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.

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

Author

V. Raman, Dec 18 2013

Keywords

• nonn
• easy

Comments

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)

Links

• Eric Weisstein's World of Mathematics, Squarefree Part.

Crossrefs

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.

Programs

PARI

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

PARI

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

PARI

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

Python

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

Formula

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