A070432 -id:A070432 - cp's OEIS frontend

A048152 Triangular array T read by rows: T(n,k) = k^2 mod n, for 1 <= k <= n, n >= 1.

0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 4, 4, 1, 0, 1, 4, 3, 4, 1, 0, 1, 4, 2, 2, 4, 1, 0, 1, 4, 1, 0, 1, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, 0, 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0

Offset: 1

Clark Kimberling

			Rows:
  0;
  1, 0;
  1, 1, 0;
  1, 0, 1, 0;
  1, 4, 4, 1, 0;
  1, 4, 3, 4, 1, 0;

T. D. Noe, Rows n = 1..100 of triangle, flattened
Eric Weisstein's World of Mathematics, Quadratic Residue

Cf. A060036.
Cf. A225126 (central terms).
Cf. A070430 (row 5), A070431 (row 6), A053879 (row 7), A070432 (row 8), A008959 (row 10), A070435 (row 12), A070438 (row 15), A070422 (row 20).
Cf. A046071 (in ascending order, without zeros and duplicates).
Cf. A063987 (for primes, in ascending order, without zeros and duplicates).

a048152 n k = a048152_tabl !! (n-1) !! (k-1)
a048152_row n = a048152_tabl !! (n-1)
a048152_tabl = zipWith (map . flip mod) [1..] a133819_tabl
-- Reinhard Zumkeller, Apr 29 2013

Flatten[Table[PowerMod[k,2,n],{n,15},{k,n}]] (* Harvey P. Dale, Jun 20 2011 *)

T(n,k) = A133819(n,k) mod n, k = 1..n. - Reinhard Zumkeller, Apr 29 2013

T(n,k) = (T(n,k-1) + (2k+1)) mod n. - Andrés Ventas, Apr 06 2021

A070430 a(n) = n^2 mod 5.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, May 12 2002

Equivalently n^6 mod 5. - Zerinvary Lajos, Nov 06 2009

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

Cf. A053879, A070431, A070432.

Table[Mod[n^2,5],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)
PowerMod[Range[0, 200], 2, 5] (* G. C. Greubel, Mar 22 2016 *)

a(n)=n^2%5 \\ Charles R Greathouse IV, Sep 28 2015

[power_mod(n,2,5)for n in range(0, 101)] # Zerinvary Lajos, Nov 06 2009

[power_mod(n,6,5)for n in range(0, 101)] # Zerinvary Lajos, Nov 06 2009

From R. J. Mathar, Apr 20 2010: (Start)
a(n) = a(n-5).
G.f.: -x*(1+x)*(x^2+3*x+1) / ( (x-1)*(1+x+x^2+x^3+x^4) ). (End)

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

Original entry on oeis.org

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

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

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.

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

A109753 n^3 mod 8; the periodic sequence {0,1,0,3,0,5,0,7}.

Original entry on oeis.org

0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0, 1, 0, 3, 0, 5, 0, 7, 0

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Aug 11 2005

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

Cf. n mod 8 = A010877; n^2 mod 8 = A070432.

a(n)=n^3%8 \\ Charles R Greathouse IV, Apr 06 2016

[power_mod(n,3,8 ) for n in range(0, 105)] # - Zerinvary Lajos, Oct 29 2009

G.f. = (x+3x^3+5x^5+7x^7)/(1-x^8)

cp's OEIS Frontend

A048152 Triangular array T read by rows: T(n,k) = k^2 mod n, for 1 <= k <= n, n >= 1.

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A070430 a(n) = n^2 mod 5.

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

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

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

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

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A109753 n^3 mod 8; the periodic sequence {0,1,0,3,0,5,0,7}.

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