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 10 results.

A003596 Numbers of the form 2^i * 11^j.

Original entry on oeis.org

1, 2, 4, 8, 11, 16, 22, 32, 44, 64, 88, 121, 128, 176, 242, 256, 352, 484, 512, 704, 968, 1024, 1331, 1408, 1936, 2048, 2662, 2816, 3872, 4096, 5324, 5632, 7744, 8192, 10648, 11264, 14641, 15488, 16384, 21296, 22528, 29282, 30976, 32768
Offset: 1

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Author

Keywords

Comments

A204455(11*a(n)) = 11, and only for these numbers. - Wolfdieter Lang, Feb 04 2012

Crossrefs

Programs

  • GAP
    Filtered([1..33000],n->PowerMod(22,n,n)=0); # Muniru A Asiru, Mar 19 2019
    
  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a003596 n = a003596_list !! (n-1)
    a003596_list = f $ singleton (1,0,0) where
       f s = y : f (insert (2 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
    
  • Magma
    [n: n in [1..2*10^5] | PrimeDivisors(n) subset [2, 11]]; // Vincenzo Librandi, Jun 27 2016
    
  • Mathematica
    fQ[n_] := PowerMod[22,n,n]==0; Select[Range[40000], fQ] (* Vincenzo Librandi, Feb 04 2012 *)
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim)\log(11),N=11^n;while(N<=lim,listput(v,N);N<<=1));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
    
  • Python
    from sympy import integer_log
    def A003596(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//11**i).bit_length() for i in range(integer_log(x,11)[0]+1))
        return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Formula

The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} mu(22*n)*x^n/(1 - x^n), where mu(n) is the Möbius function A008683. Cf. with the formula of Hanna in A051037. - Peter Bala, Mar 18 2019
Sum_{n>=1} 1/a(n) = (2*11)/((2-1)*(11-1)) = 11/5. - Amiram Eldar, Sep 23 2020
a(n) ~ exp(sqrt(2*log(2)*log(11)*n)) / sqrt(22). - Vaclav Kotesovec, Sep 23 2020

A003598 Numbers of the form 5^i * 11^j.

Original entry on oeis.org

1, 5, 11, 25, 55, 121, 125, 275, 605, 625, 1331, 1375, 3025, 3125, 6655, 6875, 14641, 15125, 15625, 33275, 34375, 73205, 75625, 78125, 161051, 166375, 171875, 366025, 378125, 390625, 805255, 831875, 859375, 1771561, 1830125, 1890625
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • GAP
    Filtered([1..2*10^6],n->PowerMod(55,n,n)=0); # Muniru A Asiru, Mar 19 2019
    
  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a003598 n = a003598_list !! (n-1)
    a003598_list = f $ singleton (1,0,0) where
       f s = y : f (insert (5 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
    
  • Magma
    [n: n in [1..2*10^6] | PrimeDivisors(n) subset [5, 11]]; // Vincenzo Librandi, Jun 27 2016
    
  • Mathematica
    Take[Union[(5^#[[1]] 11^#[[2]])&/@Tuples[Range[0,20],{2}]],50]  (* Harvey P. Dale, Dec 26 2010 *)
    fQ[n_]:=PowerMod[55, n, n] == 0; Select[Range[2*10^6], fQ] (* Vincenzo Librandi, Jun 27 2016 *)
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim)\log(11),N=11^n;while(N<=lim,listput(v,N);N*=5));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
    
  • Sage
    [n for n in (1..2*10^6) if n%55 in {0, 1, 5, 11, 15, 20, 25, 45} and all(x in {5,11} for x in prime_factors(n))] # F. Chapoton, Mar 16 2020

Formula

An asymptotic formula for a(n) is roughly 1/sqrt(55)*exp(sqrt(2*log(5)*log(11)*n)). - Benoit Cloitre, Mar 08 2002
The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} mu(55*n)*x^n/(1 - x^n), where mu(n) is the Möbius function A008683. Cf. with the formula of Hanna in A051037. - Peter Bala, Mar 18 2019
Sum_{n>=1} 1/a(n) = (5*11)/((5-1)*(11-1)) = 11/8. - Amiram Eldar, Sep 23 2020
a(n) ~ exp(sqrt(2*log(5)*log(11)*n)) / sqrt(55). - Vaclav Kotesovec, Sep 23 2020

A003597 Numbers of the form 3^i*11^j.

Original entry on oeis.org

1, 3, 9, 11, 27, 33, 81, 99, 121, 243, 297, 363, 729, 891, 1089, 1331, 2187, 2673, 3267, 3993, 6561, 8019, 9801, 11979, 14641, 19683, 24057, 29403, 35937, 43923, 59049, 72171, 88209, 107811, 131769, 161051, 177147, 216513, 264627, 323433
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • GAP
    Filtered([1..324000],n->PowerMod(33,n,n)=0); # Muniru A Asiru, Mar 19 2019
  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a003597 n = a003597_list !! (n-1)
    a003597_list = f $ singleton (1,0,0) where
       f s = y : f (insert (3 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
    
  • Magma
    [n: n in [1..4*10^5] | PrimeDivisors(n) subset [3, 11]]; // Vincenzo Librandi, Jun 27 2016
    
  • Mathematica
    fQ[n_]:=PowerMod[33, n, n] == 0; Select[Range[4*10^5], fQ] (* Vincenzo Librandi, Jun 27 2016 *)
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim)\log(11),N=11^n;while(N<=lim,listput(v,N);N*=3));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
    

Formula

The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} mu(33*n)*x^n/(1 - x^n), where mu(n) is the Möbius function A008683. Cf. with the formula of Hanna in A051037. - Peter Bala, Mar 18 2019
Sum_{n>=1} 1/a(n) = (3*11)/((3-1)*(11-1)) = 33/20. - Amiram Eldar, Sep 23 2020
a(n) ~ exp(sqrt(2*log(3)*log(11)*n)) / sqrt(33). - Vaclav Kotesovec, Sep 23 2020

A003599 Numbers of the form 7^i*11^j.

Original entry on oeis.org

1, 7, 11, 49, 77, 121, 343, 539, 847, 1331, 2401, 3773, 5929, 9317, 14641, 16807, 26411, 41503, 65219, 102487, 117649, 161051, 184877, 290521, 456533, 717409, 823543, 1127357, 1294139, 1771561, 2033647, 3195731, 5021863, 5764801
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a003599 n = a003599_list !! (n-1)
    a003599_list = f $ singleton (1,0,0) where
       f s = y : f (insert (7 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
    
  • Magma
    [n: n in [1..6*10^6] | PrimeDivisors(n) subset [7, 11]]; // Vincenzo Librandi, Jun 27 2016
  • Mathematica
    Take[Union[7^#[[1]] 11^#[[2]]&/@Tuples[Range[0,9],2]],40] (* Harvey P. Dale, Mar 11 2015 *)
    fQ[n_]:=PowerMod[77, n, n] == 0; Select[Range[6 10^6], fQ] (* Vincenzo Librandi, Jun 27 2016 *)
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim)\log(11),N=11^n;while(N<=lim,listput(v,N);N*=7));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
    

Formula

The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} mu(77*n)*x^n/(1 - x^n), where mu(n) is the Möbius function A008683. Cf. with the formula of Hanna in A051037. - Peter Bala, Mar 18 2019
Sum_{n>=1} 1/a(n) = (7*11)/((7-1)*(11-1)) = 77/60. - Amiram Eldar, Sep 23 2020
a(n) ~ exp(sqrt(2*log(7)*log(11)*n)) / sqrt(77). - Vaclav Kotesovec, Sep 23 2020

A025632 Numbers of form 7^i*10^j, with i, j >= 0.

Original entry on oeis.org

1, 7, 10, 49, 70, 100, 343, 490, 700, 1000, 2401, 3430, 4900, 7000, 10000, 16807, 24010, 34300, 49000, 70000, 100000, 117649, 168070, 240100, 343000, 490000, 700000, 823543, 1000000, 1176490, 1680700, 2401000, 3430000, 4900000, 5764801, 7000000
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a025632 n = a025632_list !! (n-1)
    a025632_list = f $ singleton (1,0,0) where
       f s = y : f (insert (7 * y, i + 1, j) $ insert (10 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
    
  • Mathematica
    n = 10^6; Flatten[Table[7^i*10^j, {i, 0, Log[7, n]}, {j, 0, Log10[n/7^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *)
  • PARI
    list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 10), N=10^n; while(N<=lim, listput(v, N); N*=7)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018

Formula

Sum_{n>=1} 1/a(n) = (7*10)/((7-1)*(10-1)) = 35/27. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(7)*log(10)*n)) / sqrt(70). - Vaclav Kotesovec, Sep 25 2020
a(n) = 7^A025671(n) * 10^A025689(n). - R. J. Mathar, Jul 06 2025

A025616 Numbers of form 3^i*10^j, with i, j >= 0.

Original entry on oeis.org

1, 3, 9, 10, 27, 30, 81, 90, 100, 243, 270, 300, 729, 810, 900, 1000, 2187, 2430, 2700, 3000, 6561, 7290, 8100, 9000, 10000, 19683, 21870, 24300, 27000, 30000, 59049, 65610, 72900, 81000, 90000, 100000, 177147, 196830, 218700, 243000, 270000
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a025616 n = a025616_list !! (n-1)
    a025616_list = f $ singleton (1,0,0) where
       f s = y : f (insert (3 * y, i + 1, j) $ insert (10 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
    
  • Mathematica
    n = 10^6; Flatten[Table[3^i*10^j, {i, 0, Log[3, n]}, {j, 0, Log10[n/3^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *)
  • PARI
    list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 10), N=10^n; while(N<=lim, listput(v, N); N*=3)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018
    
  • Python
    from sympy import integer_log
    def A025616(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(integer_log(x//10**i,3)[0]+1 for i in range(integer_log(x,10)[0]+1))
        return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Formula

Sum_{n>=1} 1/a(n) = (3*10)/((3-1)*(10-1)) = 5/3. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(3)*log(10)*n)) / sqrt(30). - Vaclav Kotesovec, Sep 25 2020
a(n) = 3^A025644(n) *10^A025685(n). - R. J. Mathar, Jul 06 2025

A108779 Numbers of the form (10^i)*(11^j), with i, j >= 0.

Original entry on oeis.org

1, 10, 11, 100, 110, 121, 1000, 1100, 1210, 1331, 10000, 11000, 12100, 13310, 14641, 100000, 110000, 121000, 133100, 146410, 161051, 1000000, 1100000, 1210000, 1331000, 1464100, 1610510, 1771561, 10000000, 11000000, 12100000, 13310000
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jun 26 2005

Keywords

Crossrefs

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a108779 n = a108779_list !! (n-1)
    a108779_list = f $ singleton (1,0,0) where
       f s = y : f (insert (10 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
             where ((y, i, j), s') = deleteFindMin s
    -- Reinhard Zumkeller, May 15 2015
  • Mathematica
    n = 10^7; Flatten[Table[10^i*11^j, {i, 0, Log10[n]}, {j, 0, Log[11, n/10^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *)

Formula

Sum_{n>=1} 1/a(n) = (10*11)/((10-1)*(11-1)) = 11/9. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(10)*log(11)*n)) / sqrt(110). - Vaclav Kotesovec, Sep 25 2020

A025691 Exponent of 10 (value of j) in n-th number of form 9^i*10^j.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Offset: 1

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Crossrefs

Cf. A025635. Differs from A002262 at a(253).

A108748 Numbers of the form (9^i)*(13^j), with i, j >= 0.

Original entry on oeis.org

1, 9, 13, 81, 117, 169, 729, 1053, 1521, 2197, 6561, 9477, 13689, 19773, 28561, 59049, 85293, 123201, 177957, 257049, 371293, 531441, 767637, 1108809, 1601613, 2313441, 3341637, 4782969, 4826809, 6908733, 9979281, 14414517, 20820969
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Jun 23 2005

Keywords

Crossrefs

Programs

  • Mathematica
    n = 10^6; Flatten[Table[9^i*13^j, {i, 0, Log[9, n]}, {j, 0, Log[13, n/9^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *)

Formula

Sum_{n>=1} 1/a(n) = (9*13)/((9-1)*(13-1)) = 39/32. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(9)*log(13)*n)) / sqrt(117). - Vaclav Kotesovec, Sep 25 2020

A025747 Index of 10^n within sequence of numbers of form 9^i*10^j.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 254, 278, 303, 329, 356, 384, 413, 443, 474, 506, 539, 573, 608, 644, 681, 719, 758, 798, 839, 881, 924, 969, 1015, 1062, 1110, 1159, 1209, 1260, 1312, 1365, 1419, 1474, 1530, 1587
Offset: 0

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Author

Keywords

Comments

Positions of zeros in A025683. - R. J. Mathar, Jul 06 2025

Crossrefs

Programs

  • Maple
    N:= 10^60: # for a(0)..a(log[10](N))
    L:= sort([seq(seq(9^i*10^j, j=0..ilog10(N/9^i)),i=0..floor(log[9](N)))]):
    select(t -> L[t] mod 9 <> 0, [$1..nops(L)]); # Robert Israel, May 12 2020

Extensions

Offset corrected by Robert Israel, May 12 2020
Showing 1-10 of 10 results.