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 1-10 of 14 results. Next

A026225 Numbers of the form 3^i * (3k+1).

Original entry on oeis.org

1, 3, 4, 7, 9, 10, 12, 13, 16, 19, 21, 22, 25, 27, 28, 30, 31, 34, 36, 37, 39, 40, 43, 46, 48, 49, 52, 55, 57, 58, 61, 63, 64, 66, 67, 70, 73, 75, 76, 79, 81, 82, 84, 85, 88, 90, 91, 93, 94, 97, 100, 102, 103, 106, 108, 109, 111, 112, 115
Offset: 1

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Author

Keywords

Comments

Old name: a(n) = (1/3)*(s(n+1) - 1), where s = A026224.
Conjectures based on old name: these are numbers of the form (3*i+1)*3^j; see A182828, and they comprise the complement of A026179, except for the initial 1 in A026179.
From Peter Munn, Mar 17 2022: (Start)
Numbers with an even number of prime factors of the form 3k-1 counting repetitions.
Numbers whose squarefree part is congruent to 1 modulo 3 or 3 modulo 9.
The integers in an index 2 subgroup of the positive rationals under multiplication. As such the sequence is closed under multiplication and - where the result is an integer - under division; also for any positive integer k not in the sequence, the sequence's complement is generated by dividing by k the terms that are multiples of k.
Alternatively, the sequence can be viewed as an index 2 subgroup of the positive integers under the commutative binary operation A059897(.,.).
Viewed either way, the sequence corresponds to a subgroup of the quotient group derived in the corresponding way from A055047. (End)
The asymptotic density of this sequence is 1/2. - Amiram Eldar, Apr 03 2022
Is this A026140 shifted right? - R. J. Mathar, Jun 24 2025

Crossrefs

Elements of array A182828 in ascending order.
Union of A055041 and A055047.
Other subsequences: A007645 (primes), A352274.
Symmetric difference of A003159 and A225838; of A007417 and A189716.

Programs

  • Mathematica
    a[b_] := Table[Mod[n/b^IntegerExponent[n, b], b], {n, 1, 160}]
    p[b_, d_] := Flatten[Position[a[b], d]]
    p[3, 1]  (* A026225 *)
    p[3, 2] (* A026179 without initial 1 *)
    (* Clark Kimberling, Oct 19 2016 *)
  • PARI
    isok(m) = core(m) % 3 == 1 || core(m) % 9 == 3; \\ Peter Munn, Mar 17 2022
    
  • Python
    from sympy import integer_log
    def A026225(n):
        def f(x): return n+x-sum(((x//3**i)-1)//3+1 for i in range(integer_log(x,3)[0]+1))
        m, k = n, f(n)
        while m != k: m, k = k, f(k)
        return m # Chai Wah Wu, Feb 15 2025

Formula

From Peter Munn, Mar 17 2022: (Start)
{a(n) : n >= 1} = {m : A001222(A343430(m)) == 0 (mod 2)}.
{a(n) : n >= 1} = {A055047(m) : m >= 1} U {3*A055047(m) : m >= 1}.
{a(n) : n >= 1} = {A352274(m) : m >= 1} U {A352274(m)/10 : m >= 1, 10 divides A352274(m)}. (End)

Extensions

New name from Peter Munn, Mar 17 2022

A189715 Numbers k such that A156595(k-1) = 0; complement of A189716.

Original entry on oeis.org

1, 4, 6, 7, 9, 10, 13, 15, 16, 19, 22, 24, 25, 28, 31, 33, 34, 36, 37, 40, 42, 43, 46, 49, 51, 52, 54, 55, 58, 60, 61, 63, 64, 67, 69, 70, 73, 76, 78, 79, 81, 82, 85, 87, 88, 90, 91, 94, 96, 97, 100, 103, 105, 106, 109, 112, 114, 115, 117, 118, 121, 123, 124, 127, 130, 132, 133, 135, 136, 139, 141, 142, 144, 145, 148, 150, 151, 154, 157, 159
Offset: 1

Views

Author

Clark Kimberling, Apr 26 2011

Keywords

Comments

See A156595.
Numbers whose squarefree part is congruent modulo 9 to 1, 4, 6 or 7. - Peter Munn, May 17 2020
The asymptotic density of this sequence is 1/2. - Amiram Eldar, Mar 08 2021

Crossrefs

Programs

  • Mathematica
    t = Nest[Flatten[# /. {0->{0,1,1}, 1->{0,1,0}}] &, {0}, 5] (*A156595*)
    f[n_] := t[[n]]
    Flatten[Position[t, 0]] (*A189715*)
    Flatten[Position[t, 1]] (*A189716*)
    s[n_] := Sum[f[i], {i, 1, n}]; s[0] = 0;
    Table[s[n], {n, 1, 120}] (*A189717*)
    f[p_, e_] := (p^Mod[e, 2]); sqfpart[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[160], MemberQ[{1, 4, 6, 7}, Mod[sqfpart[#], 9]] &] (* Amiram Eldar, Mar 08 2021 *)
  • Python
    from sympy import integer_log
    def A189715(n):
        def f(x): return n+x-sum(((m:=x//9**i)-1)//9+(m-4)//9+(m-6)//9+(m-7)//9+4 for i in range(integer_log(x,9)[0]+1))
        m, k = n, f(n)
        while m != k: m, k = k, f(k)
        return m # Chai Wah Wu, Feb 15 2025

Extensions

Name enhanced by 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

A055041 Numbers of the form 3^(2i+1)*(3*j+1).

Original entry on oeis.org

3, 12, 21, 27, 30, 39, 48, 57, 66, 75, 84, 93, 102, 108, 111, 120, 129, 138, 147, 156, 165, 174, 183, 189, 192, 201, 210, 219, 228, 237, 243, 246, 255, 264, 270, 273, 282, 291, 300, 309, 318, 327, 336, 345, 351, 354, 363, 372, 381, 390, 399
Offset: 1

Views

Author

N. J. A. Sloane, Jun 01 2000

Keywords

Comments

The numbers not of the form x^2+y^2+6z^2.
Numbers whose squarefree part is congruent to 3 modulo 9. Compare with A329575. - Peter Munn, May 17 2020
The asymptotic density of this sequence is 1/8. - Amiram Eldar, Mar 08 2021

Crossrefs

Intersection of A145204 and A189716.
Complement of A055040 with respect to A145204\{0}.
Complement of A055048 with respect to A189716.

Programs

  • Mathematica
    f[p_, e_] := (p^Mod[e, 2]); sqfpart[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[400], Mod[sqfpart[#], 9] == 3 &] (* Amiram Eldar, Mar 08 2021 *)
  • Python
    from sympy import integer_log
    def A055041(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)*3 # Chai Wah Wu, Feb 14 2025

Formula

a(n) = A055047(n) * 3. - Peter Munn, May 17 2020

A055040 Numbers of the form 3^(2i+1)*(3*j+2).

Original entry on oeis.org

6, 15, 24, 33, 42, 51, 54, 60, 69, 78, 87, 96, 105, 114, 123, 132, 135, 141, 150, 159, 168, 177, 186, 195, 204, 213, 216, 222, 231, 240, 249, 258, 267, 276, 285, 294, 297, 303, 312, 321, 330, 339, 348, 357, 366, 375, 378, 384, 393, 402, 411
Offset: 1

Views

Author

N. J. A. Sloane, Jun 01 2000

Keywords

Comments

Numbers not of the form x^2+y^2+3z^2.
Numbers whose squarefree part is congruent to 6 modulo 9. - Peter Munn, May 17 2020
The asymptotic density of this sequence is 1/8. - Amiram Eldar, Mar 08 2021

Crossrefs

Equals 3*A055048(n).
Intersection of A145204 and A189715.
Complement of A055041 with respect to A145204\{0}.
Complement of A055047 with respect to A189715.
Cf. A007913.

Programs

  • Haskell
    a055040 n = a055040_list !! (n-1)
    a055040_list = map (* 3) a055048_list
    -- Reinhard Zumkeller, Apr 07 2012
    
  • Mathematica
    max = 500; Select[ Union[ Flatten[ Table[3^(2*i + 1)*(3*j + 2), {i, 0, Ceiling[ Log[max/6]/Log[9]]}, {j, 0, Ceiling[(max/9^i - 6)/9]}]]], # <= max &] (* Jean-François Alcover, Oct 13 2011 *)
  • Python
    from sympy import integer_log
    def A055040(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-2)//3+1 for i in range(integer_log(x,9)[0]+1))
        return bisection(f,n,n)*3 # Chai Wah Wu, Feb 14 2025

Formula

G.f.: [x(x+2)(x^2+x+1)(x^7+x^3+1)]/(x^11-x^10-x+1) (conjectured).

A055048 Numbers of the form 9^i*(3*j+2).

Original entry on oeis.org

2, 5, 8, 11, 14, 17, 18, 20, 23, 26, 29, 32, 35, 38, 41, 44, 45, 47, 50, 53, 56, 59, 62, 65, 68, 71, 72, 74, 77, 80, 83, 86, 89, 92, 95, 98, 99, 101, 104, 107, 110, 113, 116, 119, 122, 125, 126, 128, 131, 134, 137, 140, 143, 146, 149, 152, 153, 155
Offset: 1

Views

Author

N. J. A. Sloane, Jun 01 2000

Keywords

Comments

The numbers not of the form x^2+3y^2+3z^2.
Numbers whose squarefree part is congruent to 2 modulo 3. - Peter Munn, May 17 2020
The asymptotic density of this sequence is 3/8. - Amiram Eldar, Mar 08 2021

Crossrefs

Intersection of A007417 and A189716.
Complement of A055047 with respect to A007417.
Complement of A055041 with respect to A189716.

Programs

  • Haskell
    a055048 n = a055048_list !! (n-1)
    a055048_list = filter (s 0) [1..] where
       s t u | m > 0  = even t && m == 2
             | m == 0 = s (t + 1) u' where (u',m) = divMod u 3
    -- Reinhard Zumkeller, Apr 07 2012
    
  • Mathematica
    max = 200; Select[ Union[ Flatten[ Table[ 9^i*(3*j + 2), {i, 0, Ceiling[Log[max]/Log[9]]}, {j, 0, Ceiling[( max/9^i - 2)/3]}]]], # <= max &] (* Jean-François Alcover, Oct 13 2011 *)
  • PARI
    is(n)=n/=9^valuation(n, 9); n%3==2 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013
    
  • Python
    from sympy import integer_log
    def A055048(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-2)//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) = A055040(n)/3. - Peter Munn, May 17 2020

A352272 Numbers whose squarefree part is congruent to 1 modulo 6.

Original entry on oeis.org

1, 4, 7, 9, 13, 16, 19, 25, 28, 31, 36, 37, 43, 49, 52, 55, 61, 63, 64, 67, 73, 76, 79, 81, 85, 91, 97, 100, 103, 109, 112, 115, 117, 121, 124, 127, 133, 139, 144, 145, 148, 151, 157, 163, 169, 171, 172, 175, 181, 187, 193, 196, 199, 205, 208, 211, 217, 220, 223, 225, 229
Offset: 1

Views

Author

Peter Munn, Mar 10 2022

Keywords

Comments

Numbers of the form 4^i * 9^j * (6k+1), i, j, k >= 0.
Closed under multiplication.
The sequence forms a subgroup of the positive integers under the commutative operation A059897(.,.), one of 8 subgroups of the form {k : A007913(k) == 1 (mod m)} - in each case m is a divisor of 24. A059897 has a relevance to squarefree parts that arises from the identity A007913(k*n) = A059897(A007913(k), A007913(n)), where A007913(n) is the squarefree part of n.
The subgroup has 8 cosets, which partition the positive integers as follows. For each i in {1, 5}, j in {1, 2, 3, 6} there is a coset {m^2 * (6k+i) * j : m >= 1, k >= 0}. See the table in the examples.
None of the 8 cosets have been entered into the database previously, but many subgroups of the quotient group (which are formed of certain combinations of cosets) are represented among earlier OEIS sequences, including 6 of the 7 subgroups of index 2 (which combine 4 cosets). This sequence can therefore be defined as the intersection of pairs or triples of these sequences in many combinations (see the cross-references). See also the table in the example section of A352273 (the coset that includes 5).

Examples

			The squarefree part of 9 is 1, which is congruent to 1 (mod 6), so 9 is in the sequence.
The squarefree part of 14 is 14, which is congruent to 2 (mod 6), so 14 is not in the sequence.
The squarefree part of 52 = 2^2 * 13 is 13, which is congruent to 1 (mod 6), so 52 is in the sequence.
The 8 cosets described in the initial comments (forming a partition of the positive integers) are shown as rows of the following table. The first half of the table corresponds to (6k+i) with i=1; the second half to i=5, with row 5 being A352273.
   1,  4,   7,   9,  13,  16,  19,  25,  28,  31,  36, ...
   2,  8,  14,  18,  26,  32,  38,  50,  56,  62,  72, ...
   3, 12,  21,  27,  39,  48,  57,  75,  84,  93, 108, ...
   6, 24,  42,  54,  78,  96, 114, 150, 168, 186, 216, ...
   5, 11,  17,  20,  23,  29,  35,  41,  44,  45,  47, ...
  10, 22,  34,  40,  46,  58,  70,  82,  88,  90,  94, ...
  15, 33,  51,  60,  69,  87, 105, 123, 132, 135, 141, ...
  30, 66, 102, 120, 138, 174, 210, 246, 264, 270, 282, ...
The product of two positive integers is in this sequence if and only if they are in the same coset. The asymptotic density of cosets (containing) 1 and 5 is 1/4; of cosets 2 and 10 is 1/8; of cosets 3 and 15 is 1/12; of cosets 6 and 30 is 1/24.
		

Crossrefs

Intersection of any 2 of A055047, A339690 and A352274.
Intersection of any 4 sets chosen from A003159, A007417, A026225, A036668, A189715 and A225837 (in most cases, only 3 sets are needed - specifically if the pairwise intersections of the 3 sets differ from each other).
Closure of A084089 under multiplication by 9.
Other subsequences: A000290\{0}, A016921, A229848 (apparently, with 55 the first difference).
A334832 lists equivalent sequences modulo other divisors of 24.

Programs

  • PARI
    isok(m) = core(m) % 6 == 1;
    
  • Python
    from itertools import count
    def A352272(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 count(0):
                i2 = 9**i
                if i2>x:
                    break
                for j in count(0,2):
                    k = i2<x:
                        break
                    c -= (x//k-1)//6+1
            return c
        return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

Formula

{a(n) : n >= 1} = {m >= 1 : A007913(m) == 1 (mod 6)}.
{a(n) : n >= 1} = A334832 U A334832/7 U A334832/13 U A334832/19 where A334832/k denotes {A334832(m)/k : m >= 1, k divides A334832(m)}.
Using the same denotation, {a(n) : n >= 1} = A352273/5 = {A307151(A352273(m)) : m >= 1}.

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

Views

Author

Peter Munn, Jun 15 2020

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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).

Programs

  • Mathematica
    Select[Range[850], Mod[Sqrt[#] /. (c_ : 1)*a_^(b_ : 0) :> (c*a^b)^2, 24] == 1 &] (* Michael De Vlieger, Jun 24 2020 *)
  • PARI
    isok(m) = core(m) % 24 == 1; \\ Peter Munn, Jun 21 2020
    
  • Python
    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

Formula

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

A352273 Numbers whose squarefree part is congruent to 5 modulo 6.

Original entry on oeis.org

5, 11, 17, 20, 23, 29, 35, 41, 44, 45, 47, 53, 59, 65, 68, 71, 77, 80, 83, 89, 92, 95, 99, 101, 107, 113, 116, 119, 125, 131, 137, 140, 143, 149, 153, 155, 161, 164, 167, 173, 176, 179, 180, 185, 188, 191, 197, 203, 207, 209, 212, 215, 221, 227, 233, 236, 239, 245, 251
Offset: 1

Views

Author

Peter Munn, Mar 10 2022

Keywords

Comments

Numbers of the form 4^i * 9^j * (6k+5), i, j, k >= 0.
1/5 of each multiple of 5 in A352272.
The product of any two terms is in A352272.
The product of a term of this sequence and a term of A352272 is a term of this sequence.
The positive integers are usefully partitioned as {A352272, 2*A352272, 3*A352272, 6*A352272, {a(n)}, 2*{a(n)}, 3*{a(n)}, 6*{a(n)}}. There is a table in the example section giving sequences formed from unions of the parts.
The parts correspond to the cosets of A352272 considered as a subgroup of the positive integers under the operation A059897(.,.). Viewed another way, the parts correspond to the intersection of the integers with the cosets of the multiplicative subgroup of the positive rationals generated by the terms of A352272.
The asymptotic density of this sequence is 1/4. - Amiram Eldar, Apr 03 2022

Examples

			The squarefree part of 11 is 11, which is congruent to 5 (mod 6), so 11 is in the sequence.
The squarefree part of 15 is 15, which is congruent to 3 (mod 6), so 15 is not in the sequence.
The squarefree part of 20 = 2^2 * 5 is 5, which is congruent to 5 (mod 6), so 20 is in the sequence.
The table below lists OEIS sequences that are unions of the cosets described in the initial comments, and indicates the cosets included in each sequence. A352272 (as a subgroup) is denoted H, and this sequence (as a coset) is denoted H/5, in view of its terms being one fifth of the multiples of 5 in A352272.
             H    2H    3H    6H    H/5  2H/5  3H/5  6H/5
A003159      X           X           X           X
A036554            X           X           X           X
.
A007417      X     X                 X     X
A145204\{0}              X     X                 X     X
.
A026225      X           X                 X           X
A026179\{1}        X           X     X           X
.
A036668      X                 X     X                 X
A325424            X     X                 X     X
.
A055047      X                             X
A055048            X                 X
A055041                  X                             X
A055040                        X                 X
.
A189715      X                 X           X     X
A189716            X     X           X                 X
.
A225837      X     X     X     X
A225838                              X     X     X     X
.
A339690      X                       X
A329575                  X                       X
.
A352274      X           X
(The sequence groupings in the table start with the subgroup of the quotient group of H, followed by its cosets.)
		

Crossrefs

Intersection of any three of A003159, A007417, A189716 and A225838.
Intersection of A036668 and A055048.
Complement within A339690 of A352272.
Closure of A084088 under multiplication by 9.
Other subsequences: A033429\{0}, A016969.
Other sequences in the example table: A036554, A145204, A026179, A026225, A325424, A055040, A055041, A055047, A189715, A225837, A329575, A352274.

Programs

  • Mathematica
    q[n_] := Module[{e2, e3}, {e2, e3} = IntegerExponent[n, {2, 3}]; EvenQ[e2] && EvenQ[e3] && Mod[n/2^e2/3^e3, 6] == 5]; Select[Range[250], q] (* Amiram Eldar, Apr 03 2022 *)
  • PARI
    isok(m) = core(m) % 6 == 5;
    
  • Python
    from itertools import count
    def A352273(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 count(0):
                i2 = 9**i
                if i2>x: break
                for j in count(0,2):
                    k = i2<x: break
                    c -= (x//k-5)//6+1
            return c
        return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

Formula

{a(n) : n >= 1} = {m >= 1 : A007913(m) == 5 (mod 6)}.
{a(n) : n >= 1} = A334832/5 U A334832/11 U A334832/17 U A334832/23 where A334832/k denotes {A334832(m)/k : m >= 1, k divides A334832(m)}.
Using the same notation, {a(n) : n >= 1} = A352272/5 = {A307151(A352272(m)) : m >= 1}.
{A225838(n) : n >= 1} = {m : m = a(j)*k, j >= 1, k divides 6}.

A233998 Values of n such that numbers of the form x^2+n*y^2 for some integers x, y cannot have prime factor of 5 raised to an odd power.

Original entry on oeis.org

2, 3, 7, 8, 12, 13, 17, 18, 22, 23, 27, 28, 32, 33, 37, 38, 42, 43, 47, 48, 50, 52, 53, 57, 58, 62, 63, 67, 68, 72, 73, 75, 77, 78, 82, 83, 87, 88, 92, 93, 97, 98, 102, 103, 107, 108, 112, 113, 117, 118, 122, 123, 127, 128, 132, 133, 137, 138, 142, 143, 147, 148, 152, 153, 157, 158
Offset: 1

Views

Author

V. Raman, Dec 18 2013

Keywords

Comments

Equivalently, this sequence is the union of numbers of the form 25^n*(5*n+2) and numbers of the form 25^n*(5*n+3).

Crossrefs

Programs

  • PARI
    is(n)=n/=25^valuation(n, 25); n%5==2||n%5==3 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013
    
  • Python
    from sympy import integer_log
    def A233998(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(((m:=x//25**i)-2)//5+(m-3)//5+2 for i in range(integer_log(x,25)[0]+1))
        return bisection(f,n,n) # Chai Wah Wu, Mar 19 2025

Formula

a(n) = 2.4 n + O(log n). - Charles R Greathouse IV, Dec 19 2013
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