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 1-2 of 2 results.

A022544 Numbers that are not the sum of 2 squares.

Original entry on oeis.org

3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, 27, 28, 30, 31, 33, 35, 38, 39, 42, 43, 44, 46, 47, 48, 51, 54, 55, 56, 57, 59, 60, 62, 63, 66, 67, 69, 70, 71, 75, 76, 77, 78, 79, 83, 84, 86, 87, 88, 91, 92, 93, 94, 95, 96, 99, 102, 103, 105, 107, 108, 110, 111, 112, 114, 115, 118, 119, 120, 123, 124, 126, 127, 129, 131, 132, 133, 134, 135, 138, 139, 140, 141, 142, 143, 147, 150, 151, 152, 154, 155, 156, 158, 159, 161, 163, 165, 166, 167, 168, 171, 172, 174, 175, 176, 177, 179, 182, 183, 184, 186, 187, 188, 189, 190, 191, 192, 195, 198, 199
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Conjecture: if k is not the sum of 2 squares then sigma(k) == 0 (mod 4) (the converse does not hold, as demonstrated by the entries in A025303). - Benoit Cloitre, May 19 2002
Numbers having some prime factor p == 3 (mod 4) to an odd power. sigma(n) == 0 (mod 4) because of this prime factor. Every k == 3 (mod 4) is a term. First differences are always 1, 2, 3 or 4, each occurring infinitely often. - David W. Wilson, Mar 09 2005
Complement of A000415 in the nonsquare positive integers A000037. - Max Alekseyev, Jan 21 2010
Integers with an equal number of 4k+1 and 4k+3 divisors. - Ant King, Oct 05 2010
A000161(a(n)) = 0; A070176(a(n)) > 0; A046712 is a subsequence. - Reinhard Zumkeller, Feb 04 2012, Aug 16 2011
There are arbitrarily long runs of consecutive terms. Record runs start at 3, 6, 21, 75, ... (A260157). - Ivan Neretin, Nov 09 2015
From Klaus Purath, Sep 04 2023: (Start)
There are no squares in this sequence.
There are also no numbers of the form n^2 + 1 (A002522) or n^2 + 4 (A087475).
Every term a(n) raised to an odd power belongs to the sequence just as every product of an odd number of terms. This is also true for all integer sequences represented by the indefinite binary quadratic forms a(n)*x^2 - y^2. These sequences also do not contain squares. (End)

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.

Links

Crossrefs

Complement of A001481; subsequence of A111909.
Cf. A018825, A025284, A000404, A007692.

Programs

  • Haskell
    import Data.List (elemIndices)
a022544 n = a022544_list !! (n-1)
a022544_list = elemIndices 0 a000161_list
-- Reinhard Zumkeller, Aug 16 2011
  • Magma
    [n: n in [0..160] | NormEquation(1, n) eq false]; // Vincenzo Librandi, Jan 15 2017
  • Mathematica
    Select[Range[199], Length[PowersRepresentations[ #, 2, 2]] == 0 &] (* Ant King, Oct 05 2010 *)
Select[Range[200],SquaresR[2,#]==0&] (* Harvey P. Dale, Apr 21 2012 *)
  • PARI
    for(n=0,200,if(sum(i=0,n,sum(j=0,i,if(i^2+j^2-n,0,1)))==0,print1((n),",")))
  • PARI
    is(n)=if(n%4==3, return(1)); my(f=factor(n)); for(i=1,#f~, if(f[i,1]%4==3 && f[i,2]%2, return(1))); 0 \\ Charles R Greathouse IV, Sep 01 2015
  • Python
    def aupto(lim):
  squares = [k*k for k in range(int(lim**.5)+2) if k*k <= lim]
  sum2sqs = set(a+b for i, a in enumerate(squares) for b in squares[i:])
  return sorted(set(range(lim+1)) - sum2sqs)
print(aupto(199)) # Michael S. Branicky, Mar 06 2021
  • Python
    from itertools import count, islice
from sympy import factorint
def A022544_gen(): # generator of terms
    return filter(lambda n:any(p & 3 == 3 and e & 1 for p, e in factorint(n).items()),count(0))
A022544_list = list(islice(A022544_gen(),30)) # Chai Wah Wu, Jun 28 2022

Formula

Limit_{n->oo} a(n)/n = 1.

Extensions

More terms from Benoit Cloitre, May 19 2002

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Links

Crossrefs

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

Programs

  • Haskell
    a042963 n = a042963_list !! (n-1)
a042963_list = [x | x <- [0..], mod x 4 `elem` [1,2]]
-- Reinhard Zumkeller, Feb 14 2012
  • Magma
    [ n : n in [1..165] | n mod 4 eq 1 or n mod 4 eq 2 ]; // Vincenzo Librandi, Jan 25 2011
  • Maple
    A046923:=n->(n mod 2) + 2n - 2; seq(A046923(n), n=1..100); # Wesley Ivan Hurt, Oct 10 2013
  • Mathematica
    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 *)
  • PARI
    a(n)=2*n-1-(n-1)%2 \\ Jianing Song, Oct 06 2018; adapted to offset by Michel Marcus, Sep 09 2022
  • PARI
    apply( A042963(n)=n*2-2+n%2, [1..99]) \\ M. F. Hasler, Oct 17 2022

Formula

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

Extensions

Offset corrected by Reinhard Zumkeller, Feb 14 2012
More terms by David Lovler, Jul 19 2022
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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