cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 24 results. Next

A001481 Numbers that are the sum of 2 squares.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49, 50, 52, 53, 58, 61, 64, 65, 68, 72, 73, 74, 80, 81, 82, 85, 89, 90, 97, 98, 100, 101, 104, 106, 109, 113, 116, 117, 121, 122, 125, 128, 130, 136, 137, 144, 145, 146, 148, 149, 153, 157, 160
Offset: 1

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Keywords

Comments

Numbers n such that n = x^2 + y^2 has a solution in nonnegative integers x, y.
Closed under multiplication. - David W. Wilson, Dec 20 2004
Also, numbers whose cubes are the sum of 2 squares. - Artur Jasinski, Nov 21 2006 (Cf. A125110.)
Terms are the squares of smallest radii of circles covering (on a square grid) a number of points equal to the terms of A057961. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Apr 16 2007. [Comment corrected by T. D. Noe, Mar 28 2008]
Numbers with more 4k+1 divisors than 4k+3 divisors. If a(n) is a member of this sequence, then so too is any power of a(n). - Ant King, Oct 05 2010
A000161(a(n)) > 0; A070176(a(n)) = 0. - Reinhard Zumkeller, Feb 04 2012, Aug 16 2011
Numbers that are the norms of Gaussian integers. This sequence has unique factorization; the primitive elements are A055025. - Franklin T. Adams-Watters, Nov 25 2011
These are numbers n such that all of n's odd prime factors congruent to 3 modulo 4 occur to an even exponent (Fermat's two-squares theorem). - Jean-Christophe Hervé, May 01 2013
Let's say that an integer n divides a lattice if there exists a sublattice of index n. Example: 2, 4, 5 divide the square lattice. The present sequence without 0 is the sequence of divisors of the square lattice. Say that n is a "prime divisor" if the index-n sublattice is not contained in any other sublattice except the original lattice itself. Then A055025 (norms of Gaussian primes) gives the "prime divisors" of the square lattice. - Jean-Christophe Hervé, May 01 2013
For any i,j > 0 a(i)*a(j) is a member of this sequence, since (a^2 + b^2)*(c^2 + d^2) = (a*c + b*d)^2 + (a*d - b*c)^2. - Boris Putievskiy, May 05 2013
The sequence is closed under multiplication. Primitive elements are in A055025. The sequence can be split into 3 multiplicatively closed subsequences: {0}, A004431 and A125853. - Jean-Christophe Hervé, Nov 17 2013
Generalizing Jasinski's comment, same as numbers whose odd powers are the sum of 2 squares, by Fermat's two-squares theorem. - Jonathan Sondow, Jan 24 2014
By the 4 squares theorem, every nonnegative integer can be expressed as the sum of two elements of this sequence. - Franklin T. Adams-Watters, Mar 28 2015
There are never more than 3 consecutive terms. Runs of 3 terms start at 0, 8, 16, 72, ... (A082982). - Ivan Neretin, Nov 09 2015
Conjecture: barring the 0+2, 0+4, 0+8, 0+16, ... sequence, the sum of 2 distinct terms in this sequence is never a power of 2. - J. Lowell, Jan 14 2022
All the areas of squares whose vertices have integer coordinates. - Neeme Vaino, Jun 14 2023
Numbers represented by the definite binary quadratic forms x^2 + 2nxy + (n^2+1)y^2 for any integer n. This sequence contains the even powers of any integer. An odd power of a number appears only if the number itself belongs to the sequence. The equation given in the comment by Boris Putievskiy 2013 is Brahmagupta's identity with n = 1. It proves that any set of numbers of the form a^2 + nb^2 is closed under multiplication. - Klaus Purath, Sep 06 2023

References

  • J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 106.
  • David A. Cox, "Primes of the Form x^2 + n y^2", Wiley, 1989.
  • L. Euler, (E388) Vollständige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 417.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.
  • G. H. Hardy, Ramanujan, pp. 60-63.
  • P. Moree and J. Cazaran, On a claim of Ramanujan in his first letter to Hardy, Expos. Math. 17 (1999), pp. 289-312.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Disjoint union of A000290 and A000415.
Complement of A022544.
A000404 gives another version. Subsequence of A091072, supersequence of A046711.
Column k=2 of A336820.

Programs

  • Haskell
    a001481 n = a001481_list !! (n-1)
    a001481_list = [x | x <- [0..], a000161 x > 0]
    -- Reinhard Zumkeller, Feb 14 2012, Aug 16 2011
    
  • Magma
    [n: n in [0..160] | NormEquation(1, n) eq true]; // Arkadiusz Wesolowski, May 11 2016
    
  • Maple
    readlib(issqr): for n from 0 to 160 do for k from 0 to floor(sqrt(n)) do if issqr(n-k^2) then printf(`%d,`,n); break fi: od: od:
  • Mathematica
    upTo = 160; With[{max = Ceiling[Sqrt[upTo]]}, Select[Union[Total /@ (Tuples[Range[0, max], {2}]^2)], # <= upTo &]]  (* Harvey P. Dale, Apr 22 2011 *)
    Select[Range[0, 160], SquaresR[2, #] != 0 &] (* Jean-François Alcover, Jan 04 2013 *)
  • PARI
    isA001481(n)=local(x,r);x=0;r=0;while(x<=sqrt(n) && r==0,if(issquare(n-x^2),r=1);x++);r \\ Michael B. Porter, Oct 31 2009
    
  • PARI
    is(n)=my(f=factor(n));for(i=1,#f[,1],if(f[i,2]%2 && f[i,1]%4==3, return(0))); 1 \\ Charles R Greathouse IV, Aug 24 2012
    
  • PARI
    B=bnfinit('z^2+1,1);
    is(n)=#bnfisintnorm(B,n) \\ Ralf Stephan, Oct 18 2013, edited by M. F. Hasler, Nov 21 2017
    
  • PARI
    list(lim)=my(v=List(),t); for(m=0,sqrtint(lim\=1), t=m^2; for(n=0, min(sqrtint(lim-t),m), listput(v,t+n^2))); Set(v) \\ Charles R Greathouse IV, Jan 05 2016
    
  • PARI
    is_A001481(n)=!for(i=2-bittest(n,0),#n=factor(n)~, bittest(n[1,i],1)&&bittest(n[2,i],0)&&return) \\ M. F. Hasler, Nov 20 2017
    
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A001481_gen(): # generator of terms
        return filter(lambda n:(lambda m:all(d & 3 != 3 or m[d] & 1 == 0 for d in m))(factorint(n)),count(0))
    A001481_list = list(islice(A001481_gen(),30)) # Chai Wah Wu, Jun 27 2022

Formula

n = square * 2^{0 or 1} * {product of distinct primes == 1 (mod 4)}.
The number of integers less than N that are sums of two squares is asymptotic to constant*N/sqrt(log(N)), hence lim_{n->infinity} a(n)/n = infinity.
Nonzero terms in expansion of Dirichlet series Product_p (1 - (Kronecker(m, p) + 1)*p^(-s) + Kronecker(m, p)*p^(-2s))^(-1) for m = -1.
a(n) ~ k*n*sqrt(log n), where k = 1.3085... = 1/A064533. - Charles R Greathouse IV, Apr 16 2012
There are B(x) = x/sqrt(log x) * (K + B2/log x + O(1/log^2 x)) terms of this sequence up to x, where K = A064533 and B2 = A227158. - Charles R Greathouse IV, Nov 18 2022

Extensions

Deleted an incorrect comment. - N. J. A. Sloane, Oct 03 2023

A014577 The regular paper-folding sequence (or dragon curve sequence). Alphabet {1,0}.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0
Offset: 0

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Author

Keywords

Comments

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n+1. E.g., n = 3, n+1 = 4 = 100_2, so a(3) = (complement of bit to left of 1) = 1. - Robert L. Brown, Nov 28 2001 [Adjusted to match offset by N. J. A. Sloane, Apr 15 2021]
To construct the sequence: start from 1,(..),0,(..),1,(..),0,(..),1,(..),0,(..),1,(..),0,... and fill undefined places with the sequence itself. - Benoit Cloitre, Jul 08 2007
This is the characteristic function of A091072 - 1. - Gary W. Adamson, Apr 11 2010
Turns (by 90 degrees) of the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See fxtbook link below. - Joerg Arndt, Apr 15 2010
Sequence can be obtained by the L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and then dropping all L's and R's (see example). - Joerg Arndt, Aug 28 2011
From Gary W. Adamson, Jun 20 2012: (Start)
One half of the infinite Farey tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are
1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...
1/2 2/3 1/3 3/4 3/5 2/5 1/4 4/5 5/7 5/8, ...
Infinite Farey tree fractions can be derived from the binary by appending a repeat of rightmost binary term to the right, then recording the number of runs to obtain the continued fraction representation. Example: 9 = 1001 which becomes 10011 which becomes [1,2,2] = 5/7. (End)
The sequence can be considered as a binomial transform operator for a target sequence S(n). Replace the first 1 in A014577 with the first term in S(n), then for successive "1" term in A014577, map the next higher term in S(n). If "0" in A014577, map the next lower term in S(n). Using the sequence S(n) = (1, 3, 5, 7, ...), we obtain (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), .... Then parse the terms into subsequences of 2^k terms, adding the terms in each string. We obtain (1, 4, 12, 32, 80, ...), the binomial transform of (1, 3, 5, 7, ...). The 8-bit string has one 1, three 5's, three 7's and one 1) as expected, or (1, 3, 3, 1) dot (1, 3, 5, 7). - Gary W. Adamson, Jun 24 2012
From Gary W. Adamson, May 29 2013: (Start)
The sequence can be generated directly from the lengths of continued fraction representations of fractions in one half of the Stern-Brocot tree (fractions between 0 and 1):
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
Record the lengths by rows then reverse rows, getting:
1,
2, 1,
2, 3, 2, 1,
2, 3, 4, 3, 2, 3, 2, 1,
...
start with "1" and if the next term is greater than the current term, record a 1, otherwise 0; getting the present sequence, the Harter-Heighway dragon curve. (End)
The paper-folding word "110110011100100111011000..." can be created by concatenating the terms of a fixed point of the morphism or string substitution rule: 00 -> 1000, 01 -> 1001, 10 -> 1100 & 11 -> 1101, beginning with "11". - Robert G. Wilson v, Jun 11 2015
From Gary W. Adamson, Jun 04 2021: (Start)
Since the Heighway dragon is composed of right angles, it can be mapped with unit trajectories (Right = 1, Left = (-1), Up = i and Down = -i) on the complex plane where i = sqrt(-1). The initial (0th) iterate is chosen in this case to be the unit line from (0,0) to (-1,0). Then follow the directions below as indicated, resulting in a reflected variant of the dragon curve shown at the Eric Weisstein link. The conjectured system of complex plane trajectories is:
0 -1
1 -1, i
2 -1, i, 1, i
3 -1, i, 1, i, 1, -i, 1, i
4 -1, i, 1, i, 1, -i, 1, i, 1, -i, -1, -i, 1, -i, 1, i
...
The conjecture succeeds through the 4th iterate. It appears that to generate the (n+1)-th row, bring down the n-th row as the left half of row (n+1). For the right half of row (n+1), bring down the n-th row but change the signs of the first half of row n. For example, to get the complex plane instructions for the third iterate of the dragon curve, bring down (-1, i, 1, i) as the left half, and the right half is (1, -i, 1, i). (End)
From Gary W. Adamson, Jun 09 2021: (Start)
Partial sums of the iterate trajectories produce a sequence of complex addresses for unit segments. Partial sums of row 4 are: -1, (-1+i), i, 2i, (1+2i), (1+i), (2+i), (2+2i), (3+2i), (3+i), (2+i), 2, 3, (3-i), (4-i), 4. (zeros are omitted with terms of the form a+0i). The reflected variant of the dragon curve has the 0th iterate from (0,0) to (1,0), and the respective addresses simply change the signs of the real terms. (End)

Examples

			1 + x + x^3 + x^4 + x^7 + x^8 + x^9 + x^12 + x^15 + x^16 + x^17 + x^19 + ...
From _Joerg Arndt_, Aug 28 2011: (Start)
Generation via string substitution:
Start: L
Rules:
  L --> L1R
  R --> L0R
  0 --> 0
  1 --> 1
-------------
0:   (#=1)
  L
1:   (#=3)
  L1R
2:   (#=7)
  L1R1L0R
3:   (#=15)
  L1R1L0R1L1R0L0R
4:   (#=31)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R
5:   (#=63)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R1L1R1L0R1L1R0L0R0L1R1L0R0L1R0L0R
Drop all L and R to obtain 1101100111001001110110001100100
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614.
  • Michel Dekking, Michel Mendes France, and Alf van der Poorten, "Folds", The Mathematical Intelligencer, 4.3 (1982): 130-138 & front cover, and 4:4 (1982): 173-181 (printed in two parts).
  • Martin Gardner, Mathematical Magic Show, New York: Vintage, pp. 207-209 and 215-220, 1978.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

Cf. A059125, A065339, A005811, A220466, A088748, A091072, A343173 (first differences), A343174 (partial sums).
The two bisections are A000035 and the sequence itself.
See A343181 for prefixes of length 2^k-1.

Programs

  • Magma
    [(1+KroneckerSymbol(-1,n))/2: n in [1..100]]; // Vincenzo Librandi, Aug 05 2015
    
  • Magma
    [Floor(1/2*(1+(-1)^(1/2*((n+1)/2^Valuation(n+1, 2)-1)))): n in [0..100]]; // Vincenzo Librandi, Aug 05 2015
    
  • Maple
    nmax:=98: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := (n+1) mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
    a014577 := proc(n) local p,s1,s2,i;
    if n=0 then return(1); fi;
    s1:=convert(n,base,2); s2:=nops(s1);
    for i from 1 to s2 do if s1[i]=1 then p:=i; break; fi; od:
    if p <= s2-1 then 1-s1[p+1]; else 1; fi; end;
    [seq(a014577(i),i=1..120)]; # N. J. A. Sloane, Apr 08 2021
    # third Maple program:
    a:= n-> 1-irem(iquo((n+1)/2^padic[ordp](n+1, 2), 2), 2):
    seq(a(n), n=0..120);  # Alois P. Heinz, Apr 08 2021
  • Mathematica
    a[n_] := Boole[EvenQ[((n+1)/2^IntegerExponent[n+1, 2]-1)/2]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Feb 16 2012, after Gary W. Adamson, updated Nov 21 2014 *)
    Table[1-(((Mod[#1,2^(#2+2)]/2^#2)&[n,IntegerExponent[n,2]])-1)/2,{n,1,100,1}] (* WolframAlpha compatible code; Robert L. Brown, Jan 06 2015 *)
    MapThread[(a[x_/;IntegerQ[(x-#1)/4]]:= #2)&,{{1,3},{1,0}}];a[x_/;IntegerQ[x/2]]:=a[x/2];a/@ Range[100] (* Bradley Klee, Aug 04 2015 *)
    (1 + JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 22 2024 *)
    Array[Boole[BitAnd[#, BitAnd[#, -#]*2] == 0] &, 100] (* Paolo Xausa, May 22 2024, after Joerg Arndt C++ code *)
  • PARI
    {a(n) = if( n%2, a(n\2), 1 - (n/2%2))} /* Michael Somos, Feb 05 2012 */
    
  • PARI
    a(n)=1/2*(1+(-1)^(1/2*((n+1)/2^valuation(n+1,2)-1))) \\ Ralf Stephan, Sep 02 2013
    
  • PARI
    a(n)=!bittest(n+1,valuation(n+1,2)+1); \\ Robert L. Brown, Jul 07 2025
    
  • Python
    def A014577(n):
        s = bin(n+1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

a(n) = (1+Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=1, b=0, S(0)=a, S(n+1) = S(n), a, F(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
From Ralf Stephan, Jul 03 2003: (Start)
a(4*n) = 1, a(4*n+2) = 0, a(2*n+1) = a(n).
a(n) = 1 - A014707(n) = 2 - A014709(n) = A014710(n) - 1. (End)
Set a=1, b=0, S(0)=a, S(n+1)=S(n), a, M(S(n)), where M(S) is S but the bit in the middle position flipped. (Proof via isomorphism of both formulas to a modified string substitution.) - Benjamin Heiland, Dec 11 2011
a(n) = 1 if A005811(n+1) > A005811(n), otherwise a(n) = 0. - Gary W. Adamson, Jun 20 2012
a((2*n+1)*2^p-1) = (n+1) mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
G.f. g(x) satisfies g(x) = x*g(x^2) + 1/(1-x^4). - Robert Israel, Jan 06 2015
a(n) = 1 - A065339(n+1) mod 2. - Peter Munn, Jun 29 2022
From A.H.M. Smeets, Mar 19 2023: (Start)
a(n) = 1 - A038189(n+1).
a(n) = A082410(n+2).
a(n) = 1 - A089013(n+1)
a(n) = (3 - A099545(n+1))/2.
a(n) = (A112347(n+1) + 1)/2.
a(n) = (A121238(n+1) + 1)/2.
a(n) = (A317335(n) + 2)/3.
a(n) = (A317336(n) + 10)/3. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Sep 14 2024

Extensions

More terms from Ralf Stephan, Jul 03 2003

A034947 Jacobi (or Kronecker) symbol (-1/n).

Original entry on oeis.org

1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, -1, 1, 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1
Offset: 1

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Author

Keywords

Comments

Also the regular paper-folding sequence.
For a proof that a(n) equals the paper-folding sequence, see Allouche and Sondow, arXiv v4. - Jean-Paul Allouche and Jonathan Sondow, May 19 2015
It appears that, replacing +1 with 0 and -1 with 1, we obtain A038189. Alternatively, replacing -1 with 0 we obtain (allowing for offset) A014577. - Jeremy Gardiner, Nov 08 2004
Partial sums = A005811 starting (1, 2, 1, 2, 3, 2, 1, 2, 3, ...). - Gary W. Adamson, Jul 23 2008
The congruence in {-1,1} of the odd part of n modulo 4 (Cf. A099545). - Peter Munn, Jul 09 2022

Examples

			G.f. = x + x^2 - x^3 + x^4 + x^5 - x^6 - x^7 + x^8 + x^9 + x^10 - x^11 - x^12 + ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • H. Cohen, Course in Computational Number Theory, p. 28.

Crossrefs

Moebius transform of A035184.
Cf. A091072 (indices of 1), A091067 (indices of -1), A371594 (indices of run starts).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

Programs

  • Magma
    [KroneckerSymbol(-1,n): n in [1..100]]; // Vincenzo Librandi, Aug 16 2016
    
  • Maple
    with(numtheory): A034947 := n->jacobi(-1,n);
  • Mathematica
    Table[KroneckerSymbol[ -1, n], {n, 0, 100}] (* Corrected by Jean-François Alcover, Dec 04 2013 *)
  • PARI
    {a(n) = kronecker(-1, n)};
    
  • PARI
    for(n=1, 81, f=factor(n); print1((-1)^sum(s=1, omega(n), f[s, 2]*(Mod(f[s, 1], 4)==3)), ", ")); \\ Arkadiusz Wesolowski, Nov 05 2013
    
  • PARI
    a(n)=direuler(p=1,n,if(p==2,1/(1-kronecker(-4, p)*X)/(1-X),1/(1-kronecker(-4, p)*X))) /* Ralf Stephan, Mar 27 2015 */
    
  • PARI
    a(n) = if(n%2==0, a(n/2), (n+2)%4-2) \\ Peter Munn, Jul 09 2022
  • Python
    def A034947(n):
        s = bin(n)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-2*int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021
    
  • Python
    def A034947(n): return -1 if n>>(-n&n).bit_length()&1 else 1 # Chai Wah Wu, Feb 26 2025
    

Formula

Multiplicative with a(2^e) = 1, a(p^e) = (-1)^(e*(p-1)/2) if p>2.
a(2*n) = a(n), a(4*n+1) = 1, a(4*n+3) = -1, a(-n) = -a(n). a(n) = 2*A014577(n-1)-1.
a(prime(n)) = A070750(n) for n > 1. - T. D. Noe, Nov 08 2004
This sequence can be constructed by starting with w = "empty string", and repeatedly applying the map w -> w 1 reverse(-w) [See Allouche and Shallit p. 182]. - N. J. A. Sloane, Jul 27 2012
a(n) = (-1)^A065339(n) = lambda(A097706(n)), where A065339(n) is number of primes of the form 4*m + 3 dividing n (counted with multiplicity) and lambda is Liouville's function, A008836. - Arkadiusz Wesolowski, Nov 05 2013 and Peter Munn, Jun 22 2022
Sum_{n>=1} a(n)/n = Pi/2, due to F. von Haeseler; more generally, Sum_{n>=1} a(n)/n^(2*d+1) = Pi^(2*d+1)*A000364(d)/(2^(2*d+2)-2)(2*d)! for d >= 0; see Allouche and Sondow, 2015. - Jean-Paul Allouche and Jonathan Sondow, Mar 20 2015
Dirichlet g.f.: beta(s)/(1-2^(-s)) = L(chi_2(4),s)/(1-2^(-s)). - Ralf Stephan, Mar 27 2015
a(n) = A209615(n) * (-1)^(v2(n)), where v2(n) = A007814(n) is the 2-adic valuation of n. - Jianing Song, Apr 24 2021
a(n) = 2 - A099545(n) == A000265(n) (mod 4). - Peter Munn, Jun 22 2022 and Jul 09 2022

A091067 Numbers whose odd part is of the form 4k+3.

Original entry on oeis.org

3, 6, 7, 11, 12, 14, 15, 19, 22, 23, 24, 27, 28, 30, 31, 35, 38, 39, 43, 44, 46, 47, 48, 51, 54, 55, 56, 59, 60, 62, 63, 67, 70, 71, 75, 76, 78, 79, 83, 86, 87, 88, 91, 92, 94, 95, 96, 99, 102, 103, 107, 108, 110, 111, 112, 115, 118, 119, 120, 123, 124, 126, 127, 131
Offset: 1

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Author

Ralf Stephan, Feb 22 2004

Keywords

Comments

Either of form 2*a(m) or 4k+3, k >= 0, 0 < m < n.
A000265(a(n)) is an element of A004767.
a(n) such that A038189(a(n)) = 1.
Numbers n such that Kronecker(-n, m) = Kronecker(m, n) for all m. - Michael Somos, Sep 22 2005
From Antti Karttunen, Feb 20-21 2015: (Start)
Gives all n for which A005811(n) - A005811(n-1) = -1, from which follows that a(n) = the least k such that A255070(k) = n.
Gives the positions of even terms in A003602. (End)
Indices of negative terms in A164677. - M. F. Hasler, Aug 06 2015
Indices of the 0's in A014577. - Gabriele Fici, Jun 02 2016
Also indices of -1 in A034947. - Jianing Song, Apr 24 2021
Conjecture: alternate definition of same sequence is that a(1)=3 and a(n) is the smallest number > a(n-1) so that no number that is the sum of at most 2 terms in this sequence is a power of 2. - J. Lowell, Jan 20 2024
The asymptotic density of this sequence is 1/2. - Amiram Eldar, Aug 31 2024

Crossrefs

Essentially one less than A060833.
Characteristic function: A038189.
Complement of A091072.
First differences are in A106836 (from its second term onward).
Sequence A246590 gives the even terms.
Gives the positions of records (after zero) for A255070 (equally, the position of the first n there).
Cf. A106837 (gives n such that both n and n+1 are terms of this sequence).
Cf. A098502 (gives n such that both n and n+2 are, but n+1 is not in this sequence).

Programs

  • Haskell
    import Data.List (elemIndices)
    a091067 n = a091067_list !! (n-1)
    a091067_list = map (+ 1) $ elemIndices 1 a014707_list
    -- Reinhard Zumkeller, Sep 28 2011
    (Scheme, with Antti Karttunen's IntSeq-library, two versions)
    (define A091067 (MATCHING-POS 1 1 (COMPOSE even? A003602)))
    (define A091067 (NONZERO-POS 1 0 A038189))
    ;; Antti Karttunen, Feb 20 2015
  • Mathematica
    Select[Range[150], Mod[# / 2^IntegerExponent[#, 2], 4] == 3 &] (* Amiram Eldar, Aug 31 2024 *)
  • PARI
    for(n=1,200,if(((n/2^valuation(n,2)-1)/2)%2,print1(n",")))
    
  • PARI
    {a(n) = local(m, c); if( n<1, 0, c=0; m=1; while( cMichael Somos, Sep 22 2005 */
    
  • PARI
    is_A091067(n)=bittest(n,valuation(n,2)+1) \\ M. F. Hasler, Aug 06 2015
    
  • PARI
    a(n) = my(t=1); n<<=1; forstep(i=logint(n,2),0,-1, if(bittest(n,i)==t, n++;t=!t)); n; \\ Kevin Ryde, Mar 21 2021
    

Formula

a(n) = A060833(n+1) - 1. [See N. Sato's Feb 12 2013 comment in A060833.]
Other identities. For all n >= 1 it holds that:
A014707(a(n) + 1) = 1. - Reinhard Zumkeller, Sep 28 2011
A055975(a(n)) < 0. - Reinhard Zumkeller, Apr 28 2012
From Antti Karttunen, Feb 20-21 2015: (Start)
a(n) = A246590(n)/2.
A255070(a(n)) = n, or equally, A236840(a(n)) = 2n.
a(n) = 1 + A255068(n-1). (End)

A014707 a(4n) = 0, a(4n+2) = 1, a(2n+1) = a(n).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0
Offset: 0

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Author

Keywords

Comments

The regular paper-folding (or dragon curve) sequence.
It appears that the sequence of run lengths is A088431. - Dimitri Hendriks, May 06 2010
Runs of three consecutive ones appear around positions n = 22, 46, 54, 86, 94, 118, 150, 174, 182, ..., or for n of the form 2^(k+3)*(4*t+3)-2, k >= 0, t >= 0. - Vladimir Shevelev, Mar 19 2011

References

  • Guy Melançon, Factorizing infinite words using Maple, MapleTech journal, Vol. 4, No. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

Equals 1 - A014577, which see for further references. Also a(n) = A038189(n+1).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410.

Programs

  • Haskell
    a014707 n = a014707_list !! n
    a014707_list = f 0 $ cycle [0,0,1,0] where
       f i (x:_:xs) = x : a014707 i : f (i+1) xs
    -- Reinhard Zumkeller, Sep 28 2011
    
  • Maple
    nmax:=92: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := n mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
    # second Maple program:
    a:= proc(n) option remember;
         `if`(n::even, irem(n/2, 2), a((n-1)/2))
        end:
    seq(a(n), n=0..92);  # Alois P. Heinz, Jun 27 2022
  • Mathematica
    a[n_ /; Mod[n, 4] == 0] = 0; a[n_ /; Mod[n, 4] == 2] = 1; a[n_ /; Mod[n, 2] == 1] := a[n] = a[(n - 1)/2]; Table[a[n],{n,0,92}] (* Jean-François Alcover, May 17 2011 *)
    (1 - JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 26 2024 *)
  • PARI
    a(n)=n+=1;my(h=bitand(n,-n));n=bitand(n,h<<1);n!=0; \\ Joerg Arndt, Apr 09 2021
  • Python
    def A014707(n):
        s = bin(n+1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return int(s[m-i-2]) if m-i-2 >= 0 else 0 # Chai Wah Wu, Apr 08 2021
    
  • Python
    def A014707(n): n+=1; h=n&-n; n=n&(h<<1); return int(n!=0)
    print([A014707(n) for n in range(93)]) # Michael S. Branicky, Mar 29 2024 after Joerg Arndt
    

Formula

a(A091072(n)-1) = 0; a(A091067(n)-1) = 1. - Reinhard Zumkeller, Sep 28 2011 [Adjusted to match offset by Peter Munn, Jul 01 2022]
a(n) = (1-Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=0, b=1, S(0)=a, S(n+1) = S(n)aF(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
a((2*n+1)*2^p-1) = n mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Aug 31 2024

Extensions

More terms from Scott C. Lindhurst (ScottL(AT)alumni.princeton.edu)

A055975 First differences of A003188 (decimal equivalent of the Gray Code).

Original entry on oeis.org

1, 2, -1, 4, 1, -2, -1, 8, 1, 2, -1, -4, 1, -2, -1, 16, 1, 2, -1, 4, 1, -2, -1, -8, 1, 2, -1, -4, 1, -2, -1, 32, 1, 2, -1, 4, 1, -2, -1, 8, 1, 2, -1, -4, 1, -2, -1, -16, 1, 2, -1, 4, 1, -2, -1, -8, 1, 2, -1, -4, 1, -2, -1, 64, 1, 2, -1, 4, 1, -2, -1, 8, 1, 2, -1, -4, 1, -2, -1, 16, 1, 2, -1, 4, 1, -2, -1, -8, 1, 2, -1, -4, 1, -2, -1, -32, 1, 2, -1, 4
Offset: 1

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Author

Alford Arnold, Jul 22 2000

Keywords

Comments

Multiplicative with a(2^e) = 2^e, a(p^e) = (-1)^((p^e-1)/2) otherwise. - Mitch Harris, May 17 2005
a(A091072(n)) > 0; a(A091067(n)) < 0. - Reinhard Zumkeller, Apr 28 2012
In the binary representation of n, clear everything left of the least significant 1 bit, and negate if the bit left of it was set originally. - Ralf Stephan, Aug 23 2013
This sequence is the trace of n in the minimal alternating binary representation of n (defined at A256696). - Clark Kimberling, Apr 07 2015

Examples

			Since A003188 is 0, 1,  3, 2, 6,  7,  5, 4, 12, 13, 15, 14, 10, ...,
sequence begins  1, 2, -1, 4, 1, -2, -1, 8,  1,  2, -1,  4, ... .
		

Crossrefs

Cf. A003188, A006519 (unsigned), A007814.
MASKTRANSi transform of A053644 (conjectural).

Programs

  • Haskell
    a055975 n = a003188 n - a003188 (n-1)
    a055975_list = zipWith (-) (tail a003188_list) a003188_list
    -- Reinhard Zumkeller, Apr 28 2012
    
  • Maple
    nmax:=100: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (-1)^(n+1)*2^p od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Jan 27 2013
  • Mathematica
    f[n_]:=BitXor[n,Floor[n/2]];Differences[Array[f,120,0]] (* Harvey P. Dale, Jul 18 2011, applying Robert G. Wilson v's program from A003188 *)
  • PARI
    a(n)=((-1)^((n/2^valuation(n,2)-1)/2)*2^valuation(n,2)) \\ Ralf Stephan
    
  • Python
    def A055975(n): return (n^(n>>1))-((n-1)^(n-1>>1)) # Chai Wah Wu, Jun 29 2022

Formula

a(2n) = 2a(n), a(2n+1) = (-1)^n. G.f. sum(k>=0, 2^k*t/(1+t^2), t=x^2^k). a(n) = 2^A007814(n) * (-1)^((n/2^A007814(n)-1)/2). - Ralf Stephan, Oct 29 2003
a((2*n-1)*2^p) = (-1)^(n+1)*2^p, p >= 0. - Johannes W. Meijer, Jan 27 2013

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Sep 05 2000

A266121 Lexicographically first injection of natural numbers beginning with a(1)=1 such that 1+(a(n)*a(n+1)) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

1, 3, 5, 4, 2, 8, 9, 7, 12, 6, 14, 24, 11, 13, 20, 16, 10, 26, 40, 17, 15, 39, 28, 19, 27, 25, 23, 48, 22, 30, 44, 31, 33, 32, 18, 36, 29, 47, 45, 52, 21, 55, 49, 84, 61, 43, 51, 53, 80, 34, 64, 37, 35, 59, 75, 117, 93, 91, 57, 41, 100, 82, 50, 104, 42, 98, 106, 90, 114, 72, 58, 144, 65, 63, 151, 56, 38, 54, 76, 71, 60
Offset: 1

Views

Author

Antti Karttunen, Dec 23 2015

Keywords

Comments

It is conjectured that this sequence is not only injective, but also surjective on N, i.e., that it is a true permutation of natural numbers.
A similar sequence, but with condition that "(a(n)*a(n+1)) must be a member of A003714" yields a sequence: 1, 2, 4, 5, 8, 9, 16, 10, 13, 20, ... (A269361) which certainly is not a bijection, because it contains only terms of A091072.
Also, with above condition and the initial value a(1) = 3 the algorithm generates A269363 which contains only terms of A091067. See also A266191.

Examples

			After the initial a(1) = 1, for obtaining the value of a(2) we try the first unused number, which is 2, but (1*2)+1 = 3, which in binary is "11", thus 2 is not qualified at this point of time. So next we try 3, and (1*3)+1 = 4, which in binary is "100", and that satisfies our criterion (no adjacent 1-bits), thus we set a(2) = 3.
For a(3), we test with the least unused numbers 2, 4, 5, etc., yielding products (3*2)+1 = 7 = "111", (3*4)+1 = 13 = "1101" and (3*5)+1 = 16 = "10000" in binary, and only 5 satisfies the criterion, thus we set a(3) = 5.
		

Crossrefs

Left inverse: A266122 (also the right inverse if this sequence is a permutation of natural numbers).
Cf. also A266191 and A266117 for similar permutations.

Extensions

Minor typo in the description corrected by Antti Karttunen, Feb 25 2016

A256696 R(k), the minimal alternating binary representation of k, concatenated for k = 0, 1, 2,....

Original entry on oeis.org

0, 1, 2, 4, -1, 4, 8, -4, 1, 8, -2, 8, -1, 8, 16, -8, 1, 16, -8, 2, 16, -8, 4, -1, 16, -4, 16, -4, 1, 16, -2, 16, -1, 16, 32, -16, 1, 32, -16, 2, 32, -16, 4, -1, 32, -16, 4, 32, -16, 8, -4, 1, 32, -16, 8, -2, 32, -16, 8, -1, 32, -8, 32, -8, 1, 32, -8, 2, 32
Offset: 0

Views

Author

Clark Kimberling, Apr 09 2015

Keywords

Comments

Suppose that b = (b(0), b(1), ... ) is an increasing sequence of positive integers satisfying b(0) = 1 and b(n+1) <= 2*b(n) for n >= 0. Let B(n) be the least b(m) >= n. Let R(0) = 1, and for n > 0, let R(n) = B(n) - R(B(n) - n). The resulting sum of the form R(n) = B(n) - B(m(1)) + B(m(2)) - ... + ((-1)^k)*B(k) is the minimal alternating b-representation of n. The sum B(n) + B(m(2)) + ... is the positive part of R(n), and the sum B(m(1)) + B(m(3)) + ... , the nonpositive part of R(n). The number ((-1)^k)*B(k) is the trace of n.
If b(n) = 2^n, the sum R(n) is the minimal alternating binary representation of n.
A055975 = trace of n, for n >= 1.
A091072 gives the numbers having positive trace.
A091067 gives the numbers having negative trace.
A072339 = number of terms in R(n).
A073122 = sum of absolute values of the terms in R(n).

Examples

			R(0) = 0
R(1) = 1
R(2) = 2
R(3) = 4 - 1
R(4) = 4
R(9) = 8 - 4 + 1
R(11) = 16 - 8 + 4 - 1
		

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1981, Vol. 2 (2nd ed.), p. 196, Exercise 27.

Crossrefs

Programs

  • Mathematica
    z = 100; b[n_] := 2^n; bb = Table[b[n], {n, 0, 40}];
    s[n_] := Table[b[n + 1], {k, 1, b[n]}];
    h[0] = {1}; h[n_] := Join[h[n - 1], s[n - 1]];
    g = h[10]; r[0] = {0};
    r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]]
    u = Flatten[Table[r[n], {n, 0, z}]]

A269361 Lexicographically first injection of natural numbers beginning with a(1)=1 such that a(n)*a(n+1) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 16, 10, 13, 20, 17, 32, 18, 29, 36, 33, 40, 26, 21, 49, 42, 52, 25, 41, 50, 82, 57, 45, 53, 80, 34, 64, 37, 113, 74, 125, 84, 61, 72, 58, 73, 65, 68, 69, 128, 66, 129, 132, 133, 77, 114, 81, 117, 90, 104, 85, 98, 89, 93, 100, 105, 161, 106, 97, 169, 122, 137, 157, 138, 136, 121, 141, 149, 221, 153, 109, 160, 116, 144
Offset: 1

Views

Author

Antti Karttunen, Feb 25 2016

Keywords

Comments

The sequence is conjectured to be a permutation of A091072.

Crossrefs

A269366 a(1) = 1, a(2n) = A269361(1+a(n)), a(2n+1) = A269363(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 12, 16, 15, 9, 11, 10, 22, 13, 24, 18, 38, 40, 48, 33, 46, 20, 14, 32, 27, 17, 19, 25, 44, 29, 28, 50, 75, 21, 30, 72, 71, 73, 70, 133, 139, 113, 76, 129, 91, 42, 35, 36, 23, 37, 54, 45, 51, 26, 43, 49, 39, 82, 62, 128, 107, 80, 56, 53, 83, 114, 140, 109, 214, 52, 59, 34, 47, 149, 150, 141, 123, 221, 111, 121
Offset: 1

Views

Author

Antti Karttunen, Feb 25 2016

Keywords

Comments

This sequence can be represented as a binary tree. Each left hand child is produced as A269361(1+n), and each right hand child as A269363(n), when the parent node contains n:
|
...................1...................
2 3
4......../ \........6 5......../ \........7
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
8 12 16 15 9 11 10 22
13 24 18 38 40 48 33 46 20 14 32 27 17 19 25 44
etc.
An example of (suspected) "entanglement permutation" where the other pair of complementary sequences is generated by a greedy algorithm.
Sequence is not only injective, but also surjective on N (thus a permutation of natural numbers) provided that A269361 is surjective on A091072 and A269363 is surjective on A091067.

Crossrefs

Left inverse: A269365 (also right inverse, if this sequence is a permutation of natural numbers).
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