = a(n-a(n-2)) for n>=2. Example: a(5) = a(5-a(3)) = a(5-a(3-a(1))) = a(5-a(3-3)) = a(5-a(0)) = a(5-1) = a(4) = a(4-a(2)) = a(4-3) = a(1) = 3. [Richard Choulet, Mar 15 2010; edited by Klaus Brockhaus, Nov 21 2010]
a(n) = 1 + 2*sgn(n mod 3). - Wesley Ivan Hurt, Jul 02 2016
a(n) = 3/gcd(n,3). - Wesley Ivan Hurt, Jul 11 2016

Keywords cofr, cons added by Klaus Brockhaus, Apr 20 2010

Minor edits, crossref added by Klaus Brockhaus, May 03 2010

A171476 a(n) = 6a(n-1) - 8a(n-2) for n > 1, a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 28, 120, 496, 2016, 8128, 32640, 130816, 523776, 2096128, 8386560, 33550336, 134209536, 536854528, 2147450880, 8589869056, 34359607296, 137438691328, 549755289600, 2199022206976, 8796090925056, 35184367894528

Offset: 0

Klaus Brockhaus, Dec 09 2009

Binomial transform of A048473; second binomial transform of A151821; third binomial transform of A010684; fourth binomial transform of A084633 without second term 0; fifth binomial transform of A168589.
Inverse binomial transform of A081625; second inverse binomial transform of A081626; third inverse binomial transform of A081627.
Partial sums of A010036.
Essentially first differences of A006095.
a(n) = A109241(n) converted from binary to decimal. - Robert Price, Jan 19 2016
a(n) is the area enclosed by a Hilbert curve with unit length segments after n iterations, when the start and end points are joined. - Jennifer Buckley, Apr 23 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jennifer Buckley, The Area Enclosed by a Hilbert Curve with Unit Length Segments
Gordon Hamilton, Cookie Monster Counts Cookies, Youtube video, 2010.
Index entries for linear recurrences with constant coefficients, signature (6,-8).

Cf. A006516 (2^(n-1)*(2^n-1)), A020522 (4^n-2^n), A048473 (2*3^n-1), A151821 (powers of 2, omitting 2 itself), A010684 (repeat 1, 3), A084633 (inverse binomial transform of repeated odd numbers), A168589 ((2-3^n)*(-1)^n), A081625 (2*5^n-3^n), A081626 (2*6^n-4^n), A081627 (2*7^n-5^n), A010036 (sum of 2^n, ..., 2^(n+1)-1), A006095 (Gaussian binomial coefficient [n, 2] for q=2), A171472, A171473.

[2*4^n-2^n: n in [0..30]]; // Vincenzo Librandi, Jul 17 2011

LinearRecurrence[{6,-8},{1,6},30] (* Harvey P. Dale, Aug 02 2020 *)

m=23; v=concat([1, 6], vector(m-2)); for(n=3, m, v[n]=6*v[n-1]-8*v[n-2]); v

a(n) = Sum_{k=1..2^n-1} k.
a(n) = 2*4^n - 2^n.
G.f.: 1/((1-2*x)*(1-4*x)).
a(n) = A006516(n+1).
a(n) = 4*a(n-1) + 2^n for n > 0, a(0)=1. - Vincenzo Librandi, Jul 17 2011
a(n) = Sum_{k=0..n} 2^(n+k). - Bruno Berselli, Aug 07 2013
a(n) = A020522(n+1)/2. - Hussam al-Homsi, Jun 06 2021
E.g.f.: exp(2*x)*(2*exp(2*x) - 1). - Stefano Spezia, Dec 10 2021

A112086 a(n) = the period of the first differences of the n-th row of A112060 (or A112070), or 0 if that row does not have a periodic first difference.

Original entry on oeis.org

2, 4, 6, 16, 72, 420, 3240

Offset: 1

Antti Karttunen, Aug 28 2005

These values have been computed empirically. An independent recomputation or a mathematical proof would be welcome. The initial terms factored: 2, 2*2, 2*3, 2*2*2*3*3, 2*2*7*3*5, 2*2*2*3*3*3*3*5, ...

These are the periods of A010684, A112132, A112133, A112134, A112135, A112136, A112137, etc. (Periods of A112138 & A112139 not computed yet.) If we sum the period length prefixes of these sequences, as Sum_{i=1..a(1)} A010684(i), Sum_{i=1..a(2)} A112132(i), Sum_{i=1..a(3)} A112133(i), etc., we get the sequence 4, 12, 60, 420, 4620, 60060, 1021020, ... (cf. A097250) and when doubled, it yields: 8, 24, 120, 840, 9240, 120120, 2042040, ... (cf. A066631 and A102476).

A153284 a(n) = n + Sum_{j=1..n-1} (-1)^j * a(j) for n >= 2, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1

Offset: 1

Walter Carlini, Dec 23 2008

Row sums of triangle A153860. - Gary W. Adamson, Jan 03 2009

1 followed by interleaving of A000012 and A010701. - Klaus Brockhaus, Jan 04 2009

			a(1)=1, a(2)=2-a(1)=2-1=1, a(3)=3+a(2)-a(1)=3+1-1=3, a(4)=4-a(3)+a(2)-a(1)=4-3+1-1=1, a(5)=5+1-3+1-1=3, a(6)=6-3+1-3+1-1=1, a(7)=7+1-3+1-3+1-1, etc.

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

Equals A010684 with the addition of the leading term of 1
The first sequence of a family that includes A153285 and A153286
Cf. A153860.
Cf. A000012 (all 1's sequence), A010701 (all 3's sequence). - Klaus Brockhaus, Jan 04 2009

S:=[ 1 ]; for n in [2..105] do Append(~S, n + &+[ (-1)^j*S[j]: j in [1..n-1] ]); end for; S; // Klaus Brockhaus, Jan 04 2009

a(n)=1 if n is 1 or even; a(n)=3 if n is odd other than 1.

G.f.: x*(1 + x + 2*x^2)/((1+x)*(1-x)). - Klaus Brockhaus, Jan 04 2009 and Oct 15 2009

A334006 Triangle read by rows: T(n,k) = (the number of nonnegative bases m < n such that m^k == m (mod n))/(the number of nonnegative bases m < n such that -m^k == m (mod n)) for nonnegative k < n, n >= 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 5, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 7, 1, 3, 1, 3, 1, 1, 4, 1, 5, 1, 5, 1, 5, 1, 9, 1, 3, 1, 3, 1, 7, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 11, 1, 3, 1, 3, 1, 3, 1, 3, 1, 1, 6, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 13, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 7, 1, 3, 1, 3

Offset: 1

Juri-Stepan Gerasimov, Apr 12 2020

If the sum of proper divisors of q in row q <= q, then q are 1, 2, 3, 4, 5, 8, 16, 17, 32, 64, 128, 256, 257, ...(union of Fermat primes and powers of 2).

			Triangle T(n,k) begins:
  n\k| 0   1  2  3  4   5  6  7  8   9 10 11 12  13 14 15 16
  ---+------------------------------------------------------
   1 | 1;
   2 | 1,  1;
   3 | 1,  3, 1;
   4 | 1,  2, 1, 3;
   5 | 1,  5, 1, 1, 1;
   6 | 1,  3, 1, 3, 1,  3;
   7 | 1,  7, 1, 3, 1,  3, 1;
   8 | 1,  4, 1, 5, 1,  5, 1, 5;
   9 | 1,  9, 1, 3, 1,  3, 1, 7, 1;
  10 | 1,  5, 1, 1, 1,  5, 1, 1, 1,  5;
  11 | 1, 11, 1, 3, 1,  3, 1, 3, 1,  3, 1;
  12 | 1,  6, 1, 9, 1,  9, 1, 9, 1,  9, 1, 9;
  13 | 1, 13, 1, 1, 1,  5, 1, 1, 1,  5, 1, 1, 1;
  14 | 1,  7, 1, 3, 1,  3, 1, 7, 1,  3, 1, 3, 1, 7;
  15 | 1, 15, 1, 3, 1, 15, 1, 3, 1, 15, 1, 3, 1, 15, 1;
  16 | 1,  8, 1, 5, 1,  9, 1, 5, 1,  9, 1, 5, 1,  9, 1, 5;
  17 | 1, 17, 1, 1, 1,  1, 1, 1, 1,  1, 1, 1, 1,  1, 1, 1, 1;
  ...
For (n, k) = (7, 3), there are three nonnegative values of m < n such that m^3 == m (mod 7) (namely 0, 1, and 6) and one nonnegative value of m < n such that -m^3 == m (mod 7) (namely 0), so T(7,3) = 3/1 = 3.

Cf. A000012, A000079, A002322, A010684, A019434, A023506, A026741, A182816, A333570, A337454, A337632, A337633, A337820, A337910.

[[#[m: m in [0..n-1] | m^k mod n eq m]/#[m: m in [0..n-1] | -m^k mod n eq m]: k in [0..n-1]]: n in [1..17]];

T(n, k) = sum(m=0, n-1, Mod(m, n)^k == m)/sum(m=0, n-1, -Mod(m, n)^k == m);
matrix(7, 7, n, k, k--; if (k>=n, 0, T(n,k))) \\ to see the triangle \\ Michel Marcus, Apr 17 2020

Name corrected by Peter Kagey, Sep 12 2020

cp's OEIS Frontend

A206787 Sum of the odd squarefree divisors of n.

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A112030 a(n) = (2 + (-1)^n) * (-1)^floor(n/2).

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A100320 A Catalan transform of (1 + 2*x)/(1 - 2*x).

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A010696 Periodic sequence: Repeat 2,6.

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A084101 Expansion of (1+x)^2/((1-x)*(1+x^2)).

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A169609 Period 3: repeat [1, 3, 3].

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A100320 A Catalan transform of (1 + 2x)/(1 - 2x).

A171476 a(n) = 6a(n-1) - 8a(n-2) for n > 1, a(0)=1, a(1)=6.