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

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

Author

Jianing Song, Sep 16 2018

Keywords

• nonn

Comments

{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.

Links

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

Crossrefs

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).

Programs

PARI

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

Python

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

Formula

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