A234000 -id:A234000 - cp's OEIS frontend

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

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

N. J. A. Sloane and J. H. Conway

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

			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.

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.

T. D. Noe, Table of n, a(n) for n = 1..10000
David S. Bettes, Letter to N. J. A. Sloane, Nov 05 1976
Richard T. Bumby, Sums Of Four Squares
International Union of Crystallography, Cubic structures.
Shuo Li, The characteristic sequence of the integers that are the sum of two squares is not morphic, arXiv:2404.08822 [math.NT], 2024.
Louis J. Mordell, A new Waring's problem with squares of linear forms, Quart. J. Math., 1 (1930), 276-288 (see p. 283).
Saburô Uchiyama, A five-square theorem, Publ. Res. Math. Sci., Vol 13, Number 1 (1977), 301-305.
Steve Waterman, Missing numbers formula
Eric Weisstein's World of Mathematics, Square Number
Wikipedia, Lagrange's four-square theorem.
Index entries for sequences related to sums of squares

Complement of A000378.
Cf. A002828, A055039, A072401, A125084, A134738, A134739, A055045, A055046, A234000.
Cf. A000118 (ways to write n as sum of 4 squares), A025427.

a004215 n = a004215_list !! (n-1)
a004215_list = filter ((== 4) . a002828) [1..]
-- Reinhard Zumkeller, Feb 26 2015

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

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 *)

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

isA004215(n)= n\4^valuation(n,4)%8==7 \\ M. F. Hasler, Mar 18 2011

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

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

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

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

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

Additional comments from Jud McCranie, Mar 19 2000

A023105 Number of distinct quadratic residues mod 2^n.

Original entry on oeis.org

1, 2, 2, 3, 4, 7, 12, 23, 44, 87, 172, 343, 684, 1367, 2732, 5463, 10924, 21847, 43692, 87383, 174764, 349527, 699052, 1398103, 2796204, 5592407, 11184812, 22369623, 44739244, 89478487, 178956972, 357913943, 715827884, 1431655767, 2863311532

Offset: 0

David W. Wilson

Number of distinct n-digit suffixes of base 2 squares.
a(n) counts the elements of A234000 smaller than 2^n plus the zero: a(7)=23 counts the elements of {0, 1, 4, 9, ..., 113, 121}, for example. - R. J. Mathar, Oct 11 2014
Conjecture: a(n) = 2 + (the number of A004215 entries < 2^n), for n>0. - Tilman Neumann, Sep 20 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
John Greene and James A. Sellers, Extending recent parity results of Nyirenda and Mugwangwavari for partitions with initial repetitions, Integers (2025), Vol. 25, Art. No. A32. See p. 5.
Lee Hae-hwang, Sequences of Growing Networks
W. D. Stangl, Counting Squares in Z_n, Mathematics Magazine, pp. 285-289, Vol. 69 No. 4 (October 1996).
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A000224, A001045, A004315, A005578, A234000.

a 0 = 1
a 1 = 2
a n | even n = 2*a(n-1)-2
a n | odd  n = 2*a(n-1)-1
-- James Spahlinger, Oct 07 2012

[Floor((2^n+10)/6): n in [0..30]]; // Vincenzo Librandi, Apr 21 2012

CoefficientList[Series[(1-3*x^2-x^3)/((1-x)*(1+x)*(1-2*x)),{x,0,35}],x] (* Vincenzo Librandi, Apr 21 2012 *)
LinearRecurrence[{2,1,-2},{1,2,2,3},40] (* Harvey P. Dale, Mar 05 2016 *)

a(n)=(2^n+10)\6 \\ Charles R Greathouse IV, Apr 21 2012

def A023105(n): return ((1<Chai Wah Wu, Aug 22 2023

[(2^n +9 -(-1)^n -3*bool(n==0))/6 for n in (0..30)] # G. C. Greubel, Aug 10 2022

a(n) = floor( (2^n+10)/6 ).
a(n) = (2^n + 9 - (-1)^n)/6 for n > 0. - David S. Dodson, Jan 06 2013
G.f.: (1-3*x^2-x^3)/((1-x)*(1+x)*(1-2*x)). - Colin Barker, Mar 08 2012
a(0)=1, a(1)=2. a(n) = 2*a(n-1)-2 if n is even, a(n) = 2*a(n-1)-1 if n is odd. - Vincenzo Librandi, Apr 21 2012
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 0. - Joerg Arndt, Apr 21 2012
a(0)=1, a(1)=2, a(n+2) = a(n+1) + A001045(n) for n >= 1. - Lee Hae-hwang, Jun 16 2014
a(n) = A000224(2^n). - R. J. Mathar, Oct 10 2014
a(n) = A005578(n-1) + 1, n > 0. - Carl Joshua Quines, Jul 17 2019
E.g.f.: (exp(2*x) + 9*exp(x) - 3 - exp(-x))/6. - G. C. Greubel, Aug 10 2022

A191257 a(n) = A067368(n)/2.

Original entry on oeis.org

1, 3, 5, 7, 8, 9, 11, 13, 15, 17, 19, 21, 23, 24, 25, 27, 29, 31, 33, 35, 37, 39, 40, 41, 43, 45, 47, 49, 51, 53, 55, 56, 57, 59, 61, 63, 64, 65, 67, 69, 71, 72, 73, 75, 77, 79, 81, 83, 85, 87, 88, 89, 91, 93, 95, 97, 99, 101, 103, 104, 105, 107, 109, 111, 113, 115, 117, 119, 120, 121, 123, 125, 127, 129, 131, 133, 135, 136, 137, 139, 141, 143

Offset: 1

Clark Kimberling, May 28 2011

nonn

From Jianing Song, Sep 21 2018: (Start)
Numbers n such that A191255(n) = 0 or 3. Previous definition was numbers n such that A191255(2*n) = 1, that is, numbers of the form 2^(3t)*s where s is an odd number.
{+-a(n)} gives all nonzero cubes modulo all powers of 2, that is, nonzero cubes over the 2-adic integers. So this sequence is closed under multiplication. (End)
The old entry had the conjecture that a(n) = A067368(n)/2. Jianing Song, Sep 21 2018 showed that this is true, and gave us the simpler definition that we have now used. The conjecture is correct because {a(n)} lists the numbers of the form 2^(3t)*s, and {A067368(n)} lists the numbers of the form 2^(3t+1)*s, where s is an odd number. Note also that a(n) = A213258(n)/4.
The asymptotic density of this sequence is 4/7. - Amiram Eldar, May 31 2024

Recto Rex M. Calingasan and Alexander Vincent B. Policarpio, On the zeros of the OEIS A191257 zeta function, AIP Conference Proceedings 1905, 030011 (2017).

Cf. A067368, A191255, A213258.
Perfect powers over the 2-adic integers:
Squares: positive: A234000; negative: A004215 (negated);
Cubes: this sequence;
Fourth powers: positive: A319281; negative: A319282 (negated).

t = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 2}, 2 -> {0, 3},
      3 -> {0, 1}}] &, {0}, 9] (* A191255 *)
Flatten[Position[t, 0]] (* A005408, the odds *)
a = Flatten[Position[t, 1]] (* A067368 *)
b = Flatten[Position[t, 2]] (* A213258 *)
a/2  (* A191257 *)
b/4  (* a/2 *)

isok(n) = valuation(2*n, 2)%3==1; \\ Altug Alkan, Sep 21 2018

def A191257(n):
    def f(x): return n+x-sum(((x>>i)-1>>1)+1 for i in range(0,x.bit_length(),3))
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, Feb 17 2025

Name corrected by Altug Alkan, Apr 03 2018

New name from Jianing Song, Sep 21 2018

A055046 Numbers of the form 4^i(8j+3).

Original entry on oeis.org

3, 11, 12, 19, 27, 35, 43, 44, 48, 51, 59, 67, 75, 76, 83, 91, 99, 107, 108, 115, 123, 131, 139, 140, 147, 155, 163, 171, 172, 176, 179, 187, 192, 195, 203, 204, 211, 219, 227, 235, 236, 243, 251, 259, 267, 268, 275, 283, 291, 299, 300, 304, 307, 315, 323, 331, 332

Offset: 1

N. J. A. Sloane, Jun 01 2000

Numbers not of the form x^2+y^2+5z^2.

Also values of n such that numbers of the form x^2+n*y^2 for some integers x, y cannot have prime factor of 2 raised to an odd power. - V. Raman, Dec 18 2013

Paolo Xausa, Table of n, a(n) for n = 1..10000
L. J. Mordell, A new Waring's problem with squares of linear forms, Quart. J. Math., 1 (1930), 276-288 (see p. 283).

Cf. A004215, A055043, A055045, A055047, A233998, A233999, A234000.

A055046Q[k_] := Mod[k/4^IntegerExponent[k, 4], 8] == 3;
Select[Range[500], A055046Q] (* Paolo Xausa, Mar 20 2025 *)

is(n)=n/=4^valuation(n,4); n%8==3 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013

from itertools import count, islice
def A055046_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:not (m:=(~n&n-1).bit_length())&1 and (n>>m)&7==3,count(max(startvalue,1)))
A055046_list = list(islice(A055046_gen(),30)) # Chai Wah Wu, Jul 09 2022

def A055046(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))-3>>3)+1 for i in range(x.bit_length()>>1))
    return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

a(n) = 6n + O(log n). - Charles R Greathouse IV, Dec 19 2013

a(n) = A055043(n)/2. - Chai Wah Wu, Mar 19 2025

A055045 Numbers of the form 4^i(8j+5).

Original entry on oeis.org

5, 13, 20, 21, 29, 37, 45, 52, 53, 61, 69, 77, 80, 84, 85, 93, 101, 109, 116, 117, 125, 133, 141, 148, 149, 157, 165, 173, 180, 181, 189, 197, 205, 208, 212, 213, 221, 229, 237, 244, 245, 253, 261, 269, 276, 277, 285, 293, 301, 308, 309, 317, 320, 325, 333, 336, 340, 341

Offset: 1

N. J. A. Sloane, Jun 01 2000

Numbers not of the form x^2+2y^2+6z^2.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
L. E. Dickson, Integers represented by positive ternary quadratic forms, Bull. Amer. Math. Soc. 33 (1927), 63-70. See Theorem XI.
L. J. Mordell, A new Waring's problem with squares of linear forms, Quart. J. Math., 1 (1930), 276-288 (see p. 283).

Cf. A004215, A055042, A055046, A234000.

a055045 n = a055045_list !! (n-1)
a055045_list = filter ((== 5) . (flip mod 8) . f) [1..] where
   f x = if r == 0 then f x' else x  where (x', r) = divMod x 4
-- Reinhard Zumkeller, Jan 02 2014

A055045Q[k_] := Mod[k/4^IntegerExponent[k, 4], 8] == 5;
Select[Range[500], A055045Q] (* Paolo Xausa, Mar 20 2025 *)

is(n)=n/=4^valuation(n,4); n%8==5 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013

from itertools import count, islice
def A055045_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:not (m:=(~n&n-1).bit_length())&1 and (n>>m)&7==5, count(max(startvalue, 1)))
A055045_list = list(islice(A055045_gen(), 30)) # Chai Wah Wu, Jul 09 2022

def A055045(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))-5>>3)+1 for i in range(x.bit_length()>>1))
    return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

a(n) = 6n + O(log n). - Charles R Greathouse IV, Dec 19 2013

a(n) = A055042(n)/2. - Chai Wah Wu, Mar 19 2025

A319281 Numbers of the form 16^i(16j + 1).

Original entry on oeis.org

1, 16, 17, 33, 49, 65, 81, 97, 113, 129, 145, 161, 177, 193, 209, 225, 241, 256, 257, 272, 273, 289, 305, 321, 337, 353, 369, 385, 401, 417, 433, 449, 465, 481, 497, 513, 528, 529, 545, 561, 577, 593, 609, 625, 641, 657, 673, 689, 705, 721, 737, 753, 769, 784

Offset: 1

Jianing Song, Sep 16 2018

nonn

{a(n)} gives all positive fourth powers modulo all powers of 2, that is, positive fourth powers over 2-adic integers. So this sequence is closed under multiplication.

Jianing Song, Table of n, a(n) for n = 1..10002 (all terms <= 150000)

A158057 is a proper subsequence.
Perfect powers over 2-adic integers:
Squares: positive: A234000; negative: A004215 (negated);
Cubes: A191257;
Fourth powers: positive: this sequence; negative: A319282 (negated).

isA319281(n)= n\16^valuation(n, 16)%16==1

def A319281(n):
    if n<3: return 15*n-14
    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-1+x-sum((((x>>(i<<2))-1)>>4)+1 for i in range(x.bit_length()>>2))
    return bisection(f,n,n) # Chai Wah Wu, Feb 17 2025

a(n) = 15*n + O(log(n)).

A334832 Numbers whose squarefree part is congruent to 1 (mod 24).

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 73, 81, 97, 100, 121, 144, 145, 169, 193, 196, 217, 225, 241, 256, 265, 289, 292, 313, 324, 337, 361, 385, 388, 400, 409, 433, 441, 457, 481, 484, 505, 529, 553, 576, 577, 580, 601, 625, 649, 657, 673, 676, 697, 721, 729, 745, 769, 772, 784, 793, 817, 841

Offset: 1

Peter Munn, Jun 15 2020

Closed under multiplication.
The sequence forms a subgroup of the positive integers under the commutative operation A059897(.,.). A059897 has a relevance to squarefree parts that arises from the identity A007913(k*m) = A059897(A007913(k), A007913(m)), where A007913(n) is the squarefree part of n.
The subgroup is one of 8 A059897(.,.) subgroups of the form {k : A007913(k) == 1 (mod m)}. The list seems complete, in anticipation of proof that such sets form subgroups only when m is a divisor of 24 (based on the property described by A. G. Astudillo in A018253). This sequence might be viewed as primitive with respect to the other 7, as the latter correspond to subgroups of its quotient group, in the sense that each one (as a set) is also a union of cosets described below. The 7 include A003159 (m=2), A055047 (m=3), A277549 (m=4), A234000 (m=8) and the trivial A000027 (m=1).
The subgroup has 32 cosets. For each i in {1, 5, 7, 11, 13, 17, 19, 23}, j in {1, 2, 3, 6} there is a coset {n : n = k^2 * (24m + i) * j, k >= 1, m >= 0}. The divisors of 2730 = 2*3*5*7*13 form a transversal. (11, clearly not such a divisor, is in the same coset as 35 = 11 + 24; 17, 19, 23 are in the same cosets as 65, 91, 455 respectively.)
The asymptotic density of this sequence is 1/16. - Amiram Eldar, Mar 08 2021

			The squarefree part of 26 is 26, which is congruent to 2 (mod 24), so 26 is not in the sequence.
The squarefree part of 292 = 2^2 * 73 is 73, which is congruent to 1 (mod 24), so 292 is in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Quotient Group.
Eric Weisstein's World of Mathematics, Right Transversal.

A subgroup under A059897, defined using A007913.
Subsequences: A000290\{0}, A103214, A107008.
Equivalent sequences modulo other members of A018253: A000027 (1), A003159 (2), A055047 (3), A277549 (4), A352272(6), A234000 (8).

Select[Range[850], Mod[Sqrt[#] /. (c_ : 1)*a_^(b_ : 0) :> (c*a^b)^2, 24] == 1 &] (* Michael De Vlieger, Jun 24 2020 *)

isok(m) = core(m) % 24 == 1; \\ Peter Munn, Jun 21 2020

from sympy import integer_log
def A334832(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<<(j<<1))-1)//24+1 for i in (9**k for k in range(integer_log(x,9)[0]+1)) for j in range((x//i>>1).bit_length()+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 21 2025

{a(n)} = {n : n = k^2 * (24m + 1), k >= 1, m >= 0}.

A233999 Values of n such that numbers of the form x^2+n*y^2 for some integers x, y cannot have prime factor of 7 raised to an odd power.

Original entry on oeis.org

1, 2, 4, 8, 9, 11, 15, 16, 18, 22, 23, 25, 29, 30, 32, 36, 37, 39, 43, 44, 46, 49, 50, 51, 53, 57, 58, 60, 64, 65, 67, 71, 72, 74, 78, 79, 81, 85, 86, 88, 92, 93, 95, 98, 99, 100, 102, 106, 107, 109, 113, 114, 116, 120, 121, 123, 127, 128, 130, 134, 135, 137, 141, 142, 144, 148, 149

Offset: 1

V. Raman, Dec 18 2013

Equivalently, numbers of the form 49^n*(7m+1), 49^n*(7m+2), or 49^n*(7m+4). [Corrected by Charles R Greathouse IV, Jan 12 2017]
From Peter Munn, Feb 08 2024: (Start)
Numbers whose squarefree part is congruent to a (nonzero) quadratic residue modulo 7.
The integers in a subgroup of the positive rationals under multiplication.  As such the sequence is closed under multiplication and - where the result is an integer - under division. The subgroup has index 4 and is generated by the primes congruent to a quadratic residue (1, 2 or 4) modulo 7, the square of 7, and 3 times the other primes; that is a generator corresponding to each prime: 2, 3*3, 3*5, 7^2, 11, 3*13, 3*17, 3*19, 23, 29, 3*31, ... .
(End)

Eric Weisstein's World of Mathematics, Squarefree Part.

Cf. A055046, A233998.
Numbers whose squarefree part is congruent to a coprime quadratic residue modulo k: A003159 (k=2), A055047 (k=3), A277549 (k=4), A352272 (k=6), A234000 (k=8), A334832 (k=24).
First differs from A047350 by including 49.

is(n)=n/=49^valuation(n, 49); n%7==1||n%7==2||n%7==4 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013

is_A233999(n)=bittest(22,n/49^valuation(n, 49)%7) \\ - M. F. Hasler, Jan 02 2014

list(lim)=my(v=List(),t,u); forstep(k=1,lim\=1,[1,2,4], listput(v,k)); for(e=1,logint(lim,49), u=49^e; for(i=1,#v, t=u*v[i]; if(t>lim, break); listput(v,t))); Set(v) \\ Charles R Greathouse IV, Jan 12 2017

from sympy import integer_log
def A233999(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):
        c = n+x
        for i in range(integer_log(x,49)[0]+1):
            m = x//49**i
            c -= (m-1)//7+(m-2)//7+(m-4)//7+3
        return c
    return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

a(n) = 16n/7 + O(log n). - Charles R Greathouse IV, Jan 12 2017

A319282 Numbers of the form 16^i(16j + 15).

Original entry on oeis.org

15, 31, 47, 63, 79, 95, 111, 127, 143, 159, 175, 191, 207, 223, 239, 240, 255, 271, 287, 303, 319, 335, 351, 367, 383, 399, 415, 431, 447, 463, 479, 495, 496, 511, 527, 543, 559, 575, 591, 607, 623, 639, 655, 671, 687, 703, 719, 735, 751, 752, 767, 783, 799, 815

Offset: 1

Jianing Song, Sep 16 2018

nonn

{-a(n)} gives all negative fourth powers modulo all powers of 2, that is, negative fourth powers over 2-adic integers.

Jianing Song, Table of n, a(n) for n = 1..9999 (all terms <= 150000)

A125169 is a proper subsequence.
Perfect powers over 2-adic integers:
Squares: positive: A234000; negative: A004215 (negated);
Cubes: A191257;
Fourth powers: positive: A319281; negative: this sequence (negated).

isA319282(n)= n\16^valuation(n, 16)%16==15

def A319282(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<<2))-15)>>4)+1 for i in range(x.bit_length()>>2))
    return bisection(f,n,n) # Chai Wah Wu, Feb 17 2025

a(n) = 15*n + O(log(n)).

A055044 Numbers of the form 2^(2i+1)(8j+1).

Original entry on oeis.org

2, 8, 18, 32, 34, 50, 66, 72, 82, 98, 114, 128, 130, 136, 146, 162, 178, 194, 200, 210, 226, 242, 258, 264, 274, 288, 290, 306, 322, 328, 338, 354, 370, 386, 392, 402, 418, 434, 450, 456, 466, 482, 498, 512, 514, 520, 530, 544, 546, 562, 578

Offset: 1

N. J. A. Sloane, Jun 01 2000

The asymptotic density of this sequence is 1/12. - Amiram Eldar, Mar 29 2025

Amiram Eldar, Table of n, a(n) for n = 1..10000
L. J. Mordell, A new Waring's problem with squares of linear forms, Quart. J. Math., 1 (1930), 276-288 (see p. 283).

Cf. A234000.

With[{max = 600}, Flatten[Table[2^(2*i + 1)*(8*j + 1), {i, 0, (Log2[max] - 1)/2}, {j, 0, Floor[(max/2^(2*i + 1) - 1)/8]}]] // Sort] (* Amiram Eldar, Mar 29 2025 *)

def A055044(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)+1)-1>>3)+1 for i in range(x.bit_length()+1>>1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 19 2025

a(n) = 2*A234000(n). - Chai Wah Wu, Mar 19 2025

cp's OEIS Frontend

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

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A023105 Number of distinct quadratic residues mod 2^n.

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A191257 a(n) = A067368(n)/2.

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A055046 Numbers of the form 4^i*(8*j+3).

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A055045 Numbers of the form 4^i*(8*j+5).

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A319281 Numbers of the form 16^i*(16*j + 1).

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A055046 Numbers of the form 4^i(8j+3).

A055045 Numbers of the form 4^i(8j+5).

A319281 Numbers of the form 16^i(16j + 1).

A319282 Numbers of the form 16^i(16j + 15).

A055044 Numbers of the form 2^(2i+1)(8j+1).