A070433 -id:A070433 - cp's OEIS frontend

A056991 Numbers with digital root 1, 4, 7 or 9.

1, 4, 7, 9, 10, 13, 16, 18, 19, 22, 25, 27, 28, 31, 34, 36, 37, 40, 43, 45, 46, 49, 52, 54, 55, 58, 61, 63, 64, 67, 70, 72, 73, 76, 79, 81, 82, 85, 88, 90, 91, 94, 97, 99, 100, 103, 106, 108, 109, 112, 115, 117, 118, 121, 124, 126, 127, 130, 133, 135, 136, 139, 142

Offset: 1

Eric W. Weisstein

All squares are members (see A070433).
May also be defined as: possible sums of digits of squares. - Zak Seidov, Feb 11 2008
First differences are periodic: 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, 3, 3, 2, 1, ... - Zak Seidov, Feb 11 2008
Minimal n with corresponding sum-of-digits(n^2) are: 1, 2, 4, 3, 8, 7, 13, 24, 17, 43, 67, 63, 134, 83, 167, 264, 314, 313, 707, 1374, 836, 1667, 2236, 3114, 4472, 6833, 8167, 8937, 16667, 21886, 29614, 60663, 41833, 74833, 89437, 94863, 134164, 191833.
a(n) is the set of all m such that 9k+m can be a perfect square (quadratic residues of 9 including the trivial case of 0). - Gary Detlefs, Mar 19 2010
From Klaus Purath, Feb 20 2023: (Start)
The sum of digits of any term belongs to the sequence. Also the products of any terms belong to the sequence.
This is the union of A017173, A017209, A017245 and A008591.
Positive integers of the forms x^2 + (2*m+1)*x*y + (m^2+m-2)*y^2, for integers m.
This sequence is closed under multiplication. (End)

R. J. Mathar, Table of n, a(n) for n = 1..22222
H. I. Okagbue, M. O. Adamu, S. A. Iyase, and A. A. Opanuga, Sequence of Integers Generated by Summing the Digits of their Squares, Indian Journal of Science and Technology, Vol 8(15), DOI: 10.17485/ijst/2015/v8i15/69912, July 2015.
Eric Weisstein's World of Mathematics, Square Number
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A000290, A008591, A017173, A017209, A017245, A056992, A070433.

For complement see A268226.

seq( 3*(n-floor(n/4)) - (3-I^n-(-I)^n-(-1)^n)/2, n=1..63); # Gary Detlefs, Mar 19 2010

LinearRecurrence[{1,0,0,1,-1},{1,4,7,9,10},70] (* Harvey P. Dale, Aug 29 2015 *)

forstep(n=1,1e3,[3,3,2,1],print1(n", ")) \\ Charles R Greathouse IV, Sep 21 2012

From R. J. Mathar, Feb 14 2008: (Start)
O.g.f.: x*(2x+1)*(x^2+x+1)/((-1+x)^2*(x+1)*(x^2+1)).
a(n) = a(n-4) + 9. (End)
a(n) = 3*(n - floor(n/4)) - (3 - i^n - (-i)^n - (-1)^n)/2, where i = sqrt(-1). - Gary Detlefs, Mar 19 2010
a(n) = a(n-1)+a(n-4)-a(n-5). - Wesley Ivan Hurt, May 27 2021
a(n) = 3*n - floor(n/4) - 2*floor((n+3)/4). - Ridouane Oudra, Jan 21 2024
E.g.f.: (cos(x) + (9*x - 1)*cosh(x) - 3*sin(x) + (9*x - 2)*sinh(x))/4. - Stefano Spezia, Feb 21 2024

Edited by N. J. A. Sloane, May 16 2008 at the suggestion of R. J. Mathar

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

A375166 Nonsquares congruent to {0, 1, 4, 7} modulo 9.

Original entry on oeis.org

7, 10, 13, 18, 19, 22, 27, 28, 31, 34, 37, 40, 43, 45, 46, 52, 54, 55, 58, 61, 63, 67, 70, 72, 73, 76, 79, 82, 85, 88, 90, 91, 94, 97, 99, 103, 106, 108, 109, 112, 115, 117, 118, 124, 126, 127, 130, 133, 135, 136, 139, 142, 145, 148, 151, 153, 154, 157, 160, 162

Offset: 1

Stefano Spezia, Aug 05 2024

Squares are congruent to {0, 1, 4, 7} modulo 9, but the reverse is not always true since there are nonsquares that have the same congruence property. See Beiler.

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 140.

Intersection of A000037 and A056991.

Cf. A000290, A056992, A070433.

Select[Range[0,162], !IntegerQ[Sqrt[#]] && MemberQ[{0,1,4,7}, Mod[#,9]] &]

from itertools import count, islice
from sympy.ntheory.primetest import is_square
def A375166_gen(): # generator of terms
    for i in count(0,9):
        for j in (0,1,4,7):
            if not is_square(i+j): yield i+j
A375166_list = list(islice(A375166_gen(),40)) # Chai Wah Wu, Jun 05 2025

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A056991 Numbers with digital root 1, 4, 7 or 9.

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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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A375166 Nonsquares congruent to {0, 1, 4, 7} modulo 9.

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