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.

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A303639 Number of ways to write n as a^2 + b^2 + binomial(2*c+1,c) + binomial(2*d+1,d), where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 2, 1, 2, 3, 3, 3, 3, 4, 2, 2, 2, 3, 4, 4, 5, 2, 4, 1, 2, 3, 3, 5, 3, 5, 1, 3, 1, 1, 6, 3, 8, 3, 6, 2, 4, 4, 2, 7, 5, 6, 2, 5, 2, 4, 5, 4, 8, 4, 7, 2, 4, 1, 3, 6, 4, 7, 3, 5, 2, 4, 2, 4, 9, 5, 6, 2, 6, 4, 5, 4, 7, 5, 2
Offset: 1

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Author

Zhi-Wei Sun, Apr 27 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is similar to the author's conjecture in A303540.
It has been verified that a(n) > 0 for all n = 2..6*10^8.

Examples

			a(9) = 1 with 9 = 1^2 + 2^2 + binomial(2*0+1,0) + binomial(2*1+1,1).
a(2530) = 1 with 2530 = 0^2 + 49^2 + binomial(2*1+1,1) + binomial(2*4+1,4).
a(3258) = 1 with 3258 = 22^2 + 52^2 + binomial(2*3+1,3) + binomial(2*3+1,3).
a(5300) = 1 with 5300 = 10^2 + 59^2 + binomial(2*1+1,1) + binomial(2*6+1,6).
a(13453) = 1 with 13453 = 51^2 + 104^2 + binomial(2*0+1,0) + binomial(2*3+1,3).
a(20964) = 1 with 20964 = 13^2 + 138^2 + binomial(2*3+1,3) + binomial(2*6+1,6).
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    c[n_]:=c[n]=Binomial[2n+1,n];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],Do[If[SQ[n-c[k]-c[j]-x^2],r=r+1],{x,0,Sqrt[(n-c[k]-c[j])/2]}]],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A000415 Numbers that are the sum of 2 but no fewer nonzero squares.

Original entry on oeis.org

2, 5, 8, 10, 13, 17, 18, 20, 26, 29, 32, 34, 37, 40, 41, 45, 50, 52, 53, 58, 61, 65, 68, 72, 73, 74, 80, 82, 85, 89, 90, 97, 98, 101, 104, 106, 109, 113, 116, 117, 122, 125, 128, 130, 136, 137, 145, 146, 148, 149, 153, 157, 160, 162, 164, 170, 173, 178, 180, 181
Offset: 1

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Author

Keywords

Comments

Only these numbers can occur as discriminants of quintic polynomials with solvable Galois group F20. - Artur Jasinski, Oct 25 2007
Complement of A022544 in the nonsquare positive integers A000037. - Max Alekseyev, Jan 21 2010
Nonsquare positive integers D such that Pell equation y^2 - D*x^2 = -1 has rational solutions. - Max Alekseyev, Mar 09 2010
Nonsquares for which all 4k+3 primes in the integer's canonical form occur with even multiplicity. - Ant King, Nov 02 2010

References

  • E. Grosswald, Representation of Integers as Sums of Squares, Springer-Verlag, New York Inc., (1985), p.15. - Ant King, Nov 02 2010

Crossrefs

Programs

  • Mathematica
    c = {}; Do[Do[k = a^2 + b^2; If[IntegerQ[Sqrt[k]], Null, AppendTo[c,k]], {a, 1, 100}], {b, 1, 100}]; Union[c] (* Artur Jasinski, Oct 25 2007 *)
    Select[Range[181],Length[PowersRepresentations[ #,2,2]]>0 && !IntegerQ[Sqrt[ # ]] &] (* Ant King, Nov 02 2010 *)
  • PARI
    is(n)=my(f=factor(n)); for(i=1, #f[, 1], if(f[i, 2]%2 && f[i, 1]%4==3, return(0))); !issquare(n) \\ Charles R Greathouse IV, Feb 07 2017
    
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A000415_gen(startvalue=2): # generator of terms >= startvalue
        for n in count(max(startvalue,2)):
            f = factorint(n).items()
            if any(e&1 for p,e in f if p&3<3) and not any(e&1 for p,e in f if p&3==3):
                yield n
    A000415_list = list(islice(A000415_gen(),20)) # Chai Wah Wu, Aug 01 2023

Formula

{ A000404 } minus { A134422 }. - Artur Jasinski, Oct 25 2007

Extensions

More terms from Arlin Anderson (starship1(AT)gmail.com)

A300441 Number of the integers 4^k*(4*u(m)^2+1) (k,m = 0,1,2,...) such that n^2 - 4^k*(4*u(m)^2+1) can be written as the sum of two squares, where u(0) = 0, u(1) = 1, and u(j+1) = 4*u(j) - u(j-1) for j = 1,2,3,....

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 1, 4, 3, 3, 2, 3, 1, 3, 1, 3, 4, 4, 3, 5, 3, 3, 2, 4, 3, 3, 1, 5, 3, 3, 1, 6, 3, 4, 4, 5, 4, 4, 3, 6, 5, 4, 3, 5, 3, 4, 2, 5, 4, 5, 3, 4, 3, 5, 1, 5, 5, 3, 3, 3, 3, 5, 1, 5, 6, 3, 3, 6, 4, 4, 4, 6, 5, 5, 4, 6, 4, 5, 3
Offset: 1

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Author

Zhi-Wei Sun, Mar 05 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k*m (k = 0,1,2,... and m = 1, 7).
This curious conjecture indicates that any positive square can be written as (2^k)^2 + (2^(k+1)*u(m))^2 + x^2 + y^2 with k,m,x,y nonnegative integers. In the 2017 JNT paper, the author proved that each n = 1,2,3,... can be written as 4^k*(1+4*x^2+y^2)+z^2 with k,x,y,z nonnegative integers.
We have verified that a(n) > 0 for all n = 1..10^7.

Examples

			a(1) = 1 since 1^2 - 4^0*(4*u(0)^2+1) = 1 is 1^2 + 0^2.
a(5) = 3 since 5^2 - 4^0*(4*u(1)^2+1) = 20 = 4^2 + 2^2, 5^2 - 4^1*(4*u(1)^2+1) = 5 = 2^2 + 1^2, and 5^2 - 4^2*(4*u(0)^2+1) = 9 = 3^2 + 0^2.
a(7) = 1 since 7^2 - 4^1*(4*u(0)^2+1) = 45 = 6^2 + 3^2.
		

Crossrefs

Programs

  • Mathematica
    u[0]=0;
    u[1]=1;
    u[n_]:=u[n]=4u[n-1]-u[n-2];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[If[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1,1,0],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
    tab={};Do[r=0;Do[m=0;Label[cc];If[4u[m]^2+1>n^2/4^k,Goto[bb]];If[QQ[n^2-4^k*(4u[m]^2+1)],r=r+1,m=m+1;Goto[cc]];
    Label[bb],{k,0,Log[2,n]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A020668 Numbers of the form x^2 + 4*y^2.

Original entry on oeis.org

0, 1, 4, 5, 8, 9, 13, 16, 17, 20, 25, 29, 32, 36, 37, 40, 41, 45, 49, 52, 53, 61, 64, 65, 68, 72, 73, 80, 81, 85, 89, 97, 100, 101, 104, 109, 113, 116, 117, 121, 125, 128, 136, 137, 144, 145, 148, 149, 153, 157, 160, 164, 169, 173, 180, 181, 185, 193, 196, 197, 200, 205, 208
Offset: 1

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Author

Keywords

Comments

x^2 + 4y^2 has discriminant -16.
Numbers that can be expressed as both the sum of two squares and the difference of two squares; the intersection of sequences A001481 and A042965. - T. D. Noe, Feb 05 2003
A004531(n) is nonzero if and only if n is of the form x^2 + 4*y^2. - Michael Somos, Jan 05 2012
These are the sum of two squares that are congruent to 0 or 1 (mod 4), and thus that are also the difference of two squares. - Jean-Christophe Hervé, Oct 25 2015

Crossrefs

Cf. A001481, A004531, A042965, A097269. For primes see A002144.

Programs

  • Magma
    [n: n in [0..208] | NormEquation(4, n) eq true]; // Arkadiusz Wesolowski, May 11 2016
  • Mathematica
    Select[Range[0, 300], SquaresR[2, #] != 0 && Mod[#, 4] != 2&] (* Jean-François Alcover, May 13 2017 *)
  • PARI
    for(n=0, 1e3, if(if( n<1, n==0, 2 * qfrep([ 1, 0; 0, 4], n)[n]) != 0, print1(n, ", "))) \\ Altug Alkan, Oct 29 2015
    

Formula

Complement of A097269 in A001481. - Jean-Christophe Hervé, Oct 25 2015

A035251 Positive numbers of the form x^2 - 2y^2 with integers x, y.

Original entry on oeis.org

1, 2, 4, 7, 8, 9, 14, 16, 17, 18, 23, 25, 28, 31, 32, 34, 36, 41, 46, 47, 49, 50, 56, 62, 63, 64, 68, 71, 72, 73, 79, 81, 82, 89, 92, 94, 97, 98, 100, 103, 112, 113, 119, 121, 124, 126, 127, 128, 136, 137, 142, 144, 146, 151, 153, 158, 161, 162, 164, 167, 169, 175, 178
Offset: 1

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Author

Keywords

Comments

x^2 - 2y^2 has discriminant 8. - N. J. A. Sloane, May 30 2014
A positive number n is representable in the form x^2 - 2y^2 iff every prime p == 3 or 5 (mod 8) dividing n occurs to an even power.
Indices of nonzero terms in expansion of Dirichlet series Product_p (1-(Kronecker(m,p)+1)*p^(-s)+Kronecker(m,p)*p^(-2s))^(-1) for m=2 (A035185). [amended by Georg Fischer, Sep 03 2020]
Also positive numbers of the form 2x^2 - y^2. If x^2 - 2y^2 = n, 2(x+y)^2 - (x+2y)^2 = n. - Franklin T. Adams-Watters, Nov 09 2009
Except 2, prime numbers in this sequence have the form p=8k+-1. According to the first comment, prime factors of the forms (8k+-3),(8k+-5) occur in x^2 - 2y^2 in even powers. If x^2 - 2y^2 is a prime number, those powers must be 0. Only factors 8k+-1 remain. Example: 137=8*17+1. - Jerzy R Borysowicz, Nov 04 2015
The product of any two terms of the sequence is a term too. A proof follows from the identity: (a^2-2b^2)(c^2-2d^2) = (2bd+ac)^2 - 2(ad+bc)^2. Example: 127*175 has form x^2-2y^2, with x=9335, y=6600. - Jerzy R Borysowicz, Nov 28 2015
Primitive terms (not a product of earlier terms that are greater than 1 in the sequence) are A055673 except 1. - Charles R Greathouse IV, Sep 10 2016
Positive numbers of the form u^2 + 2uv - v^2. - Thomas Ordowski, Feb 17 2017
For integer numbers z, a, k and z^2+a^2>0, k>=0: z^(4k) + a^4 is in A035251 because z^(4k) + a^4 = (z^(2k) + a^2)^2 - 2(a*z^k)^2. Assume 0^0 = 1. Examples: 3^4 + 1^4 = 82, 3^8+4^4=6817. - Jerzy R Borysowicz, Mar 09 2017
Numbers that are the difference between two legs of a Pythagorean right triangle. - Michael Somos, Apr 02 2017

Examples

			The (x,y) pairs, with minimum x, that solve the equation are (1,0), (2,1), (2,0), (3,1), (4,2), (3,0), (4,1), (4,0), (5,2), (6,3), (5,1), (5,0), (6,2), (7,3), (8,4), (6,1), (6,0), (7,2), (8,3), (7,1), (7,0), (10,5), (8,2), ... If the positive number is a perfect square, y=0 yields a trivial solution. - _R. J. Mathar_, Sep 10 2016
		

Crossrefs

Primes: A038873.
Complement of A232531. - Thomas Ordowski and Altug Alkan, Feb 09 2017

Programs

  • Maple
    filter:= proc(n) local F;
      F:= select(t -> t[1] mod 8 = 3 or t[1] mod 8 = 5, ifactors(n)[2]);
      map(t -> t[2],F)::list(even);
    end proc:
    select(filter, [$1..1000]); # Robert Israel, Dec 01 2015
  • Mathematica
    Reap[For[n = 1, n < 200, n++, r = Reduce[x^2 - 2 y^2 == n, {x, y}, Integers]; If[r =!= False, Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 31 2016 *)
  • PARI
    select(x -> x, direuler(p=2,201,1/(1-(kronecker(2,p)*(X-X^2))-X)), 1) \\ Fixed by Andrey Zabolotskiy, Jul 30 2020
    
  • PARI
    {a(n) = my(m, c); if( n<1, 0, c=0; m=0; while( cMichael Somos, Aug 17 2006 */
    
  • PARI
    is(n)=#bnfisintnorm(bnfinit(z^2-2),n) \\ Ralf Stephan, Oct 14 2013
    
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A035251_gen(): # generator of terms
        return filter(lambda n:all(not((2 < p & 7 < 7) and e & 1) for p, e in factorint(n).items()),count(1))
    A035251_list = list(islice(A035251_gen(),30)) # Chai Wah Wu, Jun 28 2022

Extensions

Better description from Sharon Sela (sharonsela(AT)hotmail.com), Mar 10 2002

A232499 Number of unit squares, aligned with a Cartesian grid, completely within the first quadrant of a circle centered at the origin ordered by increasing radius.

Original entry on oeis.org

1, 3, 4, 6, 8, 10, 11, 13, 15, 17, 19, 20, 22, 24, 26, 28, 30, 33, 35, 37, 39, 41, 45, 47, 48, 50, 52, 54, 56, 60, 62, 64, 66, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 89, 90, 94, 96, 98, 102, 104, 106, 108, 110, 112, 114, 115, 117, 119, 123, 125, 127, 129, 131
Offset: 1

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Author

Rajan Murthy and Vale Murthy, Nov 24 2013

Keywords

Comments

The interval between terms reflects the number of ways a square integer can be partitioned into the sum of two square integers in an ordered pair. As examples, the increase from a(1) to a(2) from 1 to 3 is due to the inclusion of (1,2) and (2,1); and the increase from a(2) to a(3) is due to the inclusion of (2,2). Larger intervals occur when there are more combinations, such as, between a(17) and a(18) when (1,7), (7,1), and (5,5) are included.

Examples

			When radius of the circle exceeds 2^(1/2), one square is completely within the circle until the radius reaches 5^(1/2) when three squares are completely within the circle.
		

Crossrefs

First differences are in A229904.
The first differences must be odd at positions given in A024517 by proof by symmetry as r^2=2*n^2 is on the x=y line.
The radii corresponding to the terms are given by the square roots of A000404.
Cf. A237707 (3-dimensional analog), A239353 (4-dimensional analog).

Programs

  • Mathematica
    (* An empirical solution *) terms = 100; f[r_] := Sum[Floor[Sqrt[r^2 - n^2]], {n, 1, Floor[r]}]; Clear[g]; g[m_] := g[m] = Union[Table[f[Sqrt[s]], {s, 2, m }]][[1 ;; terms]]; g[m = dm = 4*terms]; g[m = m + dm]; While[g[m] != g[m - dm], Print[m]; m = m + dm]; A232499 = g[m]  (* Jean-François Alcover, Mar 06 2014 *)

A057653 Odd numbers of form x^2 + y^2.

Original entry on oeis.org

1, 5, 9, 13, 17, 25, 29, 37, 41, 45, 49, 53, 61, 65, 73, 81, 85, 89, 97, 101, 109, 113, 117, 121, 125, 137, 145, 149, 153, 157, 169, 173, 181, 185, 193, 197, 205, 221, 225, 229, 233, 241, 245, 257, 261, 265, 269, 277, 281, 289, 293, 305, 313, 317, 325, 333, 337
Offset: 1

Views

Author

N. J. A. Sloane, Oct 15 2000

Keywords

Comments

Numbers with only odd prime factors and such that all prime factors congruent to 3 modulo 4 occur to an even exponent. - Jean-Christophe Hervé, Oct 24 2015
Odd terms of A020668. - Altug Alkan, Nov 19 2015
Also one half of the numbers that are the sum of two odd squares (without multiplicity). See A097269 for twice the numbers. - Wolfdieter Lang, Jan 12 2017

Crossrefs

Odd members of A001481.
Odd members of A020668.
Complement of A084109 in 4k+1 numbers (A016813).
Cf. A016754 (odd squares), A097269.

Programs

  • Maple
    readlib(issqr): for n from 1 to 1001 by 2 do for k from 0 to floor(sqrt(n)) do if issqr(n-k^2) then printf(`%d,`,n); break fi; od:od:
  • Mathematica
    fQ[n_] := Length@ Catch@ Do[If[IntegerQ@ Sqrt[n - k^2], Throw[{k, Sqrt[n - k^2]}], Nothing], {k, Floor[Sqrt@ n]^2}] != 0; Select[Range[1, 340, 2], fQ] (* Michael De Vlieger, Nov 13 2015 *)
  • PARI
    isok(n)=my(f=factor(n)); for(i=1, #f[, 1], if(f[i, 2]%2 && f[i, 1]%4==3, return(0))); 1;
    for(n=1, 1e3, if(isok(n) && n%2==1, print1(n", "))) \\ Altug Alkan, Nov 13 2015
    
  • PARI
    for(n=0, 1e3, if(if( n<1, n==0, 2 * qfrep([ 1, 0; 0, 4], n)[n]) != 0 && n%2==1, print1(n, ", "))) \\ Altug Alkan, Nov 19 2015
    
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A057653_gen(): # generator of terms
        return filter(lambda n:all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(n).items()),count(1,2))
    A057653_list = list(islice(A057653_gen(),30)) # Chai Wah Wu, Jun 28 2022

Formula

n = odd square * {product of distinct primes == 1 (mod 4)}.
a(n) = A097269(n)/2. - Wolfdieter Lang, Jan 12 2017

Extensions

More terms from James Sellers, Oct 16 2000

A118882 Numbers which are the sum of two squares in two or more different ways.

Original entry on oeis.org

25, 50, 65, 85, 100, 125, 130, 145, 169, 170, 185, 200, 205, 221, 225, 250, 260, 265, 289, 290, 305, 325, 338, 340, 365, 370, 377, 400, 410, 425, 442, 445, 450, 481, 485, 493, 500, 505, 520, 530, 533, 545, 565, 578, 580, 585, 610, 625, 629, 650, 676, 680
Offset: 1

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Author

Keywords

Comments

Numbers whose prime factorization includes at least two primes (not necessarily distinct) congruent to 1 mod 4 and any prime factor congruent to 3 mod 4 has even multiplicity. Products of two values in A004431.
Squares of distances that are the distance between two points in the square lattice in two or more nontrivially different ways. A quadrilateral with sides a,b,c,d has perpendicular diagonals iff a^2+c^2 = b^2+d^2. This sequence is the sums of the squares of opposite sides of such quadrilaterals, excluding kites (a=b,c=d), but including right triangles (the degenerate case d=0).

Examples

			50 = 7^2 + 1^2 = 5^2 + 5^2, so 50 is in the sequence.
		

Crossrefs

Programs

  • Haskell
    import Data.List (findIndices)
    a118882 n = a118882_list !! (n-1)
    a118882_list = findIndices (> 1) a000161_list
    -- Reinhard Zumkeller, Aug 16 2011
    
  • Mathematica
    Select[Range[1000], Length[PowersRepresentations[#, 2, 2]] > 1&] (* Jean-François Alcover, Mar 02 2019 *)
  • Python
    from itertools import count, islice
    from math import prod
    from sympy import factorint
    def A118882_gen(startvalue=1): # generator of terms >= startvalue
        for n in count(max(startvalue,1)):
            f = factorint(n)
            if 1>1):
                yield n
    A118882_list = list(islice(A118882_gen(),30)) # Chai Wah Wu, Sep 09 2022

Formula

A000161(a(n)) > 1. [Reinhard Zumkeller, Aug 16 2011]

A300510 Number of ways to write n^2 as 4^k*(m^2+1) + x^2 + y^2, where m is 1 or 2, and k,x,y are nonnegative integers with x <= y.

Original entry on oeis.org

0, 1, 2, 1, 3, 3, 3, 1, 4, 4, 5, 3, 5, 4, 6, 1, 3, 4, 5, 4, 7, 6, 5, 3, 8, 6, 6, 4, 5, 7, 7, 1, 5, 4, 11, 4, 7, 5, 6, 4, 6, 8, 5, 6, 12, 5, 5, 3, 6, 9, 8, 6, 7, 6, 10, 4, 7, 7, 6, 7, 5, 9, 9, 1, 8, 5, 10, 4, 9, 11, 9, 4, 11, 7, 12, 5, 8, 7, 7, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 07 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1; in other words, for any integer n > 1 there is a nonnegative integer k such that either n^2 - 2*4^k or n^2 - 5*4^k can be written as the sum of two squares. Moreover, a(n) = 1 only for n = 2^k with k > 0.
This conjecture is stronger than the first conjecture in A300448. We have verified that a(n) > 0 for all n = 2..5*10^7.
Consider positive integers c not divisible by 4 such that for any integer n > 1 there is a nonnegative integer k for which n^2 - 2*4^k or n^2 - c*4^k can be written as the sum of two squares. Our computation for n up to 3*10^7 shows that the only candidates for values of c smaller than 160 are 5, 17, 18, 26, 29, 41, 45, 65, 74, 89, 98, 101, 113, 122, 125, 146, 149, 153. These numbers have the form 9^a*(3*b+2) with a and b nonnegative integers and the p-adic order of 3*b+2 is even for any prime p == 3 (mod 4). For n = 42211965 there is no nonnegative integer k such that n^2 - 2*4^k or n^2 - 162*4^k can be written as the sum of two squares.
Qing-Hu Hou at Tianjin Univ. reported that he had verified a(n) > 0 for n up to 10^9. - Zhi-Wei Sun, Mar 14 2018
Qing-Hu Hou found that 29, 65, 113 should be excluded from the candidates. In fact, for c = 29, 65, 113 there is no nonnegative integer k such that N(c)^2 - 2*4^k or N(c)^2 - c*4^k can be written as the sum of two squares, where N(29) = 51883659, N(65) = 56173837 and N(113) = 65525725. - Zhi-Wei Sun, Mar 23 2018
a(n) > 0 for 1 < n < 6*10^9. - Giovanni Resta, Jun 14 2019

Examples

			a(1) = 0 since 1^2 - 4^k*(m^2+1) < 0 for k = 0,1,2,... and m = 1, 2.
a(2) = 1 since 2^2 = 4^0*(1^2+1) + 1^2 + 1^2.
a(3) = 2 since 3^2 = 4^0*(2^2+1) + 0^2 + 2^2 = 4^1*(1^2+1) + 0^2 + 1^2.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[QQ[n^2-4^k*(m^2+1)],Do[If[SQ[n^2-4^k(m^2+1)-x^2],r=r+1],{x,0,Sqrt[(n^2-4^k(m^2+1))/2]}]],{m,1,2},{k,0,Log[4,n^2/(m^2+1)]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A001983 Numbers that are the sum of 2 distinct squares: of form x^2 + y^2 with 0 <= x < y.

Original entry on oeis.org

1, 4, 5, 9, 10, 13, 16, 17, 20, 25, 26, 29, 34, 36, 37, 40, 41, 45, 49, 50, 52, 53, 58, 61, 64, 65, 68, 73, 74, 80, 81, 82, 85, 89, 90, 97, 100, 101, 104, 106, 109, 113, 116, 117, 121, 122, 125, 130, 136, 137, 144, 145, 146, 148, 149, 153, 157, 160, 164
Offset: 1

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Author

Keywords

Comments

This sequence lists the values of A000404(n)/2 when A000404(n) is an even number. In other words, sequence lists integers n that are the average of two nonzero squares. - Altug Alkan, May 26 2016

Crossrefs

Cf. A000404, subsequence of A001481, A004435 (complement), A025435, A004431.
Union of A000290 and A004431 excluding 0.

Programs

  • Haskell
    a001983 n = a001983_list !! (n-1)
    a001983_list = [x | x <- [0..], a025435 x > 0]
    -- Reinhard Zumkeller, Dec 20 2013
    
  • Mathematica
    upto=200;max=Floor[Sqrt[upto]];s=Total/@((Subsets[Range[0,max], {2}])^2);Union[Select[s,#<=upto&]]  (* Harvey P. Dale, Apr 01 2011 *)
    selQ[n_] := Select[ PowersRepresentations[n, 2, 2], 0 <= #[[1]] < #[[2]] &] != {}; Select[Range[200], selQ] (* Jean-François Alcover, Oct 03 2013 *)
  • PARI
    list(lim)=my(v=List()); for(x=0,sqrtint(lim\4), for(y=x+1, sqrtint(lim\1-x^2), listput(v, x^2+y^2))); Set(v) \\ Charles R Greathouse IV, Feb 07 2017

Formula

A025435(a(n)) > 0. - Reinhard Zumkeller, Dec 20 2013
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