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.

A055047 Numbers of the form 9^i*(3*j+1).

Original entry on oeis.org

1, 4, 7, 9, 10, 13, 16, 19, 22, 25, 28, 31, 34, 36, 37, 40, 43, 46, 49, 52, 55, 58, 61, 63, 64, 67, 70, 73, 76, 79, 81, 82, 85, 88, 90, 91, 94, 97, 100, 103, 106, 109, 112, 115, 117, 118, 121, 124, 127, 130, 133, 136, 139, 142, 144, 145, 148, 151, 154, 157, 160, 163
Offset: 1

Views

Author

N. J. A. Sloane, Jun 01 2000

Keywords

Comments

The numbers not of the form 2x^2+3y^2+3z^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 3 raised to an odd power. - V. Raman, Dec 18 2013
Numbers whose squarefree part is congruent to 1 modulo 3. - Peter Munn, May 17 2020

Crossrefs

Intersection of A007417 and A189715.
Complement of A055048 with respect to A007417.
Complement of A055040 with respect to A189715.

Programs

  • Mathematica
    A055047Q[k_] := Mod[k/9^IntegerExponent[k, 9], 3] == 1;
    Select[Range[300], A055047Q] (* Paolo Xausa, Mar 20 2025 *)
  • PARI
    is(n)=n/=9^valuation(n,9); n%3==1 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013
    
  • Python
    from sympy import integer_log
    def A055047(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//9**i-1)//3+1 for i in range(integer_log(x,9)[0]+1))
        return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

Formula

a(n) = 8n/3 + O(log n). - Charles R Greathouse IV, Dec 19 2013
a(n) = A055041(n)/3. - Peter Munn, May 17 2020

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

Views

Author

N. J. A. Sloane, Jun 01 2000

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    A055046Q[k_] := Mod[k/4^IntegerExponent[k, 4], 8] == 3;
    Select[Range[500], A055046Q] (* Paolo Xausa, Mar 20 2025 *)
  • PARI
    is(n)=n/=4^valuation(n,4); n%8==3 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013
    
  • Python
    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
    
  • Python
    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

Formula

a(n) = 6n + O(log n). - Charles R Greathouse IV, Dec 19 2013
a(n) = A055043(n)/2. - Chai Wah Wu, Mar 19 2025

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

Views

Author

V. Raman, Dec 18 2013

Keywords

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)

Crossrefs

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
Showing 1-3 of 3 results.