A145768 -id:A145768 - cp's OEIS frontend

A145829 Square root of squares in A145768 (XOR of squares of the numbers 1..n).

1, 4, 1, 15, 0, 16, 20, 31, 16, 49, 65, 224, 96, 96, 337, 144, 720, 400, 945, 625, 928, 828, 367, 928, 1889, 624, 2609, 3568, 3568, 2064, 10273, 1040, 545, 12384, 12639, 56800, 25812, 15119, 36, 864, 144383, 146463, 195440, 61391, 61072, 61072, 58128, 25872

Offset: 1

M. F. Hasler, Oct 20 2008

Reinhard Zumkeller, Table of n, a(n) for n = 1..63

a(n) = A000196( A145828(n)) = A000196( A145768( A145827(n))); A145828 = { a(n)^2 } = A145768 intersect A000290.

a145829 n = a145829_list !! (n-1)
a145829_list = map a000196 $ filter ((== 1) . a010052) $ tail a145768_list
-- Reinhard Zumkeller, Nov 09 2012

an = 0; Reap[ For[i = 1, i <= 10^6, i++, an = BitXor[an, i^2]; If[IntegerQ[r = Sqrt[an]], Print[r]; Sow[r]]]][[2, 1]] (* Jean-François Alcover, Oct 11 2013, translated from Pari *)

an=0; for( i=1,10^4, an=bitxor(an,i^2); issquare(an,&an) && print1(an","))

from itertools import count, islice
from sympy import integer_nthroot
def A145829gen(): # generator of terms
    m = 0
    for n in count(1):
        m ^= n**2
        a, b = integer_nthroot(m,2)
        if b: yield a
A145829_list = list(islice(A145829gen(),20)) # Chai Wah Wu, Dec 16 2021

A145828 Squares in A145768 (XOR of squares of the numbers 1...n).

Original entry on oeis.org

0, 1, 16, 1, 225, 0, 256, 400, 961, 256, 2401, 4225, 50176, 9216, 9216, 113569, 20736, 518400, 160000, 893025, 390625, 861184, 685584, 134689, 861184, 3568321, 389376, 6806881, 12730624, 12730624, 4260096, 105534529

Offset: 1

M. F. Hasler, Oct 20 2008

Chai Wah Wu, Table of n, a(n) for n = 1..97

Equals A145768 intersect A000290; a(n) = A145768(A145827(n)) = A145829(n)^2.

Reap[For[Sow[x=0]; k=1, k <= 10^4, k++, x = BitXor[x, k^2]; If[IntegerQ[ Sqrt[x]], Sow[x]]]][[2, 1]] (* Jean-François Alcover, Nov 25 2015 *)

an=0; for( i=1,10^4, an=bitxor(an,i^2); issquare(an) && print1(an","))

{a(n) = my(x, m, k); while( mMichael Somos, Aug 05 2014 */

from gmpy2 import is_square
filter(is_square, [reduce(lambda x,y:x^y, [x**2 for x in range(n)]) for n in range(1,10**4)]) # Chai Wah Wu, Aug 05 2014

Edited to start at 0 to match A145768 by Chai Wah Wu, Aug 05 2014

A145830 Indices for which A145768 (XOR of squares of the numbers 1...n) is a power of 2.

Original entry on oeis.org

1, 7, 9, 16, 47, 63

Offset: 1

M. F. Hasler, Oct 20 2008

Next term is > 10^7.
a(7) > 1.5*10^9. [Jon E. Schoenfield, Jan 14 2009]
a(7) > 10^11. [Sean A. Irvine, Aug 12 2010]

A145830 = A145768 intersect A000079. A145768(a(n)) = 2^A145831(n); See also A145827-A145829.

Position[ FoldList[ BitXor, 0, Range[10^6]^2], n_ /; IntegerQ[Log[2, n]]] - 1 // Flatten (* Jean-François Alcover, Sep 30 2013 *)

an=0; for( i=1,10^6, an=bitxor(an,i^2); an & an==1<

A145831 Log_2 ( A145768( A145830(n))).

Original entry on oeis.org

0, 4, 0, 8, 9, 8

Offset: 1

M. F. Hasler, Oct 20 2008

A145830 = A145768 intersect A000079. A145768(A145830(n)) = 2^a(n). See also A145827-A145829.

an=0; for( i=1,10^7, an=bitxor(an,i^2); an & an==1<

A014601 Numbers congruent to 0 or 3 mod 4.

Original entry on oeis.org

0, 3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 44, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 67, 68, 71, 72, 75, 76, 79, 80, 83, 84, 87, 88, 91, 92, 95, 96, 99, 100, 103, 104, 107, 108, 111, 112, 115, 116, 119, 120, 123, 124

Offset: 0

Eric Rains (rains(AT)caltech.edu)

Discriminants of orders in imaginary quadratic fields (negated). [Comment corrected by Christopher E. Thompson, Dec 11 2016]
Numbers such that Langford-Skolem problem has a solution - see A014552.
Complement of A042963. - Reinhard Zumkeller, Oct 04 2004
Also called skew amenable numbers; a number k is skew amenable if there exist a set {a(i)} of integers satisfying the relations k = Sum_{i=1..k} a(i) = -Product_{i=1..k} a(i). Thus we have 8 = 1 + 1 + 1 + 1 + 1 + 1 - 2 + 4 = -(1*1*1*1*1*1*(-2)*4). - Lekraj Beedassy, Jan 07 2005
Possible nonpositive discriminants of quadratic equation a*x^2 + b*x + c or discriminants of binary quadratic forms a*x^2 + b*x*y + c*y^2. - Artur Jasinski, Apr 28 2008
Also, disregarding the 0 term, positive integers m such that, equivalently,
  (i) +-1 +-2 +-... +-m is even for all choices of signs,
  (ii) +-1 +-2 +-... +-m = 0 for some choices of signs,
  (iii) for all -m <= k <= m, k = +-1 +-2 +-... +-(k-1) +-(k+1) +-(k+2) +-... +-m for at least one choice of signs. - Rick L. Shepherd, Oct 29 2008
A145768(a(n)) is even. - Reinhard Zumkeller, Jun 05 2012
Multiples of 4 interleaved with 1 less than multiples of 4. - Wesley Ivan Hurt, Nov 08 2013
((2*k+0) + (2*k+1) + ... + (2*k+m-1) + (2*k+m)) is even if and only if m = a(n) for some n where k is any nonnegative integer. - Gionata Neri, Jul 24 2015
Numbers whose binary reflected Gray code (A014550) ends with 0. - Amiram Eldar, May 17 2021

			G.f. = 3*x + 4*x^2 + 7*x^3 + 8*x^4 + 11*x^5 + 12*x^6 + 15*x^7 + 16*x^8 + ...

H. Cohen, Course in Computational Alg. No. Theory, Springer, 1993, pp. 514-5.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973, p. 108.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. F. Barger, Solution to problem 10454: Amenable Numbers, Amer. Math. Monthly, Vol. 105, No. 4 (April 1998), p. 368.
Steven R. Finch, Class number theory [Cached copy, with permission of the author]
Heiko Harborth, Solution of Steinhaus's problem with plus and minus signs, Journal of Combinatorial Theory, Series A, Volume 12, Issue 2 (March 1972), Pages 253-259.
Mickaël Launay, Les routes de Numland, L’énigme maths du "Monde" n°2. In French.
Michael Penn, The most beautiful arrangement of numbers, YouTube video, 2024.
Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004676, A014550, A042948, A079896.

Cf. A274406. - Bruno Berselli, Jun 26 2016

a014601 n = a014601_list !! n
a014601_list = [x | x <- [0..], mod x 4 `elem` [0, 3]]
-- Reinhard Zumkeller, Jun 05 2012

[n: n in [0..200]|n mod 4 in {0,3}]; // Vincenzo Librandi, Dec 24 2010

A014601:=n->3*n-2*floor(n/2); seq(A014601(k), k=0..100); # Wesley Ivan Hurt, Nov 08 2013

aa = {}; Do[Do[Do[d = b^2 - 4 a c; If[d <= 0, AppendTo[aa, -d]], {a, 0, 50}], {b, 0, 50}], {c, 0, 50}]; Union[aa] (* Artur Jasinski, Apr 28 2008 *)
Select[Range[0, 124], Or[Mod[#, 4] == 0, Mod[#, 4] == 3] &] (* Ant King, Nov 18 2010 *)
CoefficientList[Series[2 x/(1 - x)^2 + (1/(1 - x) + 1/(1 + x)) x/2, {x, 0, 100}], x] (* Vincenzo Librandi, May 18 2014 *)
a[ n_] := 2 n + Mod[n, 2]; (* Michael Somos, Jul 24 2015 *)

{a(n) = 2*n + n%2}; /* Michael Somos, Dec 27 2010 */

a(n) = (n + 1)*2 + 1 - n mod 2. - Reinhard Zumkeller, Apr 21 2003
A014494(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004
a(n) = Sum_{k=1..n} (2 - (-1)^k). - William A. Tedeschi, Mar 20 2008
A139131(a(n)) = A078636(a(n)). - Reinhard Zumkeller, Apr 10 2008
From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.
G.f.: x*(3+x)/((1+x)*(x-1)^2). (End)
a(n) = 2*n + (n mod 2). - Paolo Valzasina (p.valzasina(AT)gmail.com), Nov 24 2009
a(n) = (4*n - (-1)^n + 1)/2. - Bruno Berselli, Oct 06 2010
a(n) = 4*n - a(n-1) - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = -A042948(-n) for all n in Z. - Michael Somos, Dec 27 2010
G.f.: 2*x / (1 - x)^2 + (1 / (1 - x) + 1 / (1 + x)) * x/2. - Michael Somos, Dec 27 2010
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0) = 3 and b(k) = 2^(k+1) for k > 0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((4/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = 3n - 2*floor(n/2). - Wesley Ivan Hurt, Nov 08 2013
a(n) = A042948(n+1) - 1 for all n in Z. - Michael Somos, Jul 24 2015
a(n) + a(n+1) = A004767(n) for all n in Z. - Michael Somos, Jul 24 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2)/4 - Pi/8. - Amiram Eldar, Dec 05 2021
E.g.f.: ((4*x + 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 04 2022

A042963 Numbers congruent to 1 or 2 mod 4.

Original entry on oeis.org

1, 2, 5, 6, 9, 10, 13, 14, 17, 18, 21, 22, 25, 26, 29, 30, 33, 34, 37, 38, 41, 42, 45, 46, 49, 50, 53, 54, 57, 58, 61, 62, 65, 66, 69, 70, 73, 74, 77, 78, 81, 82, 85, 86, 89, 90, 93, 94, 97, 98, 101, 102, 105, 106, 109, 110, 113, 114, 117, 118, 121, 122, 125, 126, 129, 130, 133, 134, 137, 138

Offset: 1

N. J. A. Sloane

Complement of A014601. - Reinhard Zumkeller, Oct 04 2004
Let S(x) = (1, 2, 2, 2, ...). Then A042963 = ((S(x))^2 + S(x^2))/2 = ((1, 4, 8, 12, 16, 20, ...) + (1, 0, 2, 0, 2, 0, 2, ...))/2 = (1, 2, 5, 6, 9, 10, ...). - Gary W. Adamson, Jan 03 2011
(a(n)*(a(n) + 1 + 4*k))/2 is odd, for k >= 0. - Gionata Neri, Jul 19 2015
Equivalent to the following variation on Fermat's Diophantine m-tuple: 1 + the product of any two distinct terms is not a square; this sequence, which we'll call sequence S, is produced by the following algorithm. At the start, S is initially empty. At stage n, starting at n = 1, the algorithm checks whether there exists a number m already in the sequence, such that mn+1 is a perfect square. If such a number m is found, then n is not added to the sequence; otherwise, n is added. Then n is incremented to n + 1, and we repeat the procedure. Proof by Clark R. Lyons: We prove by strong induction that n is in the sequence S if and only if n == 1 (mod 4) or n == 2 (mod 4). Suppose now that this holds for all k < n. In case 1, either n == 1 (mod 4) or n == 2 (mod 4), and we wish to show that n does indeed enter the sequence S. That is, we wish to show that there does not exist m < n, already in the sequence at this point such that mn+1 is a square. By the inductive hypothesis m == 1 (mod 4) or m == 2 (mod 4). This means that both m and n are one of 1, 2, 5, or 6 mod 8. Using a multiplication table mod 8, we see that this implies mn+1 is congruent to one of 2, 3, 5, 6, or 7 mod 8. But we also see that mod 8, a perfect square is congruent to 0, 1, or 4. Thus mn+1 is not a perfect square, so n is added to the sequence. In case 2, n == 0 (mod 4) or n == 3 (mod 4), and we wish to show that n is not added to the sequence. That is, we wish to show that there exists m < n already in the sequence such that mn+1 is a perfect square. For this we let m = n - 2, which is positive since n >= 3. By the inductive hypothesis, since m == 1 (mod 4) or m == 2 (mod 4) and m < n, m is already in the sequence. And we have m*n + 1 = (n - 2)*n + 1 = n^2 - 2*n + 1 = (n - 1)^2, so mn+1 is indeed a perfect square, and so n is not added to the sequence. Thus n is added to the sequence if and only if n == 1 (mod 4) or n == 2 (mod 4). This completes the proof. - Robert C. Lyons, Jun 30 2016
Also the number of maximal cliques in the (n + 1) X (n + 1) black bishop graph. - Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any two or more consecutive terms is never an integer. (For opposite property see A005408.) - Ivan Neretin, Dec 21 2017
Numbers whose binary reflected Gray code (A014550) ends with 1. - Amiram Eldar, May 17 2021
Also: append its negated last bit to n-1. - M. F. Hasler, Oct 17 2022

David Lovler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Reinhard Zumkeller)
Eric Weisstein's World of Mathematics, Black Bishop Graph.
Eric Weisstein's World of Mathematics, Maximal Clique.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A153284 (first differences), A014848 (partial sums).
Cf. A014550, A046712 (subsequence).
Union of A016813 and A016825.

a042963 n = a042963_list !! (n-1)
a042963_list = [x | x <- [0..], mod x 4 `elem` [1,2]]
-- Reinhard Zumkeller, Feb 14 2012

[ n : n in [1..165] | n mod 4 eq 1 or n mod 4 eq 2 ]; // Vincenzo Librandi, Jan 25 2011

A046923:=n->(n mod 2) + 2n - 2; seq(A046923(n), n=1..100); # Wesley Ivan Hurt, Oct 10 2013

Select[Range[109], Or[Mod[#, 4] == 1, Mod[#, 4] == 2] &] (* Ant King, Nov 17 2010 *)
Table[(4 n - 3 - (-1)^n)/2, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{1, 1, -1}, {1, 2, 5}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + x + 2 x^2)/((-1 + x)^2 (1 + x)), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n)=2*n-1-(n-1)%2 \\ Jianing Song, Oct 06 2018; adapted to offset by Michel Marcus, Sep 09 2022

apply( A042963(n)=n*2-2+n%2, [1..99]) \\ M. F. Hasler, Oct 17 2022

a(n) = 1 + A042948(n-1). [Corrected by Jianing Song, Oct 06 2018]
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(1 + x + 2*x^2)/((1 - x)^2*(1 + x)).
a(n) = a(n-1) + 2 + (-1)^n, a(0) = 1. (End) [This uses offset 0. - Jianing Song, Oct 06 2018]
A014493(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004, Feb 14 2012
a(n) = Sum_{k=0..n} (A001045(k) mod 4). - Paul Barry, Mar 12 2004
A145768(a(n)) is odd. - Reinhard Zumkeller, Jun 05 2012
a(n) = A005843(n-1) + A059841(n-1). - Philippe Deléham, Mar 31 2009 [Corrected by Jianing Song, Oct 06 2018]
a(n) = 4*n - a(n-1) - 5 for n > 1. [Corrected by Jerzy R Borysowicz, Jun 09 2023]
From Ant King, Nov 17 2010: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) = (4*n - 3 - (-1)^n)/2. (End)
a(n) = (n mod 2) + 2*n - 2. - Wesley Ivan Hurt, Oct 10 2013
A163575(a(n)) = n - 1. - Reinhard Zumkeller, Jul 22 2014
E.g.f.: 2 + (2*x - 1)*sinh(x) + 2*(x - 1)*cosh(x). - Ilya Gutkovskiy, Jun 30 2016
E.g.f.: 2 + (2*x - 1)*exp(x) - cosh(x). - David Lovler, Jul 19 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 + log(2)/4. - Amiram Eldar, Dec 05 2021

Offset corrected by Reinhard Zumkeller, Feb 14 2012

More terms by David Lovler, Jul 19 2022

A003815 a(0) = 0, a(n) = a(n-1) XOR n.

Original entry on oeis.org

0, 1, 3, 0, 4, 1, 7, 0, 8, 1, 11, 0, 12, 1, 15, 0, 16, 1, 19, 0, 20, 1, 23, 0, 24, 1, 27, 0, 28, 1, 31, 0, 32, 1, 35, 0, 36, 1, 39, 0, 40, 1, 43, 0, 44, 1, 47, 0, 48, 1, 51, 0, 52, 1, 55, 0, 56, 1, 59, 0, 60, 1, 63, 0, 64, 1, 67, 0

Offset: 0

Marc LeBrun

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

Cf. A003816.

Cf. A077140, A145768. - M. F. Hasler, Oct 20 2008

an = 0; Reap[ For[i = 0, i <= 100, i++, an = BitXor[an, i]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Oct 11 2013, translated from PARI *)
CoefficientList[Series[x (1 + 3 x - x^2 + x^3)/((1 - x^4) (1 - x^2)), {x, 0, 100}], x] (* Vincenzo Librandi, Oct 12 2013 *)
nxt[{n_,a_}]:={n+1,BitXor[n+1,a]}; NestList[nxt,{0,0},70][[All,2]] (* Harvey P. Dale, Mar 10 2019 *)
{#,1,#+1,0}[[1+Mod[#,4]]]&/@Range[0,100] (* Federico Provvedi, May 10 2021 *)

print1(an=0); for( i=1,100, print1(",",an=bitxor(an,i))) \\ M. F. Hasler, Oct 20 2008

a(n) = n + (-1)^n*a(n-1). - Vladeta Jovovic, Mar 13 2003
a(0)=0, a(4n+1)=1, a(4n+2)=4n+3, a(4n+3)=0, a(4n+4)=4n+4, n >= 0.
a(n) = f(n,0) with f(n,x) = x if n=0, otherwise f(n-1,x+n) if x is even, otherwise f(n-1,x-n). - Reinhard Zumkeller, Oct 09 2007
a(n) = abs(A077140(n)) for n > 0. - Reinhard Zumkeller, Oct 09 2007
G.f.: x*(1+3*x-x^2+x^3)/((1-x^4)*(1-x^2)). - Vincenzo Librandi, Oct 12 2013
a(n) = (1 + n + n*(-1)^n + (-1)^floor((n-1)/2))/2. - Wesley Ivan Hurt, May 08 2021

A193232 Bitwise XOR of first n triangular numbers.

Original entry on oeis.org

0, 1, 2, 4, 14, 1, 20, 8, 44, 1, 54, 116, 58, 97, 8, 112, 248, 97, 202, 116, 166, 65, 188, 424, 132, 449, 158, 484, 114, 449, 16, 480, 1008, 449, 914, 484, 894, 449, 804, 40, 796, 65, 966, 116, 938, 1953, 920, 2032, 872, 1953, 858, 1652, 790, 1665, 844, 1352

Offset: 0

Franklin T. Adams-Watters, Jul 18 2011

nonn

John Tyler Rascoe, Table of n, a(n) for n = 0..10000

Cf. A000217, A145768, A003815.

a:= proc(n) option remember; `if`(n=0, 0,
      Bits[Xor](a(n-1), n*(n+1)/2))
    end:
seq(a(n), n=0..55);  # Alois P. Heinz, Feb 19 2023

Module[{nn=60,trs},trs=Accumulate[Range[nn]];Table[BitXor@@Take[trs,n],{n,0,nn}]] (* Harvey P. Dale, Dec 15 2017 *)

al(n) = local(m); vector(n,k,m=bitxor(m,k*(k+1)\2))

from operator import xor
from functools import reduce
def A193232(n): return reduce(xor, (x*(x+1) for x in range(n+1)))//2 # Chai Wah Wu, Dec 16 2021

cp's OEIS Frontend

A145827 Numbers n such that A145768(n) is a square.

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A145829 Square root of squares in A145768 (XOR of squares of the numbers 1..n).

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A145828 Squares in A145768 (XOR of squares of the numbers 1...n).

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A145830 Indices for which A145768 (XOR of squares of the numbers 1...n) is a power of 2.

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A145831 Log_2 ( A145768( A145830(n))).

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A014601 Numbers congruent to 0 or 3 mod 4.

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A042963 Numbers congruent to 1 or 2 mod 4.

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