A070431 -id:A070431 - cp's OEIS frontend

A070434 a(n) = n^2 mod 11.

0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1

Offset: 0

N. J. A. Sloane, May 12 2002

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

Cf. A010880, A070430, A070431, A070432.

Table[Mod[n^2,11],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)

a(n)=n^2%11 \\ Charles R Greathouse IV, Apr 06 2016

From R. J. Mathar, Apr 20 2010: (Start)
a(n) = a(n-11).
G.f.: ( -x*(1+x)*(x^8+3*x^7+6*x^6-x^5+4*x^4-x^3+6*x^2+3*x+1) ) / ( (x-1)*(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10) ). (End)
a(n) = A010880(n^2). - Michel Marcus, Mar 24 2016

A070432 Period 4: repeat [0, 1, 4, 1]; a(n) = n^2 mod 8.

Original entry on oeis.org

0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0

Offset: 0

N. J. A. Sloane, May 12 2002

Multiplicative with a(2) = 4, a(2^e) = 0 if e >= 2, a(p^e) = 1 otherwise. - David W. Wilson, Jun 12 2005

			G.f. = x + 4*x^2 + x^3 + x^5 + 4*x^6 + x^7 + x^9 + 4*x^10 + x^11 + x^13 + ...

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1). - _R. J. Mathar_, Apr 20 2010

Cf. A070430, A070431.

&cat [[0, 1, 4, 1]^^30]; // Wesley Ivan Hurt, Dec 21 2016

seq(n mod 2 + 4*floor(((n+1) mod 4)/3), n = 0..200) # Gary Detlefs, Dec 29 2011

Table[Mod[n^2, 8], {n, 0, 99}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)
Mod[Range[0, 99]^2, 8] (* Alonso del Arte, Mar 20 2015 *)

a(n)=n^2%8 \\ Charles R Greathouse IV, Oct 07 2015

From R. J. Mathar, Apr 20 2010: (Start)
a(n) = a(n-4) for n > 3.
G.f.: -x*(1+4*x+x^2) / ( (x-1)*(1+x)*(x^2+1) ). (End)
Dirichlet g.f.: zeta(s)*(1 + 4*2^(-s))*(1 - 2^(-s)). - R. J. Mathar, Mar 10 2011
a(n) = (n mod 2) + 4*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 29 2011
From Wesley Ivan Hurt, Mar 19 2015: (Start)
a(n) = (((n+1) mod 4) - 1)^2.
a(n) = (1 + (-1)^n - 2(-1)^((2n + 1 - (-1)^n)/4))^2/4. (End)
E.g.f.: 2*cosh(x) + sinh(x) - 2*cos(x). - G. C. Greubel, Mar 22 2016
a(n) = (3 + cos(n*Pi) - 4*cos(n*Pi/2))/2. - Wesley Ivan Hurt, Dec 21 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Dec 22 2016

A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 4, 9, 10, 11, 6, 13, 14, 15, 8, 17, 18, 19, 10, 21, 22, 23, 12, 25, 26, 27, 14, 29, 30, 31, 16, 33, 34, 35, 18, 37, 38, 39, 20, 41, 42, 43, 22, 45, 46, 47, 24, 49, 50, 51, 26, 53, 54, 55, 28, 57, 58, 59, 30, 61, 62, 63, 32, 65, 66, 67, 34, 69, 70, 71, 36, 73, 74, 75, 38, 77, 78, 79, 40, 81, 82, 83, 42, 85, 86, 87, 44, 89, 90, 91, 46, 93, 94, 95, 48, 97, 98, 99

Offset: 1

R. J. Mathar, Feb 25 2011

a(n) is the length of the period of the sequence k^2 mod n, k=1,2,3,4,..., i.e., the length of the period of A000035 (n=2), A011655 (n=3), A000035 (n=4), A070430 (n=5), A070431 (n=6), A053879 (n=7), A070432 (n=8), A070433 (n=9), A008959 (n=10), A070434 (n=11), A070435 (n=12) etc.
From Franklin T. Adams-Watters, Feb 24 2011: (Start)
Clearly if gcd(n,m) = 1, a(nm) = lcm(a(n),a(m)), so it suffices to establish this for prime powers.
If p is a prime, the period must divide p, but k^2 mod p is not constant, so a(p) = p.
a(p^e), e > 1, must be divisible by a(p^(e-1)), and must divide p^e. If p != 2, (p^(e-1)+1)^2 =  p^(2e-2)+2p^(e-1)+1 == 2p^(e-1)+1 (mod p^2), so a(p^e) != p^(e-1); it must then be e.
By inspection, a(4) = 2 and a(8) = 4.
This leaves a(2^e), e > 3. But then (2^(e-2)+1)^2 = 2^(2e-4)+2^(e-1)+1 == 2^(e-1)+1 (mod 2^e), so a(n) > 2^(e-2). On the other hand, (2^(e-1)+c)^2 = 2^(2e-2)+c2^e+c^2 == c^2 (mod 2^e). Hence the period is 2^(e-1). (End)

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

Cf. A000224 (size of the set of moduli of k^2 mod n), A019554, A060819, A061037, A090129, A142705, A164115, A283971.

A186646 := proc(n) if n mod 4 = 0 then n/2 ; else n ; end if; end proc ;

Flatten[Table[{n,n+1,n+2,(n+3)/2},{n,1,101,4}]] (* or *) LinearRecurrence[ {0,0,0,2,0,0,0,-1},{1,2,3,2,5,6,7,4},100] (* Harvey P. Dale, May 30 2014 *)
Table[n (7 - (-1)^n - 2 Cos[n Pi/2])/8, {n, 100}] (* Federico Provvedi , Jan 02 2018 *)

a(n)=if(n%4,n,n/2) \\ Charles R Greathouse IV, Oct 16 2015

def A186646(n): return n if n&3 else n>>1 # Chai Wah Wu, Jan 10 2023

a(n) = 2*a(n-4) - a(n-8).
a(4n) = 2n; a(4n+1) = 4n+1; a(4n+2) = 4n+2; a(4n+3) = 4n+3.
a(n) = n/A164115(n).
G.f.: x*(1 + 2*x + 3*x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6) / ( (x-1)^2*(1+x)^2*(x^2+1)^2 ).
Dirichlet g.f.: (1-2/4^s)*zeta(s-1).
A019554(n) | a(n). - Charles R Greathouse IV, Feb 24 2011
a(n) = n*(7 - (-1)^n - (-i)^n - i^n)/8, with i=sqrt(-1). - Bruno Berselli, Feb 25 2011
Multiplicative with a(p^e)=2^e if p=2 and e<=1; a(p^e)=2^(e-1) if p=2 and e>=2; a(p^e)=p^e otherwise. - David W. Wilson, Feb 26 2011
a(n) * A060819(n+2) = A142705(n+1) = A061037(2n+2). - Paul Curtz, Mar 02 2011
a(n) = n - (n/2)*floor(((n-1) mod 4)/3). - Gary Detlefs, Apr 14 2013
a(2^n) = A090129(n+1). - R. J. Mathar, Oct 09 2014
a(n) = n*(7 - (-1)^n - 2*cos(n*Pi/2))/8. - Federico Provvedi, Jan 02 2018
E.g.f.: (1/4)*x*(4*cosh(x) + sin(x) + 3*sinh(x)). - Stefano Spezia, Jan 26 2020
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

A070433 a(n) = n^2 mod 9.

Original entry on oeis.org

0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1

Offset: 0

N. J. A. Sloane, May 12 2002

Also decimal expansion of 4692347/333333333. - Enrique Pérez Herrero, Nov 27 2022

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

Cf. A000290, A010878, A053879, A070430, A070431.

Table[Mod[n^2,9],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)
PowerMod[Range[0,200],2,9] (* Harvey P. Dale, Jun 11 2011 *)

a(n)=[0,1,4,0,7,7,0,4,1][n%9+1] \\ Charles R Greathouse IV, Jun 11 2011

a(n)=n^2%9 \\ Charles R Greathouse IV, Jun 11 2011

From R. J. Mathar, Apr 20 2010: (Start)
a(n) = a(n-9).
G.f.: ( -x*(1+x)*(x^6+3*x^5-3*x^4+10*x^3-3*x^2+3*x+1) ) / ( (x-1)*(1+x+x^2)*(x^6+x^3+1) ). (End)
a(n) = A010878(A000290(n)) = A010878(n^2). - Enrique Pérez Herrero, Nov 27 2022

A174452 a(n) = n^2 mod 1000.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 24, 89, 156, 225, 296, 369, 444, 521, 600, 681, 764, 849, 936, 25, 116, 209, 304, 401, 500, 601, 704, 809, 916, 25

Offset: 0

Reinhard Zumkeller, Mar 21 2010

a(n) = A000290(n) for n < 32, but a(32) = 24;
A008959(n) = a(n) mod 10; A002015(n) = a(n) mod 100;
periodic with period 500: a(n+500)=a(n) and a(250*n+k)=a(250*n-k) for k <= 250*n;
a(n) = (n mod 1000)^2 mod 1000;
a(m*n) = a(m)*a(n) mod 1000;
A122986 gives the range of this sequence;
a(n) = n for n = 0, 1, and 376.

			Some calculations for n=982451653, to be realized by hand:
a(n) = (53^2 + 200*6*3) mod 1000 = 6409 mod 1000 = 409;
a(n) = (653^2) mod 1000 = 426409 mod 1000 = 409;
a(n) = a(n mod 500) = a(153) = 409;
a(n) = 965211250482432409 mod 1000 = 409.

Cf. A053879, A070430, A070431, A070432, A070433, A070434, A070435, A070438, A070442, A070452, A159852.

a174452 = (`mod` 1000) . (^ 2)  -- Reinhard Zumkeller, Jul 06 2011

seq(n^2 mod 1000, n=0..55); # Nathaniel Johnston, Jun 22 2011

PowerMod[Range[0,60],2,1000] (* Harvey P. Dale, Feb 08 2022 *)

a(n)=n^2%1000 \\ Charles R Greathouse IV, Apr 06 2016

a(n) = ((n mod 100)^2 + 200 * (floor(n/100) mod 10) * (n mod 10)) mod 1000.

A239418 Numbers n such that n^10 - 10 is prime.

Original entry on oeis.org

21, 201, 267, 321, 369, 459, 537, 651, 669, 699, 723, 753, 1071, 1113, 1197, 1203, 1209, 1323, 1401, 1503, 1587, 1647, 1773, 1791, 1797, 1917, 1941, 2007, 2139, 2223, 2427, 2493, 2613, 2733, 2769, 2787, 2847, 3147, 3249, 3267, 3297, 3399, 3423, 3441, 3771

Offset: 1

Derek Orr, Mar 17 2014

nonn

All of the numbers in this sequence are odd multiples of 3 and, thus, congruent to 3 (mod 6).

The tenth powers modulo 6 are 1, 4, 3, 4, 1, 0, ... (A070431). Subtracting 10 (still modulo 6), we get 3, 0, 5, 0, 3, 2, ... which means that only n = 3 mod 6 can produce a potential prime p = 5 mod 6.

			21^10 - 10 = 16679880978191 is prime. Thus, 21 is a member of this sequence.

Cf. A028870, A153974, A239413, A239414, A239415, A239416, A239417, A239347.

Select[Range[1000], PrimeQ[#^10 - 10] &] (* Alonso del Arte, Mar 18 2014 *)

is(n)=isprime(n^10-10) \\ Charles R Greathouse IV, Feb 20 2017

import sympy
from sympy import isprime
{print(n) for n in range(10**4) if isprime(n**10-10)}

A002015 a(n) = n^2 reduced mod 100.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81

Offset: 0

N. J. A. Sloane

Periodic with period 50: (0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1) and next term is 0. The period is symmetrical about the "midpoint" 25. - Zak Seidov, Oct 26 2009

A010461 gives the range of this sequence. - Reinhard Zumkeller, Mar 21 2010

Index entries for sequences related to final digits of numbers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,1).

Cf. A053879, A070430, A070431, A070432, A070433, A070434, A070435, A070438, A070442, A070452, A159852.

PowerMod[Range[0,60],2,100] (* Harvey P. Dale, Nov 28 2012 *)

a(n)=n^2%100 \\ Charles R Greathouse IV, Oct 07 2015

From Reinhard Zumkeller, Mar 21 2010: (Start)
a(n) = (n mod 10) * ((n mod 10) + 20 * ((n\10) mod 10)) mod 100.
a(n) = A174452(n) mod 100; A008959(n) = a(n) mod 10;
a(m*n) = a(m)*a(n) mod 100;
a(n) = (n mod 100)^2 mod 100;
a(n) = n for n = 0, 1, and 25. (End)

Definition rephrased at the suggestion of Zak Seidov, Oct 26 2009

A034796 a(1)=1, a(n-1) is a square mod a(n), and a(n) > a(n-1).

Original entry on oeis.org

1, 2, 7, 9, 10, 13, 17, 19, 25, 26, 34, 37, 41, 43, 49, 50, 62, 67, 73, 74, 82, 87, 89, 94, 97, 99, 105, 106, 109, 113, 121, 122, 127, 129, 130, 133, 137, 139, 145, 146, 157, 162, 167, 173, 178, 181, 185, 187, 193, 194, 206, 214, 217, 218, 223, 237, 241, 243, 249

Offset: 1

David W. Wilson

a(n) is the smallest number larger than a(n-1) such that a(n-1) is a quadratic residue mod a(n). - R. J. Mathar, Jul 27 2015

			For n=3 we have a(2)=2. 2 is not quadratic residue mod 3 because the quadratic residues mod 3 are {0,1}, see A011655. 2 is not a quadratic residue mod 4 because the quadratic residues mod 4 are {0,1}, see A000035. 2 is not a quadratic residue mod 5 because the quadratic residues mod 5 are {0,1,4}, see A070430. 2 is not a quadratic residue mod 6 because the quadratic residues mod 6 are {0,1,3,4}, see A070431. 2 is a quadratic residue mod 7 because the quadratic residues mod 7 are {0,1,2,4}, see A053879. So a(3)=7. - _R. J. Mathar_, Jul 27 2015

T. D. Noe, Table of n, a(n) for n=1..1000

A034796 := proc(n)
    option remember;
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if numtheory[quadres](procname(n-1),a) = 1 then
                return a;
            end if;
        end do:
    end  if;
end proc: # R. J. Mathar, Jul 27 2015

residueQ[n_, k_] := Length[ Select[ Range[ Floor[k/2]]^2, Mod[#, k] == n &, 1]] == 1; a[1] = 1; a[n_] := a[n] = For[k = a[n-1] + 1, True, k++, If[residueQ[a[n-1], k], Return[k]]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Aug 13 2013 *)

Clarified definition, Joerg Arndt, Aug 14 2013

A114448 Array a(n,k) = n^k (mod k) read by antidiagonals (k>=1, n>=1).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 2, 1, 0, 0, 2, 0, 3, 4, 1, 0, 1, 0, 1, 4, 3, 2, 1, 0, 0, 1, 0, 0, 4, 3, 0, 1, 0, 1, 2, 1, 1, 1, 4, 1, 8, 1, 0, 0, 0, 0, 2, 0, 5, 0, 0, 4, 1, 0, 1, 1, 1, 3, 1, 6, 1, 1, 9, 2, 1, 0, 0, 2, 0, 4, 4, 0, 0, 8, 6, 3, 4, 1, 0, 1, 0, 1, 0, 3, 1, 1, 0, 5, 4, 9, 2, 1

Offset: 1

Leroy Quet, Feb 14 2006

Alternate description: triangular array a(n, k) = n^k (mod k) read by rows (n > 1, 0 < k < n). This is equivalent because a(n, k) = a(n-k, k). - David Wasserman, Jan 25 2007

			2^6 = 64 and 64 (mod 6) is 4. So a(2,6) = 4.

Cf. A051127, A051128, A094263. Transpose of A060154. First 13 rows are A057427(n-1), A015910, A066601, A066602, A066603, A066604, A066438, A066439, A066440, A056969, A066441, A066442, A116609. First 12 columns are A000004, A000035, A010872, A000035, A010874, A070431, A010876, A000035, A021559(n+1), A008959, A010880, A070435.

a[n_, k_] := Mod[n^k, k]; Table[a[n - k + 1, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 12 2012 *)

More terms from David Wasserman, Jan 25 2007

cp's OEIS Frontend

A070434 a(n) = n^2 mod 11.

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A070432 Period 4: repeat [0, 1, 4, 1]; a(n) = n^2 mod 8.

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A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

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A070433 a(n) = n^2 mod 9.

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A174452 a(n) = n^2 mod 1000.

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A239418 Numbers n such that n^10 - 10 is prime.

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A002015 a(n) = n^2 reduced mod 100.

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