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 12 results. Next

A003595 Numbers of the form 5^i*7^j with i, j >= 0.

Original entry on oeis.org

1, 5, 7, 25, 35, 49, 125, 175, 245, 343, 625, 875, 1225, 1715, 2401, 3125, 4375, 6125, 8575, 12005, 15625, 16807, 21875, 30625, 42875, 60025, 78125, 84035, 109375, 117649, 153125, 214375, 300125, 390625, 420175, 546875, 588245, 765625, 823543, 1071875, 1500625
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Successive k such that phi(35*k) = 24*k: 35*a(n) = A033851(n). - Artur Jasinski, Nov 09 2008

Links

Crossrefs

Cf. A033851, A143207, A147571-A147580.
Cf. A003586, A003592, A003593, A003594.
Cf. A025652, A025667.

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
a003595 n = a003595_list !! (n-1)
a003595_list = f $ singleton 1 where
   f s = y : f (insert (5 * y) $ insert (7 * y) s')
               where (y, s') = deleteFindMin s
-- Reinhard Zumkeller, May 16 2015
  • Magma
    [n: n in [1..600000] | PrimeDivisors(n) subset [5,7]]; // Bruno Berselli, Sep 24 2012
  • Mathematica
    a = {}; Do[If[EulerPhi[35 k] == 24 k, AppendTo[a, k]], {k, 1, 10000}]; a (* Artur Jasinski, Nov 09 2008 *)
fQ[n_] := PowerMod[35, n, n] == 0; Select[Range[600000], fQ] (* Bruno Berselli, Sep 24 2012 *)
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim)\log(7),N=7^n;while(N<=lim,listput(v,N);N*=5));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
  • Python
    from sympy import integer_log
def A003595(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//7**i,5)[0]+1 for i in range(integer_log(x,7)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Sep 16 2024

Formula

Sum_{n>=1} 1/a(n) = (5*7)/((5-1)*(7-1)) = 35/24. - Amiram Eldar, Sep 22 2020
a(n) ~ exp(sqrt(2*log(5)*log(7)*n)) / sqrt(35). - Vaclav Kotesovec, Sep 22 2020
a(n) = 5^A025652(n) * 7^A025667(n). - R. J. Mathar, Jul 06 2025

A147571 Numbers with exactly 4 distinct prime divisors {2,3,5,7}.

Original entry on oeis.org

210, 420, 630, 840, 1050, 1260, 1470, 1680, 1890, 2100, 2520, 2940, 3150, 3360, 3780, 4200, 4410, 5040, 5250, 5670, 5880, 6300, 6720, 7350, 7560, 8400, 8820, 9450, 10080, 10290, 10500, 11340, 11760, 12600, 13230, 13440, 14700, 15120, 15750, 16800
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Successive numbers k such that EulerPhi(x)/x = m:
( Family of sequences for successive n primes )
m=1/2 numbers with exactly 1 distinct prime divisor {2} see A000079
m=1/3 numbers with exactly 2 distinct prime divisors {2,3} see A033845
m=4/15 numbers with exactly 3 distinct prime divisors {2,3,5} see A143207
m=8/35 numbers with exactly 4 distinct prime divisors {2,3,5,7} see A147571
m=16/77 numbers with exactly 5 distinct prime divisors {2,3,5,7,11} see A147572
m=192/1001 numbers with exactly 6 distinct prime divisors {2,3,5,7,11,13} see A147573
m=3072/17017 numbers with exactly 7 distinct prime divisors {2,3,5,7,11,13,17} see A147574
m=55296/323323 numbers with exactly 8 distinct prime divisors {2,3,5,7,11,13,17,19} see A147575

Links

Crossrefs

Cf. A002473, A086780, A143207, A147572, A147573, A147574, A147575, A147576, A147577, A147578, A147579, A147580.

Programs

  • Magma
    [n: n in [1..2*10^4] | PrimeDivisors(n) eq [2,3,5,7]]; // Vincenzo Librandi, Sep 15 2015
  • Mathematica
    a = {}; Do[If[EulerPhi[x]/x == 8/35, AppendTo[a, x]], {x, 1, 100000}]; a
Select[Range[20000],PrimeNu[#]==4&&Max[FactorInteger[#][[;;,1]]]<11&] (* Harvey P. Dale, Nov 05 2024 *)
  • PARI
    is(n)=n%210==0 && n==2^valuation(n,2) * 3^valuation(n,3) * 5^valuation(n,5) * 7^valuation(n,7) \\ Charles R Greathouse IV, Jun 19 2016

Formula

a(n) = 210 * A002473(n). - David A. Corneth, May 14 2019
Sum_{n>=1} 1/a(n) = 1/48. - Amiram Eldar, Nov 12 2020

A033851 Numbers whose prime factors are 5 and 7.

Original entry on oeis.org

35, 175, 245, 875, 1225, 1715, 4375, 6125, 8575, 12005, 21875, 30625, 42875, 60025, 84035, 109375, 153125, 214375, 300125, 420175, 546875, 588245, 765625, 1071875, 1500625, 2100875, 2734375, 2941225, 3828125, 4117715, 5359375, 7503125
Offset: 1

Views

Author

Jeff Burch

Keywords

Comments

Numbers k such that phi(k)/k == 24/35. - Artur Jasinski, Nov 09 2008
Subsequence of A143202. - Reinhard Zumkeller, Sep 13 2011

Links

Crossrefs

Cf. A003595, A147571-A147575, A147576-A147580. [Artur Jasinski, Nov 09 2008]
Cf. A033845, A033846, A033847, A033848, A033849, A033850, A143201, A143202.
Subsequence of A256617.

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
a033851 n = a033851_list !! (n-1)
a033851_list = f (singleton (5*7)) where
   f s = m : f (insert (5*m) $ insert (7*m) s') where
     (m,s') = deleteFindMin s
-- Reinhard Zumkeller, Sep 13 2011
  • Mathematica
    a = {}; Do[If[EulerPhi[x]/x == 24/35, AppendTo[a, x]], {x, 1, 10000}]; a (* Artur Jasinski, Nov 09 2008 *)
Take[With[{nn=10},Sort[Flatten[Table[5^i 7^j,{i,nn},{j,nn}]]]],40] (* Harvey P. Dale, Feb 09 2013 *)

Formula

a(n) = 35*A003595(n). - Artur Jasinski, Nov 09 2008
A143201(a(n)) = 3. - Reinhard Zumkeller, Sep 13 2011
Sum_{n>=1} 1/a(n) = 1/24. - Amiram Eldar, Dec 22 2020

Extensions

Offset fixed by Reinhard Zumkeller, Sep 13 2011

A147576 Numbers with exactly 3 distinct odd prime divisors {3,5,7}.

Original entry on oeis.org

105, 315, 525, 735, 945, 1575, 2205, 2625, 2835, 3675, 4725, 5145, 6615, 7875, 8505, 11025, 13125, 14175, 15435, 18375, 19845, 23625, 25515, 25725, 33075, 36015, 39375, 42525, 46305, 55125, 59535, 65625, 70875, 76545, 77175, 91875, 99225
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Numbers k such that phi(k)/k = m
( Family of sequences for successive n odd primes )
m=2/3 numbers with exactly 1 distinct prime divisor {3} see A000244
m=8/15 numbers with exactly 2 distinct prime divisors {3,5} see A033849
m=16/35 numbers with exactly 3 distinct prime divisors {3,5,7} see A147576
m=32/77 numbers with exactly 4 distinct prime divisors {3,5,7,11} see A147577
m=384/1001 numbers with exactly 5 distinct prime divisors {3,5,7,11,13} see A147578
m=6144/17017 numbers with exactly 6 distinct prime divisors {3,5,7,11,13,17} see A147579
m=3072/323323 numbers with exactly 7 distinct prime divisors {3,5,7,11,13,17,19} see A147580
m=110592/323323 numbers with exactly 8 distinct prime divisors {3,5,7,11,13,17,19,23} see A147581

Links

Crossrefs

Cf. A060735, A108347, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[x]/x == 16/35, AppendTo[a, x]], {x, 1, 100000}]; a
Select[Range[100000],EulerPhi[#]/#==16/35&] (* Harvey P. Dale, Dec 01 2013 *)

Formula

a(n) = 105 * A108347(n). - Amiram Eldar, Mar 10 2020
Sum_{n>=1} 1/a(n) = 1/48. - Amiram Eldar, Dec 22 2020

A147575 Numbers with exactly 8 distinct prime divisors {2,3,5,7,11,13,17,19}.

Original entry on oeis.org

9699690, 19399380, 29099070, 38798760, 48498450, 58198140, 67897830, 77597520, 87297210, 96996900, 106696590, 116396280, 126095970, 135795660, 145495350, 155195040, 164894730, 174594420, 184294110, 193993800, 203693490, 213393180, 232792560, 242492250, 252191940
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Successive numbers k such that EulerPhi(x)/x = m:
( Family of sequences for successive n primes )
m=1/2 numbers with exactly 1 distinct prime divisor {2} see A000079
m=1/3 numbers with exactly 2 distinct prime divisors {2,3} see A033845
m=4/15 numbers with exactly 3 distinct prime divisors {2,3,5} see A143207
m=8/35 numbers with exactly 4 distinct prime divisors {2,3,5,7} see A147571
m=16/77 numbers with exactly 5 distinct prime divisors {2,3,5,7,11} see A147572
m=192/1001 numbers with exactly 6 distinct prime divisors {2,3,5,7,11,13} see A147573
m=3072/17017 numbers with exactly 7 distinct prime divisors {2,3,5,7,11,13,17} see A147574
m=55296/323323 numbers with exactly 8 distinct prime divisors {2,3,5,7,11,13,17,19} see A147575

Links

Crossrefs

Cf. A060735, A080682, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[9699690 x] == 1658880 x, AppendTo[a, 9699690 x]], {x, 1, 100}]; a

Formula

a(n) = 9699690 * A080682(n). - Amiram Eldar, Mar 10 2020
Sum_{n>=1} 1/a(n) = 1/1658880. - Amiram Eldar, Nov 12 2020

Extensions

More terms from Amiram Eldar, Mar 10 2020

A147572 Numbers with exactly 5 distinct prime divisors {2,3,5,7,11}.

Original entry on oeis.org

2310, 4620, 6930, 9240, 11550, 13860, 16170, 18480, 20790, 23100, 25410, 27720, 32340, 34650, 36960, 41580, 46200, 48510, 50820, 55440, 57750, 62370, 64680, 69300, 73920, 76230, 80850, 83160, 92400, 97020, 101640, 103950, 110880, 113190, 115500, 124740, 127050
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Successive numbers k such that EulerPhi(x)/x = m:
( Family of sequences for successive n primes )
m=1/2 numbers with exactly 1 distinct prime divisor {2} see A000079
m=1/3 numbers with exactly 2 distinct prime divisors {2,3} see A033845
m=4/15 numbers with exactly 3 distinct prime divisors {2,3,5} see A143207
m=8/35 numbers with exactly 4 distinct prime divisors {2,3,5,7} see A147571
m=16/77 numbers with exactly 5 distinct prime divisors {2,3,5,7,11} see A147572
m=192/1001 numbers with exactly 6 distinct prime divisors {2,3,5,7,11,13} see A147573
m=3072/17017 numbers with exactly 7 distinct prime divisors {2,3,5,7,11,13,17} see A147574
m=55296/323323 numbers with exactly 8 distinct prime divisors {2,3,5,7,11,13,17,19} see A147575

Links

Crossrefs

Cf. A051038, A060735, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[x]/x == 16/77, AppendTo[a, x]], {x, 1, 100000}]; a
Select[Range[130000],FactorInteger[#][[All,1]]=={2,3,5,7,11}&] (* Harvey P. Dale, Oct 04 2020 *)
  • Python
    from sympy import integer_log, prevprime
def A147572(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 g(x,m): return sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1)) if m==3 else sum(g(x//(m**i),prevprime(m))for i in range(integer_log(x,m)[0]+1))
    def f(x): return n+x-g(x,11)
    return 2310*bisection(f,n,n) # Chai Wah Wu, Sep 16 2024

Formula

a(n) = 2310 * A051038(n). - Amiram Eldar, Mar 10 2020
Sum_{n>=1} 1/a(n) = 1/480. - Amiram Eldar, Nov 12 2020

Extensions

More terms from Amiram Eldar, Mar 10 2020

A147577 Numbers with exactly 4 distinct odd prime divisors {3,5,7,11}.

Original entry on oeis.org

1155, 3465, 5775, 8085, 10395, 12705, 17325, 24255, 28875, 31185, 38115, 40425, 51975, 56595, 63525, 72765, 86625, 88935, 93555, 114345, 121275, 139755, 144375, 155925, 169785, 190575, 202125, 218295, 259875, 266805, 280665, 282975, 317625
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Numbers k such that phi(k)/k = m
( Family of sequences for successive n odd primes )
m=2/3 numbers with exactly 1 distinct prime divisor {3} see A000244
m=8/15 numbers with exactly 2 distinct prime divisors {3,5} see A033849
m=16/35 numbers with exactly 3 distinct prime divisors {3,5,7} see A147576
m=32/77 numbers with exactly 4 distinct prime divisors {3,5,7,11} see A147577
m=384/1001 numbers with exactly 5 distinct prime divisors {3,5,7,11,13} see A147578
m=6144/17017 numbers with exactly 6 distinct prime divisors {3,5,7,11,13,17} see A147579
m=3072/323323 numbers with exactly 7 distinct prime divisors {3,5,7,11,13,17,19} see A147580
m=110592/323323 numbers with exactly 8 distinct prime divisors {3,5,7,11,13,17,19,23} see A147581

Links

Crossrefs

Cf. A060735, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[x]/x == 32/77, AppendTo[a, x]], {x, 1, 1000000}]; a
Select[Range[350000],EulerPhi[#]/#==32/77&] (* Harvey P. Dale, Mar 25 2016 *)
  • Python
    from sympy import integer_log
def A147577(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 i11 in range(integer_log(x,11)[0]+1):
            for i7 in range(integer_log(x11:=x//11**i11,7)[0]+1):
                for i5 in range(integer_log(x7:=x11//7**i7,5)[0]+1):
                    c -= integer_log(x7//5**i5,3)[0]+1
        return c
    return 1155*bisection(f,n,n) # Chai Wah Wu, Oct 22 2024

Formula

Sum_{n>=1} 1/a(n) = 1/480. - Amiram Eldar, Dec 22 2020

A147578 Numbers with exactly 5 distinct odd prime divisors {3,5,7,11,13}.

Original entry on oeis.org

15015, 45045, 75075, 105105, 135135, 165165, 195195, 225225, 315315, 375375, 405405, 495495, 525525, 585585, 675675, 735735, 825825, 945945, 975975, 1126125, 1156155, 1216215, 1366365, 1486485, 1576575, 1756755, 1816815, 1876875, 2027025, 2147145, 2207205, 2477475
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Numbers k such that phi(k)/k = m
( Family of sequences for successive n odd primes )
m=2/3 numbers with exactly 1 distinct prime divisor {3} see A000244;
m=8/15 numbers with exactly 2 distinct prime divisors {3,5} see A033849;
m=16/35 numbers with exactly 3 distinct prime divisors {3,5,7} see A147576;
m=32/77 numbers with exactly 4 distinct prime divisors {3,5,7,11} see A147577;
m=384/1001 numbers with exactly 5 distinct prime divisors {3,5,7,11,13} see A147578;
m=6144/17017 numbers with exactly 6 distinct prime divisors {3,5,7,11,13,17} see A147579;
m=3072/323323 numbers with exactly 7 distinct prime divisors {3,5,7,11,13,17,19} see A147580;
m=110592/323323 numbers with exactly 8 distinct prime divisors {3,5,7,11,13,17,19,23} see A147581.

Links

Crossrefs

Cf. A000010 (EulerPhi), A060735, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[x]/x == 384/1001, AppendTo[a, x]], {x, 1, 1000000}]; a

Formula

Sum_{n>=1} 1/a(n) = 1/5760. - Amiram Eldar, Dec 22 2020

Extensions

More terms from Amiram Eldar, Mar 11 2020

A147579 Numbers with exactly 6 distinct odd prime divisors {3,5,7,11,13,17}.

Original entry on oeis.org

255255, 765765, 1276275, 1786785, 2297295, 2807805, 3318315, 3828825, 4339335, 5360355, 6381375, 6891885, 8423415, 8933925, 9954945, 11486475, 12507495, 13018005, 14039025, 16081065, 16591575, 19144125, 19654635, 20675655, 21696675, 23228205, 25270245, 26801775
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Numbers k such that phi(k)/k = m
( Family of sequences for successive n odd primes )
m=2/3 numbers with exactly 1 distinct prime divisor {3} see A000244
m=8/15 numbers with exactly 2 distinct prime divisors {3,5} see A033849
m=16/35 numbers with exactly 3 distinct prime divisors {3,5,7} see A147576
m=32/77 numbers with exactly 4 distinct prime divisors {3,5,7,11} see A147577
m=384/1001 numbers with exactly 5 distinct prime divisors {3,5,7,11,13} see A147578
m=6144/17017 numbers with exactly 6 distinct prime divisors {3,5,7,11,13,17} see A147579
m=3072/323323 numbers with exactly 7 distinct prime divisors {3,5,7,11,13,17,19} see A147580
m=110592/323323 numbers with exactly 8 distinct prime divisors {3,5,7,11,13,17,19,23} see A147581

Links

Crossrefs

Cf. A060735, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[255255 x] == 92160 x, AppendTo[a, 255255 x]], {x, 1, 100}]; a

Formula

Sum_{n>=1} 1/a(n) = 1/92160. - Amiram Eldar, Dec 22 2020

Extensions

More terms from Amiram Eldar, Mar 11 2020

A147581 Numbers with exactly 8 distinct odd prime divisors {3,5,7,11,13,17,19,23}.

Original entry on oeis.org

111546435, 334639305, 557732175, 780825045, 1003917915, 1227010785, 1450103655, 1673196525, 1896289395, 2119382265, 2342475135, 2565568005, 2788660875, 3011753745, 3681032355, 3904125225, 4350310965, 5019589575, 5465775315, 5688868185, 6135053925, 6358146795
Offset: 1

Views

Author

Artur Jasinski, Nov 07 2008

Keywords

Comments

Numbers k such that phi(k)/k = m
( Family of sequences for successive n odd primes )
m=2/3 numbers with exactly 1 distinct prime divisor {3} see A000244
m=8/15 numbers with exactly 2 distinct prime divisors {3,5} see A033849
m=16/35 numbers with exactly 3 distinct prime divisors {3,5,7} see A147576
m=32/77 numbers with exactly 4 distinct prime divisors {3,5,7,11} see A147577
m=384/1001 numbers with exactly 5 distinct prime divisors {3,5,7,11,13} see A147578
m=6144/17017 numbers with exactly 6 distinct prime divisors {3,5,7,11,13,17} see A147579
m=3072/323323 numbers with exactly 7 distinct prime divisors {3,5,7,11,13,17,19} see A147580
m=110592/323323 numbers with exactly 8 distinct prime divisors {3,5,7,11,13,17,19,23} see A147581

Links

Crossrefs

Cf. A060735, A143207, A147571-A147575, A147576-A147580.

Programs

  • Mathematica
    a = {}; Do[If[EulerPhi[111546435 x] == 36495360 x, AppendTo[a, 111546435 x]], {x, 1, 100}]; a
  • Python
    from sympy import integer_log
def A147581(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 i23 in range(integer_log(x,23)[0]+1):
            for i19 in range(integer_log(x23:=x//23**i23,19)[0]+1):
                for i17 in range(integer_log(x19:=x23//19**i19,17)[0]+1):
                    for i13 in range(integer_log(x17:=x19//17**i17,13)[0]+1):
                        for i11 in range(integer_log(x13:=x17//13**i13,11)[0]+1):
                            for i7 in range(integer_log(x11:=x13//11**i11,7)[0]+1):
                                for i5 in range(integer_log(x7:=x11//7**i7,5)[0]+1):
                                    c -= integer_log(x7//5**i5,3)[0]+1
        return c
    return 111546435*bisection(f,n,n) # Chai Wah Wu, Oct 22 2024

Formula

Sum_{n>=1} 1/a(n) = 1/36495360. - Amiram Eldar, Dec 22 2020

Extensions

More terms from Amiram Eldar, Mar 11 2020
Showing 1-10 of 12 results. Next

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