A008959 -id:A008959 - 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

A070514 Final digit of n^4: a(n) = n^4 mod 10.

Original entry on oeis.org

0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0, 1, 6, 1, 6, 5, 6, 1, 6, 1, 0

Offset: 0

N. J. A. Sloane, May 13 2002

Decimal expansion of 538853870/3333333333. - Alexander R. Povolotsky, Mar 09 2013

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

Cf. A010879, A008959, A008960. - Doug Bell, Jun 15 2015

[n^4 mod (10): n in [0..80]]; // Vincenzo Librandi, Jun 16 2015

A070514:=n->n^4 mod 10: seq(A070514(n), n=0..100); # Wesley Ivan Hurt, Apr 01 2016

LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 1, 6, 1, 6, 5, 6, 1, 6, 1}, 100] (* Vincenzo Librandi, Jun 16 2015 *)
PowerMod[Range[0, 100], 4, 10] (* G. C. Greubel, Apr 01 2016 *)

vector(100,n,n--;n^4%10) \\ Derek Orr, Jun 16 2015

[power_mod(n,4,10)for n in range(0, 101)] # Zerinvary Lajos, Oct 30 2009

a(n) = n^k mod 10; for k > 0 where k mod 4 = 0. - Doug Bell, Jun 15 2015
From G. C. Greubel, Apr 01 2016: (Start)
a(n) = a(n-10).
a(2*n) = 6*A011558(n).
G.f.: (x +6*x^2 +x^3 +6*x^4 +5*x^5 +6*x^6 +x^7 +6*x^8 +x^9)/(1 - x^10). (End)

A033569 a(n) = (2n - 1)(3*n + 1).

Original entry on oeis.org

-1, 4, 21, 50, 91, 144, 209, 286, 375, 476, 589, 714, 851, 1000, 1161, 1334, 1519, 1716, 1925, 2146, 2379, 2624, 2881, 3150, 3431, 3724, 4029, 4346, 4675, 5016, 5369, 5734, 6111, 6500, 6901, 7314, 7739, 8176, 8625, 9086, 9559, 10044, 10541, 11050, 11571

Offset: 0

N. J. A. Sloane

For n>0, a(n) is the sum of the numbers from 2n+2 to 4n. The last digit of a(n) corresponds to the last digit of the squares mod 10 (A008959). Binomial Transform of a(n) starts: -1, 3, 28, 124, 432, 1328, 3776, 10176, 26368, ... . - Wesley Ivan Hurt, Dec 06 2014

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

Cf. A008959, A060747, A016777, A259758 (subsequence).

List([0..50], n-> (2*n-1)*(3*n+1)) # G. C. Greubel, Apr 02 2019

a033569 n = (2 * n - 1) * (3 * n + 1)
a033569_list = map a033569 [0..]
-- Reinhard Zumkeller, Jul 05 2015

[(2*n-1)*(3*n+1): n in [0..50]]; // Vincenzo Librandi, Jul 07 2012

A033569:=n->(2*n-1)*(3*n+1): seq(A033569(n), n=0..50); # Wesley Ivan Hurt, Dec 06 2014

CoefficientList[Series[(-1+7*x+6*x^2)/(1-x)^3,{x,0,50}],x] (* Vincenzo Librandi, Jul 07 2012 *)

a(n)=(2*n-1)*(3*n+1) \\ Charles R Greathouse IV, Jun 17 2017

[(2*n-1)*(3*n+1) for n in (0..50)] # G. C. Greubel, Apr 02 2019

G.f.: (-1+7*x+6*x^2)/(1-x)^3. - Vincenzo Librandi, Jul 07 2012
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Jul 07 2012
E.g.f.: (-1+5*x+6*x^2)*e^x. - Robert Israel, Dec 07 2014
a(n) = A060747(n) * A016777(n). - Reinhard Zumkeller, Jul 05 2015
Sum_{n>=0} 1/a(n) = 2/5*(log(2)-1) -sqrt(3)*Pi/30 -3*log(3)/10 = -0.6337047... - R. J. Mathar, Apr 22 2024

A325437 Final digit of primes of the form k^2 + 1.

Original entry on oeis.org

2, 5, 7, 7, 1, 7, 7, 1, 7, 7, 7, 1, 7, 7, 7, 7, 7, 1, 7, 1, 7, 1, 7, 7, 1, 7, 7, 1, 7, 1, 1, 7, 1, 7, 7, 7, 1, 7, 1, 7, 1, 1, 7, 1, 7, 1, 1, 7, 1, 7, 7, 7, 1, 1, 7, 7, 7, 1, 7, 1, 1, 7, 1, 7, 7, 7, 1, 7, 1, 7, 7, 7, 7, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 1, 7, 1, 7

Offset: 1

Martin Renner, Apr 27 2019

This sequence is presumably infinite. See 1st comment of A002496.
For k > 2, i.e., primes > 5 the final digit is always 1 or 7. Proof: Let k = 2*m - 1 odd. Then k^2 + 1 is divisible by 2, hence prime only for m = 1. Let k = 2*m even. Then k^2 + 1 = 4*m^2 + 1. The final digit of multiples of four is 4, 8, 2, 6, 0, 4, 8, 2, 6, 0, ... and of squares 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, ... (cf. A008959), hence the last digit of the product 4*m^2 is 4, 6, 6, 4, 0, ... or of the sum 4*m^2 + 1 is 5, 7, 7, 5, 1, ... (cf. A053755) and therefore for primes > 5 the final digit is 1 or 7.
Accordingly, for large k approximately one-third of the primes of the form k^2 + 1 end in 1, two-thirds end in 7.

Robert Israel, Table of n, a(n) for n = 1..10000
Edmund Landau, Gelöste und ungelöste Probleme aus der Theorie der Primzahlverteilung und der Riemannschen Zetafunktion, Jahresbericht der Deutschen Mathematiker-Vereinigung (1912), Vol. 21, page 208-228, here p. 224.
Eric Weisstein's World of Mathematics, Landau's Problems., Nr. 4.
Eric Weisstein's World of Mathematics, Near-Square Prime.

Cf. A001912, A002496, A005574.

seq(k mod 10,k=select(isprime,[2,seq(4*i^2+1,i=1..10000)]));

Mod[#,10]&/@Select[Range[1000]^2+1,PrimeQ] (* Harvey P. Dale, Jul 05 2023 *)

lista(nn) = {forprime(p=2, nn, if (issquare(p-1), print1(p % 10, ", ")););} \\ Michel Marcus, May 07 2019

a(n) = A002496(n) mod 10.

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

A210251 Residues modulo 100 of odd squares.

Original entry on oeis.org

1, 9, 21, 25, 29, 41, 49, 61, 69, 81, 89

Offset: 1

M. F. Hasler, Mar 19 2012

Range of A016754. Odd terms from A010461. See also A002015, A008959, A174452.

Also the range of A030156 and A192775 without initial term.

Mod[#,100]&/@(Range[1,55,2]^2)//Union (* Harvey P. Dale, Jul 27 2017 *)

PARI
```
vecsort(vector(12,n,(2*n-1)^2)%100,,8)
```

{1,9} + {0,1,2,3,4}*20 union {25}.

A191759 Least significant decimal digit of (2n-1)^2.

Original entry on oeis.org

1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9, 5, 9, 1, 1, 9

Offset: 1

Ant King, Jun 15 2011

Bisection of A008959.

This sequence is periodic with repeating part <1,9,5,9,1> of length five. Hence, as the members of each cycle sum to 25, the terms satisfy the fifth-order homogeneous recurrence a(n) = a(n-5) and the fourth-order inhomogeneous recurrence a(n) = 25 - a(n-1) - a(n-2) - a(n-3) - a(n-4).

			The fifth odd square number is 81 which has least significant digit 1. Hence a(5)=1.

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

Cf. A016754 (odd squares).

Mod[(2#-1)^2,10]&/@Range[50]
LinearRecurrence[{0, 0, 0, 0, 1},{1, 9, 5, 9, 1},87] (* Ray Chandler, Aug 25 2015 *)
PadRight[{},120,{1,9,5,9,1}] (* Harvey P. Dale, Aug 04 2019 *)

a(n)=[1,9,5,9,1][n%5+1] \\ Charles R Greathouse IV, Jun 15 2011

a(n) = (2n-1)^2 mod 10.
G.f.: x*(1+9*x+5*x^2+9*x^3+x^4)/(1-x^5) (note that the coefficients of x in the numerator are precisely the terms that constitute the periodic cycle of the sequence).
a(n) = 5 + 4*A080891(n+2). - R. J. Mathar, Jun 16 2011
Continued fraction of (97+17*sqrt(3077))/938. - R. J. Mathar, Jun 25 2011
a(n) = (-n^2 + n + 1) mod 10. - Arkadiusz Wesolowski, Jul 03 2012

More terms from Arkadiusz Wesolowski, Jul 03 2012

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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A070514 Final digit of n^4: a(n) = n^4 mod 10.

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A033569 a(n) = (2*n - 1)*(3*n + 1).

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A325437 Final digit of primes of the form k^2 + 1.

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A033569 a(n) = (2n - 1)(3*n + 1).