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 1-3 of 3 results.

A004215 Numbers that are the sum of 4 but no fewer nonzero squares.

Original entry on oeis.org

7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71, 79, 87, 92, 95, 103, 111, 112, 119, 124, 127, 135, 143, 151, 156, 159, 167, 175, 183, 188, 191, 199, 207, 215, 220, 223, 231, 239, 240, 247, 252, 255, 263, 271, 279, 284, 287, 295, 303, 311, 316, 319, 327, 335, 343
Offset: 1

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Author

Keywords

Comments

Lagrange's theorem tells us that each positive integer can be written as a sum of four squares.
If n is in the sequence and k is an odd positive integer then n^k is in the sequence because n^k is of the form 4^i(8j+7). - Farideh Firoozbakht, Nov 23 2006
Numbers whose cubes do not have a partition as a sum of 3 squares. a(n)^3 = A134738(n). - Artur Jasinski, Nov 07 2007
A002828(a(n)) = 4; A025427(a(n)) > 0. - Reinhard Zumkeller, Feb 26 2015
There are infinitely many adjacent pairs (for example, 128n + 111 and 128n + 112 for any n), but never a triple of consecutive integers. - Ivan Neretin, Aug 17 2017
These numbers are called "forbidden numbers" in crystallography: for a cubic crystal, no reflection with index hkl such that h^2 + k^2 + l^2 = a(n) appears in the crystal's diffraction pattern. - A. Timothy Royappa, Aug 11 2021

Examples

			15 is in the sequence because it is the sum of four squares, namely, 3^2 + 2^2 + 1^2 + 1^2, and it can't be expressed as the sum of fewer squares.
16 is not in the sequence, because, although it can be expressed as 2^2 + 2^2 + 2^2 + 2^2, it can also be expressed as 4^2.
		

References

  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 261.
  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 12.
  • E. Poznanski, 1901. Pierwiastki pierwotne liczb pierwszych. Warszawa, pp. 1-63.
  • W. Sierpiński, 1925. Teorja Liczb. pp. 1-410 (p. 125).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, entry 4181.

Crossrefs

Complement of A000378.
Cf. A000118 (ways to write n as sum of 4 squares), A025427.

Programs

  • Haskell
    a004215 n = a004215_list !! (n-1)
    a004215_list = filter ((== 4) . a002828) [1..]
    -- Reinhard Zumkeller, Feb 26 2015
    
  • Maple
    N:= 1000: # to get all terms <= N
    {seq(seq(4^i * (8*j + 7), j = 0 .. floor((N/4^i - 7)/8)), i = 0 .. floor(log[4](N)))}; # Robert Israel, Sep 02 2014
  • Mathematica
    Sort[Flatten[Table[4^i(8j + 7), {i, 0, 2}, {j, 0, 42}]]] (* Alonso del Arte, Jul 05 2005 *)
    Select[Range[120], Mod[ #/4^IntegerExponent[ #, 4], 8] == 7 &] (* Ant King, Oct 14 2010 *)
  • PARI
    isA004215(n)={ local(fouri,j) ; fouri=1 ; while( n >=7*fouri, if( n % fouri ==0, j= n/fouri -7 ; if( j % 8 ==0, return(1) ) ; ) ; fouri *= 4 ; ) ; return(0) ; } { for(n=1,400, if(isA004215(n), print1(n,",") ; ) ; ) ; } \\ R. J. Mathar, Nov 22 2006
    
  • PARI
    isA004215(n)= n\4^valuation(n,4)%8==7 \\ M. F. Hasler, Mar 18 2011
    
  • Python
    def valuation(n, b):
        v = 0
        while n > 1 and n%b == 0: n //= b; v += 1
        return v
    def ok(n): return n//4**valuation(n, 4)%8 == 7 # after M. F. Hasler
    print(list(filter(ok, range(344)))) # Michael S. Branicky, Jul 15 2021
    
  • Python
    from itertools import count, islice
    def A004215_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:not (m:=(~n&n-1).bit_length())&1 and (n>>m)&7==7,count(max(startvalue,1)))
    A004215_list = list(islice(A004215_gen(),30)) # Chai Wah Wu, Jul 09 2022
    
  • Python
    def A004215(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(((x>>(i<<1))-7>>3)+1 for i in range(x.bit_length()>>1))
        return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

Formula

a(n) = A055039(n)/2. - Ray Chandler, Jan 30 2009
Numbers of the form 4^i*(8*j+7), i >= 0, j >= 0. [A.-M. Legendre & C. F. Gauss]
Products of the form A000302(i)*A004771(j), i, j >= 0. - R. J. Mathar, Nov 29 2006
a(n) = 6*n + O(log(n)). - Charles R Greathouse IV, Dec 19 2013
Conjecture: The number of terms < 2^n is A023105(n) - 2. - Tilman Neumann, Sep 20 2020

Extensions

More terms from Arlin Anderson (starship1(AT)gmail.com)
Additional comments from Jud McCranie, Mar 19 2000

A134738 Cubes which are not the sum of three squares.

Original entry on oeis.org

343, 3375, 12167, 21952, 29791, 59319, 103823, 166375, 216000, 250047, 357911, 493039, 658503, 778688, 857375, 1092727, 1367631, 1404928, 1685159, 1906624, 2048383, 2460375, 2924207, 3442951, 3796416, 4019679, 4657463, 5359375
Offset: 1

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Author

Artur Jasinski, Nov 07 2007

Keywords

Comments

This sequence was inspired by e-mail from Ray Chandler, Nov 07 2007

Crossrefs

Programs

  • Maple
    N:= 10^10: # to get all terms <= N
    sort([seq(seq(4^(3*i) * (8*j + 7)^3, j = 0 .. floor((N^(1/3)/4^i - 7)/8)), i = 0 .. floor(log[4](N^(1/3))))]); # Robert Israel, Dec 26 2017
  • Mathematica
    b = Table[x^3, {x, 1, 300}]; a = {}; Do[Do[Do[AppendTo[a, (x^2 + y^2 + z^2)^3], {x, 0, 30}], {y, 0, 30}], {z, 0, 30}]; Union[a]; Complement[b, a] (*Artur Jasinski*)
    Select[Range[200]^3,SquaresR[3,#]==0&] (* Harvey P. Dale, Feb 03 2015 *)
  • Python
    def A134738(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(((x>>(i<<1))-7>>3)+1 for i in range(x.bit_length()>>1))
        return bisection(f,n,n)**3 # Chai Wah Wu, Feb 14 2025

Formula

a(n) = A004215(n)^3. - Ray Chandler, Jan 29 2009

Extensions

Extended by Ray Chandler, Jan 29 2009

A267189 (Cubes of positive numbers) that are not the sum of three nonzero squares.

Original entry on oeis.org

1, 8, 64, 343, 512, 3375, 4096, 12167, 21952, 29791, 32768, 59319, 103823, 166375, 216000, 250047, 262144, 357911, 493039, 658503, 778688, 857375, 1092727, 1367631, 1404928, 1685159, 1906624, 2048383, 2097152, 2460375, 2924207, 3442951, 3796416, 4019679, 4657463, 5359375, 6128487
Offset: 1

Views

Author

N. J. A. Sloane, Jan 18 2016

Keywords

Comments

This is the intersection of A004214 and A000578.
The original definition of A134739 was ambiguous - this is the second way it could have been interpreted. The other way, now the official definition of A134739, being "Cubes of (positive numbers that are not the sum of three nonzero squares)".

Crossrefs

Programs

  • PARI
    is(n) = { my(a, b) ; a=1; while(a^2+1Altug Alkan, Jan 18 2016

Extensions

More terms from Altug Alkan, Jan 18 2016
Showing 1-3 of 3 results.