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.

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A108056 Numbers of the form (7^i)*(13^j).

Original entry on oeis.org

1, 7, 13, 49, 91, 169, 343, 637, 1183, 2197, 2401, 4459, 8281, 15379, 16807, 28561, 31213, 57967, 107653, 117649, 199927, 218491, 371293, 405769, 753571, 823543, 1399489, 1529437, 2599051, 2840383, 4826809, 5274997, 5764801, 9796423, 10706059, 18193357, 19882681
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jun 02 2005

Keywords

Crossrefs

Programs

  • Mathematica
    n = 10^7; Flatten[Table[7^i*13^j, {i, 0, Log[7, n]}, {j, 0, Log[13, n/7^i]}]] // Sort (* Amiram Eldar, Sep 23 2020 *)
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim)\log(13),N=13^n;while(N<=lim,listput(v,N);N*=7));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
    
  • Python
    from sympy import integer_log
    def A108056(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: 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(integer_log(x//13**i,7)[0]+1 for i in range(integer_log(x,13)[0]+1))
        return bisection(f,n,n) # Chai Wah Wu, Oct 22 2024

Formula

Sum_{n>=1} 1/a(n) = (7*13)/((7-1)*(13-1)) = 91/72. - Amiram Eldar, Sep 23 2020
a(n) ~ exp(sqrt(2*log(7)*log(13)*n)) / sqrt(91). - Vaclav Kotesovec, Sep 23 2020

Extensions

More terms from Amiram Eldar, Sep 23 2020

A108347 Numbers of the form (3^i)*(5^j)*(7^k), with i, j, k >= 0.

Original entry on oeis.org

1, 3, 5, 7, 9, 15, 21, 25, 27, 35, 45, 49, 63, 75, 81, 105, 125, 135, 147, 175, 189, 225, 243, 245, 315, 343, 375, 405, 441, 525, 567, 625, 675, 729, 735, 875, 945, 1029, 1125, 1215, 1225, 1323, 1575, 1701, 1715, 1875, 2025, 2187, 2205, 2401, 2625, 2835, 3087
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jul 01 2005

Keywords

Comments

The Heinz numbers of the partitions into parts 2,3, and 4 (including the number 1, the Heinz number of the empty partition). We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product(p_j-th prime, j=1...r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [2,3,3,4] the Heinz number is 3*5*5*7 = 525; it is in the sequence. - Emeric Deutsch , May 21 2015
Numbers m | 105^e with integer e >= 0. - Michael De Vlieger, Aug 22 2019

Crossrefs

Programs

  • Magma
    [n: n in [1..4000] | PrimeDivisors(n) subset [3,5,7]]; // Bruno Berselli, Sep 24 2012
    
  • Maple
    with(numtheory): S := {}: for j to 3100 do if `subset`(factorset(j), {3, 5, 7}) then S := `union`(S, {j}) else end if end do: S; # Emeric Deutsch, May 21 2015
    # alternative
    isA108347 := proc(n)
          if n = 1 then
            true;
        else
            return (numtheory[factorset](n) minus {3, 5, 7} = {} );
        end if;
    end proc:
    A108347 := proc(n)
         option remember;
         if n = 1 then
            1;
        else
            for a from procname(n-1)+1 do
                if isA108347(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A108347(n),n=1..80); # R. J. Mathar, Jun 06 2024
  • Mathematica
    With[{n = 3087}, Sort@ Flatten@ Table[3^i * 5^j * 7^k, {i, 0, Log[3, n]}, {j, 0, Log[5, n/2^i]}, {k, 0, Log[7, n/(3^i*5^j)]}]] (* Michael De Vlieger, Aug 22 2019 *)
  • Python
    from sympy import integer_log
    def A108347(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: 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,7)[0]+1):
                for j in range(integer_log(m:=x//7**i,5)[0]+1):
                    c -= integer_log(m//5**j,3)[0]+1
            return c
        return bisection(f,n,n) # Chai Wah Wu, Sep 16 2024

Formula

Sum_{n>=1} 1/a(n) = (3*5*7)/((3-1)*(5-1)*(7-1)) = 35/16. - Amiram Eldar, Sep 22 2020
a(n) ~ exp((6*log(3)*log(5)*log(7)*n)^(1/3)) / sqrt(105). - Vaclav Kotesovec, Sep 23 2020

A108319 Numbers of the form (2^i)*(3^j)*(7^k), with i, j, k >= 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 27, 28, 32, 36, 42, 48, 49, 54, 56, 63, 64, 72, 81, 84, 96, 98, 108, 112, 126, 128, 144, 147, 162, 168, 189, 192, 196, 216, 224, 243, 252, 256, 288, 294, 324, 336, 343, 378, 384, 392, 432, 441, 448, 486, 504, 512, 567
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jun 30 2005

Keywords

Comments

Numbers m | 42^e with integer e >= 0. - Michael De Vlieger, Aug 22 2019
Sum_{n>=1} 1/a(n) = (2*3*7)/((2-1)*(3-1)*(7-1)) = 7/2. - Amiram Eldar, Sep 24 2020

Crossrefs

Programs

  • Mathematica
    With[{n = 567}, Sort@ Flatten@ Table[2^i * 3^j * 7^k, {i, 0, Log2@ n}, {j, 0, Log[3, n/2^i]}, {k, 0, Log[7, n/(2^i*3^j)]}]] (* Michael De Vlieger, Aug 22 2019 *)
  • PARI
    list(lim)=my(v=List(), s, t); for(i=0, logint(lim\=1, 7), t=7^i; for(j=0, logint(lim\t, 3), s=t*3^j; while(s<=lim, listput(v, s); s<<=1))); Set(v) \\ Charles R Greathouse IV, Nov 20 2024

Formula

a(n) ~ exp((6*log(2)*log(3)*log(7)*n)^(1/3)) / sqrt(42). - Vaclav Kotesovec, Sep 23 2020

A025626 Numbers of form 6^i*7^j, with i, j >= 0.

Original entry on oeis.org

1, 6, 7, 36, 42, 49, 216, 252, 294, 343, 1296, 1512, 1764, 2058, 2401, 7776, 9072, 10584, 12348, 14406, 16807, 46656, 54432, 63504, 74088, 86436, 100842, 117649, 279936, 326592, 381024, 444528, 518616, 605052, 705894, 823543, 1679616, 1959552
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    n = 10^6; Flatten[Table[6^i*7^j, {i, 0, Log[6, n]}, {j, 0, Log[7, n/6^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *)

Formula

Sum_{n>=1} 1/a(n) = (6*7)/((6-1)*(7-1)) = 7/5. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(6)*log(7)*n)) / sqrt(42). - Vaclav Kotesovec, Sep 25 2020
a(n) = 6^A025660(n) * 7^A025668(n). - R. J. Mathar, Jul 06 2025

A177805 Numbers k such that k divides 15^k - 1.

Original entry on oeis.org

1, 2, 4, 7, 8, 14, 16, 28, 32, 49, 56, 64, 98, 112, 128, 136, 196, 224, 256, 272, 343, 392, 448, 452, 512, 544, 686, 784, 812, 896, 904, 952, 1024, 1088, 1372, 1568, 1624, 1792, 1808, 1904, 2048, 2176, 2312, 2401, 2744, 3136, 3164, 3248, 3584, 3616, 3808, 4096
Offset: 1

Views

Author

Alexander Adamchuk, May 17 2010

Keywords

Comments

A000420 are the only odd terms of the sequence. - Robert Israel, Feb 25 2020

Crossrefs

Programs

A025619 Numbers of form 4^i*7^j, with i, j >= 0.

Original entry on oeis.org

1, 4, 7, 16, 28, 49, 64, 112, 196, 256, 343, 448, 784, 1024, 1372, 1792, 2401, 3136, 4096, 5488, 7168, 9604, 12544, 16384, 16807, 21952, 28672, 38416, 50176, 65536, 67228, 87808, 114688, 117649, 153664, 200704, 262144, 268912, 351232, 458752, 470596
Offset: 1

Views

Author

Keywords

Crossrefs

Subsequence of A003591.

Programs

  • Mathematica
    n=10^6; Flatten[Table[4^i*7^j, {i, 0, Log[4, n]}, {j, 0, Log[7, n/4^i]}]]//Sort (* Amiram Eldar, Sep 24 2020 *)

Formula

Sum_{n>=1} 1/a(n) = (4*7)/((4-1)*(7-1)) = 14/9. - Amiram Eldar, Sep 24 2020
a(n) ~ exp(sqrt(2*log(4)*log(7)*n)) / sqrt(28). - Vaclav Kotesovec, Sep 24 2020
a(n) = 4^A025648(n) * 7^A025666(n). - R. J. Mathar, Jul 06 2025

A316991 Numbers m such that 1 < gcd(m, 14) < m and m does not divide 14^e for e >= 0.

Original entry on oeis.org

6, 10, 12, 18, 20, 21, 22, 24, 26, 30, 34, 35, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 58, 60, 62, 63, 66, 68, 70, 72, 74, 76, 77, 78, 80, 82, 84, 86, 88, 90, 91, 92, 94, 96, 100, 102, 104, 105, 106, 108, 110, 114, 116, 118, 119, 120, 122, 124, 126, 130, 132
Offset: 1

Views

Author

Michael De Vlieger, Aug 02 2018

Keywords

Comments

Complement of A000027 and union of A003591 and A162699.
Analogous to A081062 and A105115 that apply to A120944(1) and A120944(2), respectively.
This sequence applies to A120944(3).

Examples

			6 is in the sequence since gcd(6, 14) = 2 and 6 does not divide 14^e with integer e >= 0.
2 is not in the sequence since 2 | 14.
4 is not in the sequence since 4 | 14^2.
3 and 5 are not in the sequence since they are coprime to 14.
		

Crossrefs

Programs

  • Mathematica
    With[{nn = 132, k = 14}, Select[Range@ nn, And[1 < GCD[#, k] < #, PowerMod[k, Floor@ Log2@ nn, #] != 0] &]]

A369417 Powerful numbers k with multiple distinct prime factors such that rad(k) is not a primorial, where rad(k) = A007947(k).

Original entry on oeis.org

100, 196, 200, 225, 392, 400, 441, 484, 500, 675, 676, 784, 800, 968, 1000, 1089, 1125, 1156, 1225, 1323, 1352, 1372, 1444, 1521, 1568, 1600, 1764, 1936, 2000, 2025, 2116, 2312, 2500, 2601, 2704, 2744, 2888, 3025, 3087, 3136, 3200, 3249, 3267, 3364, 3375, 3528
Offset: 1

Views

Author

Michael De Vlieger, Jan 22 2024

Keywords

Comments

Numbers k such that Omega(k) > omega(k) > 1, where all prime power factors p^m have exponents m > 1, such that squarefree kernel rad(k) not in A002110, where Omega = A001222 and omega = A001221.

Examples

			Let S = A366413 = {A120944 \ A002110}.
This sequence is the union of the following infinite sets:
S(1)^2 * A003592 = 10^2 * A003592 = {100, 200, 400, 500, 800, 1000, ...}
                 = { m*S(1)^2 : rad(m) | S(1) }.
S(2)^2 * A003591 = 14^2 * A003591 = {196, 392, 784, 1372, 1568, ...}
                 = { m*S(2)^2 : rad(m) | S(2) }.
S(3)^2 * A003593 = 15^2 * A003593 = {225, 675, 1125, 2025, 3375, ...}
                 = { m*S(3)^2 : rad(m) | S(3) }, etc.
		

Crossrefs

Programs

  • Mathematica
    With[{nn = 2^14},
      Select[
        Select[
          Rest@ Union@ Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}],
        Not@*PrimePowerQ],
      Nand[EvenQ[#],
        Union@ Differences@ PrimePi[FactorInteger[#][[All, 1]]] == {1}] &] ]

Formula

{a(n)} = { A286708 \ A055932 }.
{a(n)} = { m*s^2 : Omega(s) = omega(s) > 1, s not in A002110, rad(m) | s }.
A286708 is the union of A369374 and this sequence.

A108513 Numbers of the form (2^i)*(5^j)*(7^k), with i, j, k >= 0.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 25, 28, 32, 35, 40, 49, 50, 56, 64, 70, 80, 98, 100, 112, 125, 128, 140, 160, 175, 196, 200, 224, 245, 250, 256, 280, 320, 343, 350, 392, 400, 448, 490, 500, 512, 560, 625, 640, 686, 700, 784, 800, 875, 896, 980, 1000, 1024, 1120
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jul 05 2005

Keywords

Comments

Numbers m | 70^e with integer e >= 0. - Michael De Vlieger, Aug 22 2019

Crossrefs

Programs

  • Mathematica
    With[{n = 1120}, Sort@ Flatten@ Table[2^i * 5^j * 7^k, {i, 0, Log2@ n}, {j, 0, Log[5, n/2^i]}, {k, 0, Log[7, n/(2^i*5^j)]}]] (* Michael De Vlieger, Aug 22 2019 *)
  • PARI
    isok(n) = (n/(2^valuation(n,2)*5^valuation(n,5)*7^valuation(n,7)) == 1); \\ Michel Marcus, Oct 01 2013

Formula

Sum_{n>=1} 1/a(n) = (2*5*7)/((2-1)*(5-1)*(7-1)) = 35/12. - Amiram Eldar, Sep 23 2020
a(n) ~ exp((6*log(2)*log(5)*log(7)*n)^(1/3)) / sqrt(70). - Vaclav Kotesovec, Sep 23 2020

A025637 Exponent of 2 (value of i) in n-th number of form 2^i*7^j.

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 2, 5, 0, 3, 6, 1, 4, 7, 2, 5, 8, 0, 3, 6, 9, 1, 4, 7, 10, 2, 5, 8, 11, 0, 3, 6, 9, 12, 1, 4, 7, 10, 13, 2, 5, 8, 11, 14, 0, 3, 6, 9, 12, 15, 1, 4, 7, 10, 13, 16, 2, 5, 8, 11, 14, 0, 17, 3, 6, 9, 12, 15, 1, 18, 4, 7, 10, 13, 16, 2, 19, 5, 8, 11, 14, 0, 17, 3, 20, 6, 9, 12, 15, 1
Offset: 1

Views

Author

Keywords

Crossrefs

Cf. A003591.
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