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 117 results. Next

A071520 Number of 5-smooth numbers (A051037) <= n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 9, 10, 10, 10, 11, 12, 12, 13, 13, 14, 14, 14, 14, 15, 16, 16, 17, 17, 17, 18, 18, 19, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 24, 25, 25, 25, 25, 25, 25, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 27, 27, 28, 28
Offset: 1

Views

Author

Benoit Cloitre, Jun 02 2002

Keywords

Comments

A 5-smooth number is a number of the form 2^x*3^y*5^z (x,y,z) >= 0.

Crossrefs

Number of p-smooth numbers <= n: A070939 (p=2), A071521 (p=3), this sequence (p=5), A071604 (p=7), A071523 (p=11), A080684 (p=13), A080685 (p=17), A080686 (p=19).

Programs

  • Mathematica
    Accumulate[Table[If[Max[FactorInteger[n][[;;,1]]]<6,1,0],{n,80}]] (* Harvey P. Dale, Aug 04 2024 *)
  • PARI
    for(n=1,100,print1(sum(k=1,n,if(sum(i=4,n,if(k%prime(i),0,1)),0,1)),","))
    
  • PARI
    a(n)=-sum(k=1,n,moebius(2*3*5*k)*floor(n/k)) \\ Benoit Cloitre, Jun 14 2007
    
  • Python
    from sympy import integer_log
    def A071520(n):
        c = 0
        for i in range(integer_log(n,5)[0]+1):
            for j in range(integer_log(m:=n//5**i,3)[0]+1):
                c += (m//3**j).bit_length()
        return c # Chai Wah Wu, Sep 16 2024

Formula

a(n) = Card{ k | A051037(k) <= n }.
Asymptotically : let a = 1/(6*log(2)*log(3)*log(5)) and b = sqrt(30) then a(n) = a*log(b*n)^3 + O(log(n)).
a(n) = -Sum_{k=1,n} mu(30*k)*floor(n/k). - Benoit Cloitre, Jun 14 2007
a(n) = Sum_{i=0..floor(log_5(n))} Sum_{j=0..floor(log_3(n/5^i))} floor(log_2(2*n/(5^i*3^j))). - Ridouane Oudra, Jul 17 2020

Extensions

Title corrected by Rainer Rosenthal, Aug 30 2020

A308456 Numbers that cannot be written as a difference of 5-smooth numbers (A051037).

Original entry on oeis.org

281, 289, 353, 413, 421, 439, 443, 457, 469, 493, 541, 562, 563, 578, 581, 583, 641, 653, 661, 677, 683, 691, 701, 706, 707, 731, 733, 737, 751, 761, 769, 779, 787, 793, 803, 811, 817, 823, 826, 827, 829, 841, 842, 843, 853, 857, 867, 877, 878, 881, 883, 886
Offset: 1

Views

Author

Keywords

Comments

Terms were found by generating in sequential order the 5-smooth numbers up to some limit and collecting the differences. The first 1000 candidates k were then proved to be correct by showing that each of the following congruences holds:
{5} +- k !== {2,3} mod 205910575871,
{3} +- k !== {2,5} mod 220411358713,
{2} +- k !== {3,5} mod 3019333681,
where {a,b,...} represents the subgroup generated by a,b,... of the multiplicative subgroup modulo m. For a discussion iof this method of proof see A308247.

Examples

			281 = A308247(3) cannot be written as the difference of 5-smooth numbers. All smaller numbers can; for example, 277 = 3^4*5 - 2^7, 271 = 2^3*5^3 - 3^6.
		

Crossrefs

Cf. A051037 (5-smooth numbers).
Cf. numbers not the difference of p-smooth numbers for other values of p: A101082 (p=2), A290365 (p=3), A326318 (p=7), A326319 (p=11), A326320 (p=13).
Cf. A308247.

Programs

  • PARI
    \\ Computes the first N elements in the sequence.
    \\ At least the first 10000 are correct.
    N=100;
    \\computes the multiplicative subgroup generated
    \\by the elements of the vector L modulo m.
    SGR(L,m)={S=[1];for(l=1,length(L),z=znorder(Mod(L[l],m));T=[1];for(t=1,z,s=lift(Mod(L[l],m)^t);if(setsearch(S,s),break,T=concat(T,s);));for(t=1,length(T),S=Set(concat(S,lift(S*Mod(T[t],m))))));S}
    m1=205910575871; L1= SGR([2,3],m1); M1 = SGR([5],m1);
    m2=220411358713; L2= SGR([2,5],m2); M2 = SGR([3],m2);
    m3=  3019333681; L3= SGR([3,5],m3); M3 = SGR([2],m3);
    chkdif(k)={r=1;
       D=1;while(gcd(k/D,30)>1,D*=gcd(k/D,30));
       fordiv(D,d,
         if(vecmax(factor(k/d+1)[,1])<= 5 ,r=0);
         if(r,for(t=1,length(M1),
           if(setsearch(L1,(M1[t]+k/d)%m1),r=0;break)));
         if(r,for(t=1,length(M2),
           if(setsearch(L2,(M2[t]+k/d)%m2),r=0;break)));
         if(r,for(t=1,length(M3),
           if(setsearch(L3,(M3[t]+k/d)%m3),r=0;break)));
         if(r,for(t=1,length(M1),
           if(setsearch(L1,(M1[t]-k/d)%m1),r=0;break)));
         if(r,for(t=1,length(M2),
           if(setsearch(L2,(M2[t]-k/d)%m2),r=0;break)));
         if(r,for(t=1,length(M3),
           if(setsearch(L3,(M3[t]-k/d)%m3),r=0;break)));
         if(r==0, break)
       );
       r
    }
    for(k=1,m3,if(chkdif(k),print1(k,", ");if(N--==0, break))); print();

A188425 Position of 2^n in A051037 (5-smooth numbers).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 27, 38, 52, 68, 87, 110, 137, 167, 201, 240, 284, 332, 386, 446, 511, 582, 660, 745, 836, 934, 1041, 1155, 1277, 1407, 1545, 1692, 1849, 2015, 2190, 2376, 2571, 2777, 2994, 3222, 3461, 3711, 3974, 4249, 4535, 4834, 5147, 5473, 5811, 6164, 6531, 6911, 7306, 7716, 8142, 8582, 9039, 9511, 9999, 10503, 11024, 11563, 12119, 12692, 13283, 13892, 14519
Offset: 0

Views

Author

Zak Seidov, Mar 30 2011

Keywords

Examples

			2^0 = 1 = A051037(1), hence a(0) = 1.
a(25) = 934 because 2^25 = 33554432 = A051037(934).
		

Crossrefs

Cf. A051037, A188426, A188427 (positions of powers of 3 and 5 in A051037).

Formula

A051037(a(n)) = 2^n.
a(n) ~ c * n^3, where c = log(2)^2/(6*log(3)*log(5)) = 0.045287775775708... - Vaclav Kotesovec and Amiram Eldar, Sep 22 2024

A188426 Position of 3^n in A051037 (5-smooth numbers).

Original entry on oeis.org

1, 3, 8, 17, 31, 50, 77, 112, 157, 211, 276, 355, 448, 554, 676, 815, 973, 1149, 1344, 1561, 1801, 2064, 2351, 2663, 3003, 3371, 3766, 4190, 4647, 5135, 5657, 6212, 6802, 7430, 8094, 8797, 9539, 10322, 11147, 12014, 12925, 13881, 14883, 15933, 17031, 18178, 19375, 20624, 21925, 23281, 24690
Offset: 0

Views

Author

Zak Seidov, Mar 31 2011

Keywords

Crossrefs

Cf. A051037, A188425, A188427 (powers of 2 and 5).

Formula

A051037(a(n)) = 3^n.
a(n) ~ c * n^3, where c = log(3)^2/(6*log(2)*log(5)) = 0.180317536836709... - Vaclav Kotesovec and Amiram Eldar, Sep 22 2024

A188427 Position of 5^n in A051037 (5-smooth numbers).

Original entry on oeis.org

1, 5, 16, 37, 72, 125, 198, 295, 419, 574, 764, 991, 1259, 1572, 1931, 2342, 2808, 3332, 3918, 4569, 5288, 6078, 6943, 7887, 8913, 10024, 11224, 12515, 13902, 15387, 16975, 18668, 20470, 22385, 24416, 26567, 28840, 31239, 33768, 36429, 39226, 42164, 45244, 48470, 51846, 55375, 59061, 62907, 66917, 71094, 75441
Offset: 0

Views

Author

Zak Seidov, Mar 31 2011

Keywords

Crossrefs

Cf. A051037, A188425, A188426 (powers of 2 and 3).

Formula

A051037(a(n)) = 5^n.
a(n) ~ c * n^3, where c = log(5)^2/(6*log(2)*log(3)) = 0.566927196003501... - Vaclav Kotesovec and Amiram Eldar, Sep 22 2024

A372400 Position of 30^n among 5-smooth numbers A051037.

Original entry on oeis.org

1, 18, 83, 228, 486, 888, 1466, 2255, 3283, 4583, 6189, 8134, 10445, 13158, 16305, 19916, 24027, 28667, 33870, 39665, 46086, 53166, 60937, 69429, 78675, 88709, 99561, 111263, 123849, 137347, 151793, 167219, 183658, 201139, 219695, 239359, 260165, 282141, 305320
Offset: 0

Views

Author

Michael De Vlieger, Jun 03 2024

Keywords

Comments

Also position of 30^(n+1) in A143207.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Floor@ Log[5, 30^n/(2^i*3^j)] + 1, {i, 0, Log[2, 30^n]}, {j, 0, Log[3, 30^n/2^i]}], {n, 0, 38}]
  • PARI
    a(n)=my(t=30^n,u=5*t); sum(a=0,logint(t,5), u\=5; sum(b=0,logint(u,3), logint(u\3^b,2)+1)) \\ Charles R Greathouse IV, Sep 18 2024
  • Python
    # uses imports/function in A372401
    print(list(islice(A372401gen(p=5), 40))) # Michael S. Branicky, Jun 05 2024
    
  • Python
    from sympy import integer_log
    def A372400(n):
        c, x = 0, 30**n
        for i in range(integer_log(x,5)[0]+1):
            for j in range(integer_log(y:=x//5**i,3)[0]+1):
                c += (y//3**j).bit_length()
        return c # Chai Wah Wu, Sep 16 2024
    

Formula

a(n) = k*n^3 + (3k/2)*n^2 + O(n) where k = (log 30)^3/(6 log 2 log 3 log 5) = 5.35057081984.... - Charles R Greathouse IV, Sep 19 2024

A250089 5-smooth numbers (A051037) written in base 60, concatenating the decimal values of the sexagesimal digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 100, 104, 112, 115, 120, 121, 130, 136, 140, 148, 200, 205, 208, 215, 224, 230, 240, 242, 300, 312, 320, 336, 345, 400, 403, 410, 416, 430, 448, 500, 520, 524, 600
Offset: 1

Views

Author

Michael De Vlieger, Nov 11 2014

Keywords

Comments

Each pair of digits constitutes the decimal value of a single sexagesimal digit, as on a digital clock, eliminating the colon (:). Any leading zeros are truncated. Thus decimal 64 appears as "104" and not "0104".

Examples

			a(28) = 112 since A051037(28) = 72. 72 = 1 * 60 + 12, thus sexagesimal 1,12. Concatenating the decimal values of the sexagesimal places gives "112".
		

References

  • D. E. Knuth, Ancient Babylonian Algorithms, Communications of the ACM 15 (1972): 671-677.

Crossrefs

Programs

  • Mathematica
    a250089[n_Integer] := FromDigits /@ Map[StringJoin, If[# < 10, StringJoin["0", ToString[#]], ToString[#]] & /@ IntegerDigits[#, 60] & /@ Select[Range[n], Last@Map[First, FactorInteger@#] < 7 &], 2]; a250089[360] (* Michael De Vlieger, Nov 11 2014, after Robert G. Wilson v at A051037 *)
    With[{n = 360}, Map[FromDigits@ IntegerDigits[#, MixedRadix[ Flatten@ ConstantArray[{6, 10}, {2 Ceiling@ Log[60, n]}]]] &, Union@ Flatten@ Table[2^p1*3^p2*5^p3, {p1, 0, Log[2, n/(1)]}, {p2, 0, Log[3, n/(2^p1)]}, {p3, 0, Log[5, n/(2^p1*3^p2)]}]]] (* Version 10.2, or *)
    With[{n = 360}, FromDigits@ StringJoin@ Map[If[# < 10, StringJoin["0", ToString@ #], ToString@ #] &, IntegerDigits[#, 60]] & /@ Union@ Flatten@ Table[2^p1*3^p2*5^p3, {p1, 0, Log[2, n/(1)]}, {p2, 0, Log[3, n/(2^p1)]}, {p3, 0, Log[5, n/(2^p1*3^p2)]}]] (* Michael De Vlieger, Feb 20 2017 *)

A352219 a(n) is the least k such that A051037(n) | 60^k.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 3, 1, 3, 2, 2, 2, 2, 2, 3, 1, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 3, 4, 3, 2, 2, 3, 4, 2, 3, 2, 3, 2, 2, 5, 3, 4, 3, 3, 2, 3, 4, 2, 3, 4, 2, 4, 3, 2, 3, 5, 3, 5, 3, 3, 2, 4, 4, 4, 3, 2, 6, 3, 4, 3, 4, 3, 2, 3, 5, 3, 5
Offset: 1

Views

Author

Michael De Vlieger, Mar 08 2022

Keywords

Comments

Also, number of digits in the sexagesimal expansion of terminating unit fractions 1/A051037.

Examples

			a(1) = 0 since A051037(1) = 1 | 60^0.
a(2) = 1 since A051037(2) = 2 | 60^1; 1/2 in base 60 is represented by digit 30 after the radix point ":", i.e., :30.
a(7) = 2 since A051037(7) = 8 | 60^2; 1/8 in base 60 is :7,30, etc.
		

References

  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Chapter IX: The Representation of Numbers by Decimals, Theorem 136. 8th ed., Oxford Univ. Press, 2008, 144-145.

Crossrefs

Programs

  • Mathematica
    With[{nn = 1024}, Sort[Flatten[Table[{2^a * 3^b * 5^c, Max[Ceiling[a/2], b, c]}, {a, 0, Log2[nn]}, {b, 0, Log[3, nn/(2^a)]}, {c, 0, Log[5, nn/(2^a*3^b)]}], 2]][[All, -1]] ]

Formula

a(n) ≍ n^(1/3), with lim sup a(n)/n^(1/3) being (6*log(2)*log(3)*log(5))^(1/3)/log(3) = 1.770... where A051037(n) is a power of 3 and the lim inf being (6*log(2)*log(3)*log(5))^(1/3)/log(60) = 0.4749... where A051037(n) is a power of 60. - Charles R Greathouse IV, Mar 08 2022

A219794 First differences of 5-smooth numbers (A051037).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 2, 2, 4, 1, 2, 3, 2, 4, 4, 5, 3, 2, 4, 6, 4, 8, 3, 5, 1, 9, 6, 4, 8, 12, 5, 3, 7, 9, 6, 10, 2, 18, 12, 8, 16, 9, 15, 3, 7, 6, 14, 18, 12, 20, 4, 36, 15, 9, 16, 5, 27, 18, 30, 6, 14, 12, 28, 36, 24, 25, 15, 8, 27, 45, 9, 21, 18
Offset: 1

Views

Author

Zak Seidov, Nov 28 2012

Keywords

Comments

lim inf a(n) >= 2 by Størmer's theorem. Is lim a(n) = infinity? It would be very surprising if this were false, since then there is some k such that n and n+k are both 5-smooth for infinitely many n. - Charles R Greathouse IV, Nov 28 2012
A085152 gives all n's for which a(n) = 1. Thue-Siegel-Roth theorem gives lim a(n) = infinity. With the aid of lower bounds for linear forms in logarithms, Tijdeman showed that a(n+1)-a(n) > a(n)/(log a(n))^C for some effectively computable constant C. - Tomohiro Yamada, Apr 15 2017

Crossrefs

Programs

  • Mathematica
    Differences@ Union@ Flatten@ Table[2^i * 3^j * 5^k, {i, 0, Log2[#]}, {j, 0, Log[3, #/(2^i)]}, {k, 0, Log[5, #/(2^i*3^j)] } ] &[1000] (* Michael De Vlieger, Mar 16 2024 *)

Formula

a(n) = A051037(n+1) - A051037(n).

A353385 Irregular triangle T(n,k) with row n listing A051037(j) not divisible by 60 such that A352219(j) = n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 8, 9, 16, 18, 24, 25, 36, 40, 45, 48, 50, 72, 75, 80, 90, 100, 144, 150, 200, 225, 400, 450, 27, 32, 54, 64, 96, 108, 125, 135, 160, 192, 216, 250, 270, 288, 320, 375, 432, 500, 576, 675, 750, 800, 864, 1000, 1125, 1350, 1600
Offset: 0

Views

Author

Michael De Vlieger, Apr 15 2022

Keywords

Comments

All terms in A051037 are products T(n,k)*60^j, j >= 0.
When expressed in base 60, T(n,k) does not end in zero, yet 1/T(n,k) is a terminating fraction, regular to 60.
The first 11 terms are the proper divisors of 60.
For these reasons, the terms may be called sexagesimal "proper regular" numbers.

Examples

			For row w, plot terms m = 2^x * 3^y * 5^z at (x,y,z). Rows are labeled below the figures parenthetically for clarity. The x axis points toward the bottom right, the y axis to the bottom left, and the z axis upward. In the plot, we mark terms from previous rows by ".", and use "*" to show the origin, that is, the empty product 1:
                                                   125
                                                375   250
                                             1125   750   500
                                         3375  2250        1000
                                            6750              2000
                         25                          .           4000
                      75    50                    .     .           8000
                  225   150   100              .     .     .
                     450         200      675     .           .
                                    400     1350                 .
         5                .                          .            800
     15    10          .     .                    .     .           1600
         30    20   45     .     .              .     .     .
                      90          40      135     .           .
                                      80      270                 .
1         *                *                          *            160
       3     2          .     .                    .     .            320
          6     4    9     .     .              .     .     .
            12         18     .     8       27     .     .     .
                          36    24    16       54     .     .     .
                             72    48            108     .     .    32
                               144                  216     .    96    64
                                                       432   288   192
                                                          864   432
                                                            1728
(0)      (1)              (2)                        (3)
The terms in row w are sorted, hence row 1 has {2, 3, 4, 5, 6, 10, 12, 15, 20, 30}.
		

References

  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Chapter IX: The Representation of Numbers by Decimals, Theorem 136. 8th ed., Oxford Univ. Press, 2008, 144-145.

Crossrefs

Programs

  • Mathematica
    Block[{t, s = DeleteCases[Sort[Flatten[Table[{2^a* 3^b * 5^c, Max[Ceiling[a/2], b, c]}, {a, 0, Log2[#]}, {b, 0, Log[3, #/(2^a)]}, {c, 0, Log[5, #/(2^a*3^b)]}], 2]] &[60^3], _?(Mod[First[#], 60] == 0 &)]}, #[[1 ;; 2 + LengthWhile[Rest@ Differences[Length /@ #], # == 12 &]]] &@ Map[s[[#, 1]] &, Values@ PositionIndex[s[[All, -1]]]]] // Flatten

Formula

Row 0 contains the empty product, thus row length = 1.
For n > 0, length of row n = 12(n-1) + 10 = A017641(n-1).
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