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 1-6 of 6 results.

A028983 Numbers whose sum of divisors is even.

Original entry on oeis.org

3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 82
Offset: 1

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Author

Keywords

Comments

The even terms of this sequence are the even terms appearing in A178910. [Edited by M. F. Hasler, Oct 02 2014]
A071324(a(n)) is even. - Reinhard Zumkeller, Jul 03 2008
Sigma(a(n)) = A000203(a(n)) = A152678(n). - Jaroslav Krizek, Oct 06 2009
A083207 is a subsequence. - Reinhard Zumkeller, Jul 19 2010
Numbers k such that the number of odd divisors of k (A001227) is even. - Omar E. Pol, Apr 04 2016
Numbers k such that the sum of odd divisors of k (A000593) is even. - Omar E. Pol, Jul 05 2016
Numbers with a squarefree part greater than 2. - Peter Munn, Apr 26 2020
Equivalently, numbers whose odd part is nonsquare. Compare with the numbers whose square part is even (i.e., nonodd): these are the positive multiples of 4, A008586\{0}, and A225546 provides a self-inverse bijection between the two sets. - Peter Munn, Jul 19 2020
Also numbers whose reversed prime indices have alternating product > 1, where we define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)). Also Heinz numbers of the partitions counted by A347448. - Gus Wiseman, Oct 29 2021
Numbers whose number of middle divisors is not odd (cf. A067742). - Omar E. Pol, Aug 02 2022

Crossrefs

The complement is A028982 = A000290 U A001105.
Subsequences: A083207, A091067, A145204\{0}, A225838, A225858.
Cf. A334748 (a permutation).
Related to A008586 via A225546.
Ranks the partitions counted by A347448, complement A119620.

Programs

  • Mathematica
    Select[Range[82],EvenQ[DivisorSigma[1,#]]&] (* Jayanta Basu, Jun 05 2013 *)
  • PARI
    is(n)=!issquare(n)&&!issquare(n/2) \\ Charles R Greathouse IV, Jan 11 2013
    
  • Python
    from math import isqrt
    def A028983(n):
        def f(x): return n-1+isqrt(x)+isqrt(x>>1)
        kmin, kmax = 1,2
        while f(kmax) >= kmax:
            kmax <<= 1
        while True:
            kmid = kmax+kmin>>1
            if f(kmid) < kmid:
                kmax = kmid
            else:
                kmin = kmid
            if kmax-kmin <= 1:
                break
        return kmax # Chai Wah Wu, Aug 22 2024

Formula

a(n) ~ n. - Charles R Greathouse IV, Jan 11 2013
a(n) = n + (1 + sqrt(2)/2)*sqrt(n) + O(1). - Charles R Greathouse IV, Sep 01 2015
A007913(a(n)) > 2. - Peter Munn, May 05 2020

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

A225837 Numbers of form 2^i*3^j*(6k+1), i, j, k >= 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 16, 18, 19, 21, 24, 25, 26, 27, 28, 31, 32, 36, 37, 38, 39, 42, 43, 48, 49, 50, 52, 54, 55, 56, 57, 61, 62, 63, 64, 67, 72, 73, 74, 75, 76, 78, 79, 81, 84, 85, 86, 91, 93, 96, 97, 98, 100, 103, 104, 108, 109, 110, 111, 112
Offset: 1

Views

Author

Ralf Stephan, May 16 2013

Keywords

Comments

The asymptotic density of this sequence is 1/2. - Amiram Eldar, Apr 03 2022
From Peter Munn, Nov 16 2023: (Start)
Contains all nonzero squares.
Dividing by 5 the terms that are multiples of 5 gives its complement, A225838.
(A352272, 2*A352272, 3*A352272, 6*A352272) is a partition of the terms.
The terms form a subgroup of the positive integers under the operation A059897(.,.) and are the positive integers in an index 2 multiplicative subgroup of rationals that is generated by 2, 3 and integers congruent to 1 modulo 6. See A225857 and A352272 for further information about such subgroups.
(End)

Crossrefs

Complement of A225838.
Subsequences: A003136\{0}, A083854\{0}, A260488\{0}, A352272.
Symmetric difference of A026225 and A036554; of A036668 and A189716.

Programs

  • Magma
    [n: n in [1..200] | IsOne(d mod 6) where d is n div (2^Valuation(n,2)*3^Valuation(n,3))]; // Bruno Berselli, May 16 2013
    
  • Mathematica
    mx = 122; t = {}; Do[n = 2^i*3^j (6 k + 1); If[n <= mx, AppendTo[t, n]], {i, 0, Log[2, mx]}, {j, 0, Log[3, mx]}, {k, 0, mx/6}]; Union[t] (* T. D. Noe, May 16 2013 *)
  • PARI
    for(n=1,200,t=n/(2^valuation(n,2)*3^valuation(n,3));if((t%6==1),print1(n,",")))
    
  • Python
    from sympy import integer_log
    def A225837(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//3**i>>j)+5)//6 for i in range(integer_log(x,3)[0]+1) for j in range((x//3**i).bit_length()))
        return bisection(f,n,n) # Chai Wah Wu, Feb 02 2025

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}.

A225858 Numbers of the form 2^i*3^j*(12k+7) or 2^i*3^j*(12k+11), i, j, k >= 0.

Original entry on oeis.org

7, 11, 14, 19, 21, 22, 23, 28, 31, 33, 35, 38, 42, 43, 44, 46, 47, 55, 56, 57, 59, 62, 63, 66, 67, 69, 70, 71, 76, 79, 83, 84, 86, 88, 91, 92, 93, 94, 95, 99, 103, 105, 107, 110, 112, 114, 115, 118, 119, 124, 126, 127, 129, 131, 132, 134, 138, 139, 140
Offset: 1

Views

Author

Ralf Stephan, May 18 2013

Keywords

Comments

The asymptotic density of this sequence is 1/2. - Amiram Eldar, Nov 14 2023

Examples

			From _David A. Corneth_, Nov 11 2023: (Start)
28 = 2^2 * 7 = 2^2 * 3^0 * (12*0 + 7) is in the sequence as it meets the first form.
76 = 2^2 * 19 = 2^2 * 3^0 * (12*1 + 7) is in the sequence as it meets the first form.
15 = 3 * 5 = 2^0 * 3^1 * (12*0 + 5) is not in the sequence as it does not match any of the desired forms. (End)
		

Crossrefs

Complement of A225857. Cf. A225838.

Programs

  • Magma
    [n: n in [1..200] | d mod 4 eq 3 where d is n div (2^Valuation(n,2)*3^Valuation(n,3))]; // Bruno Berselli, May 16 2013
    
  • Mathematica
    Select[Range[140], Mod[#/Times @@ ({2, 3}^IntegerExponent[#, {2, 3}]), 4] == 3 &] (* Amiram Eldar, Nov 14 2023 *)
  • PARI
    for(n=1,200,t=n/(2^valuation(n,2)*3^valuation(n,3));if((t%4==3),print1(n,",")))
    
  • Python
    from itertools import count
    from sympy import integer_log
    def A225858(n):
        def f(x):
            c = n
            for i in range(integer_log(x,3)[0]+1):
                i2 = 3**i
                for j in count(0):
                    k = i2<x:
                        break
                    m = x//k
                    c += (m-1)//12+(m-5)//12+2
            return c
        m, k = n, f(n)
        while m != k: m, k = k, f(k)
        return m # Chai Wah Wu, Feb 24 2025

Extensions

Name clarified by Peter Munn, Nov 11 2023

A260489 a(n) = 3n - A260488(n).

Original entry on oeis.org

0, 2, 4, 5, 5, 7, 5, 7, 8, 8, 5, 7, 8, 8, 10, 8, 10, 8, 5, 7, 8, 8, 10, 8, 10, 11, 11, 8, 10, 11, 11, 8, 10, 8, 5, 7, 8, 8, 10, 8, 10, 11, 11, 8, 10, 11, 11, 13, 11, 13, 11, 8, 10, 11, 11, 13, 11, 13, 11, 8, 10, 11, 11, 8, 10, 8, 5, 7, 8, 8, 10
Offset: 0

Views

Author

Keywords

Comments

The sequence is nonnegative for at least the first 10000 terms, so the nonn keyword was used. That it is always nonnegative is a conjecture. That it is unbounded is also a conjecture.
The sequence is nonnegative for essentially the reason given in A225838 for its terms exceeding the terms of its complement. Here, the customers are labeled with numbers, m, that are powers of 2 and they each deposit 2 coins at each epoch (6k+1)*m and withdraw them separately at epoch (6k+3)*m and (6k+5)*m. The terms here are the number of coins in the vault after the n-th deposit. Simple accountancy says the sum of nonnegative bank balances is nonnegative. - Peter Munn, Nov 15 2023

Crossrefs

Showing 1-6 of 6 results.