A070434 -id:A070434 - cp's OEIS frontend

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

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

A112651 Numbers k such that k^2 == k (mod 11).

Original entry on oeis.org

0, 1, 11, 12, 22, 23, 33, 34, 44, 45, 55, 56, 66, 67, 77, 78, 88, 89, 99, 100, 110, 111, 121, 122, 132, 133, 143, 144, 154, 155, 165, 166, 176, 177, 187, 188, 198, 199, 209, 210, 220, 221, 231, 232, 242, 243, 253, 254, 264, 265, 275, 276, 286, 287, 297, 298

Offset: 1

Jeremy Gardiner, Dec 28 2005

Numbers that are congruent to {0,1} (mod 11). - Philippe Deléham, Oct 17 2011

			12 is a term because 12*12 = 144 == 1 (mod 11) and 12 == 1 (mod 11).

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

Cf. A010880 (n mod 11), A070434 (n^2 mod 11).

Cf. A005015, A030308.

m = 11 for n = 1 to 300 if n^2 mod m = n mod m then print n; next n

Select[Range[0,300],PowerMod[#,2,11]==Mod[#,11]&] (* or *) LinearRecurrence[ {1,1,-1},{0,1,11},60] (* Harvey P. Dale, Apr 19 2015 *)

a(n)=11*n/2-31/4-9*(-1)^n/4 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = 11*n - a(n-1) - 21 (with a(1)=0). - Vincenzo Librandi, Nov 13 2010
From R. J. Mathar, Oct 08 2011: (Start)
a(n) = 11*n/2 - 31/4 - 9*(-1)^n/4.
G.f.: x^2*(1+10*x) / ( (1+x)*(x-1)^2 ). (End)
a(n+1) = Sum_{k>=0} A030308(n,k)*A005015(k-1) with A005015(-1)=1. - Philippe Deléham, Oct 17 2011

Edited by N. J. A. Sloane, Aug 19 2010

Definition clarified by Harvey P. Dale, Apr 19 2015

A070436 a(n) = n^2 mod 13.

Original entry on oeis.org

0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4, 1, 0, 1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4

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 (0,0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A070434, A070435.

Table[Mod[n^2,13],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)

a(n)=n^2%13 \\ Charles R Greathouse IV, Apr 06 2016

G.f.: (x^12 +4*x^11 +9*x^10 +3*x^9 +12*x^8 +10*x^7 +10*x^6 +12*x^5 +3*x^4 +9*x^3 +4*x^2 +x)/(-x^13 +1). - Colin Barker, Aug 14 2012

a(n) = a(n-13). - G. C. Greubel, Mar 24 2016

A070437 a(n) = n^2 mod 14.

Original entry on oeis.org

0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4, 1, 0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4, 1, 0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4, 1, 0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4, 1, 0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4, 1, 0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4, 1, 0, 1, 4, 9, 2, 11, 8, 7, 8, 11, 2, 9, 4

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, 0, 0, 0, 1).

Cf. A070434, A070435, A070436.

Table[Mod[n^2,14],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)
PowerMod[Range[0,100],2,14] (* Harvey P. Dale, Dec 02 2014 *)

a(n)=n^2%14 \\ Charles R Greathouse IV, Apr 06 2016

G.f.: -x*(1 +4*x +9*x^2 +2*x^3 +11*x^4 +8*x^5 +7*x^6 +8*x^7 +11*x^8 +2*x^9 +9*x^10 +4*x^11 +x^12) / ((x-1)*(1+x^6+x^5+x^4+x^3+x^2+x)*(1+x)*(1-x+x^2-x^3+x^4-x^5+x^6)). - R. J. Mathar, Jul 27 2015

a(n) = a(n-14). - G. C. Greubel, Mar 24 2016

A096459 Triangle read by rows: T(n,k) = n^2 mod prime(k), 1<=k<=n.

Original entry on oeis.org

1, 0, 1, 1, 0, 4, 0, 1, 1, 2, 1, 1, 0, 4, 3, 0, 0, 1, 1, 3, 10, 1, 1, 4, 0, 5, 10, 15, 0, 1, 4, 1, 9, 12, 13, 7, 1, 0, 1, 4, 4, 3, 13, 5, 12, 0, 1, 0, 2, 1, 9, 15, 5, 8, 13, 1, 1, 1, 2, 0, 4, 2, 7, 6, 5, 28, 0, 0, 4, 4, 1, 1, 8, 11, 6, 28, 20, 33, 1, 1, 4, 1, 4, 0, 16, 17, 8, 24, 14, 21, 5, 0, 1, 1, 0

Offset: 1

Reinhard Zumkeller, Aug 12 2004

T(n,k)=0 iff k is a prime factor of n:
A001221(n) = number of zeros in n-th row;
T(n,1)=A000035(n);
T(n,2)=A011655(n) for n>1; T(n,3)=A070430(n) for n>2;
T(n,4)=A053879(n) for n>3; T(n,5)=A070434(n) for n>4;
T(n,6)=A070436(n) for n>5; T(n,7)=A054580(n) for n>6;
T(n,8)=A070441(n) for n>7; T(n,9)=A070445(n) for n>8;
T(n,10)=A070451(n) for n>9;
T(n,n)=A069547(n).

			Triangle begins:
1;
0, 1;
1, 0, 4;
0, 1, 1, 2;
1, 1, 0, 4, 3;
0, 0, 1, 1, 3, 10;
1, 1, 4, 0, 5, 10, 15;
......

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

Cf. A000290, A000040, A049759.

Table[Mod[n^2, Prime[k]], {n, 1, 10}, {k, 1, n}] (* G. C. Greubel, May 20 2017 *)

cp's OEIS Frontend

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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A112651 Numbers k such that k^2 == k (mod 11).

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A070436 a(n) = n^2 mod 13.

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A070437 a(n) = n^2 mod 14.

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A096459 Triangle read by rows: T(n,k) = n^2 mod prime(k), 1<=k<=n.

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