A319281 -id:A319281 - cp's OEIS frontend

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

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

A365103 Number of distinct quartic residues x^4 (mod 4^n), x=0..4^n-1.

Original entry on oeis.org

1, 2, 2, 6, 18, 70, 274, 1094, 4370, 17478, 69906, 279622, 1118482, 4473926, 17895698, 71582790, 286331154, 1145324614, 4581298450, 18325193798, 73300775186, 293203100742, 1172812402962, 4691249611846, 18764998447378

Offset: 0

Albert Mukovskiy, Aug 24 2023

nonn

a(n) = A364811(2n).
For n>=2, A319281(a(n)) == 4^n + [n mod 2 == 1].
For n>=2, a(n)=k: [ A319281(k) == 4^n + [n mod 2 == 1] ].

Index entries for linear recurrences with constant coefficients, signature (4, 1, -4).

Cf. A023105, A046631, A195637, A365104, A364811, A319281.

a[n_] = Ceiling[4^n/15] + Boole[Mod[n,2]==1]; Array[a, 24]

PARI
```
a(n) = ceil(4^n/15)+(Mod(n,2)==1);
```

def A365103(n): return len({pow(x,4,1<<(n<<1)) for x in range(1<<(n<<1))}) # Chai Wah Wu, Sep 18 2023

a(n) = ceiling(4^n/15) + (n mod 2).

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)).

A364811 Number of distinct residues x^4 (mod 2^n), x=0..2^n-1.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 6, 10, 18, 36, 70, 138, 274, 548, 1094, 2186, 4370, 8740, 17478, 34954, 69906, 139812, 279622, 559242, 1118482, 2236964, 4473926

Offset: 0

Albert Mukovskiy, Sep 14 2023

For n>=4, A319281(a(n)) == 2^n + [(n mod 4)>0].

It appears that for n>4: a(n)=2*a(n-1)-2*[(n mod 4)==2]; a(n) = ceiling(2^n/15) - [(n mod 4)==0] + 1.

Cf. A319281, A023105, A046630, A365103.

a[n_]:=CountDistinct[Table[PowerMod[x-1, 4, 2^(n-1)], {x, 1, 2^(n-1)}]]; Array[a, 24]

a(n) = #Set(vector(2^(n-1), x, Mod(x-1, 2^(n-1))^4))

def A364811(n): return len({pow(x,4,1<Chai Wah Wu, Sep 17 2023

cp's OEIS Frontend

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

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A365103 Number of distinct quartic residues x^4 (mod 4^n), x=0..4^n-1.

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

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A364811 Number of distinct residues x^4 (mod 2^n), x=0..2^n-1.

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Author

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Programs

Mathematica

PARI

Python

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