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 38 results. Next

A169581 Positions in A002260(n) and A002024(n) when canonically enumerating A038566(n)/A038567(n), the positive rational numbers <= 1.

Original entry on oeis.org

1, 2, 4, 5, 7, 9, 11, 12, 13, 14, 16, 20, 22, 23, 24, 25, 26, 27, 29, 31, 33, 35, 37, 38, 40, 41, 43, 44, 46, 48, 52, 54, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 71, 73, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 92, 94, 96, 100, 102, 104, 106, 107, 109, 112, 113
Offset: 1

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Author

Reinhard Zumkeller, Dec 02 2009

Keywords

Comments

A038566(n) = A002260(a(n)); A038567(n) = A002024(a(n));
A054521(a(n)) = 1; complement of A169582.

A182976 Denominators of fractions with the same position in A020652/A038567 and A182972/A182973.

Original entry on oeis.org

2, 3, 5, 23, 73, 143, 163, 235, 477, 1238, 4175, 4641, 7820, 11217, 25915, 37643, 95299, 576088, 1203677
Offset: 1

Views

Author

William Rex Marshall, Dec 16 2010

Keywords

Comments

The positions of the matching fractions are given in A182974.
The numerators of the matching fractions are given in A182975.
The initial (zeroth) term of A038567 is ignored.

Examples

			The matching fractions are 1/2, 1/3, 2/5, 9/23, 30/73, 59/143 ... (this is A182975/A182976).
		

References

  • S. Cook, Problem 511: An Enumeration Problem, Journal of Recreational Mathematics, Vol. 9:2 (1976-77), 137. Solution by the Problem Editor, JRM, Vol. 10:2 (1977-78), 122-123.

Crossrefs

A054427 Permutation of natural numbers: maps the fractions A038567/A038566 to the right side (n/m > 1) of Stern-Brocot tree.

Original entry on oeis.org

1, 2, 4, 3, 8, 5, 16, 7, 6, 9, 32, 17, 64, 15, 13, 12, 10, 33, 128, 14, 11, 65, 256, 31, 25, 24, 18, 129, 512, 29, 20, 257, 1024, 63, 30, 28, 49, 48, 21, 19, 34, 513, 2048, 26, 23, 1025, 4096, 127, 61, 57, 27, 97, 96, 22, 40, 36, 66, 2049, 8192, 62, 56, 41, 35, 4097
Offset: 1

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Author

Keywords

Examples

			Right side of Stern-Brocot tree: 1/1 2/1 3/2 3/1 4/3 5/3 5/2 4/1 5/4 7/5 8/5 7/4 7/3 8/3 7/2 5/1
A038567/A038566: 1/1 2/1 3/1 3/2 4/1 4/3 5/1 5/2 5/3 5/4 6/1 6/5 7/1 7/2 7/3 7/4
		

Crossrefs

Inverse permutation: A054428.

Programs

  • Maple
    A038567_A038566_to_SternBrocot_permutation := proc(u) local a,n,i; a := []; for n from 1 to u do for i from 1 to n do if (1 = igcd(n,i)) then a := [op(a),cfrac2binexp(convert((n/i),confrac))+1]; fi; od; od; RETURN(a); end; # cfrac2binexp given in A054424.

A182974 Numbers n for which A020652(n)/A038567(n) = A182972(n)/A182973(n).

Original entry on oeis.org

1, 2, 7, 158, 1617, 6211, 8058, 16765, 69093, 465988, 5297983, 6546724, 18588348, 38244610, 204136352, 430712111, 2760559191, 100878516991, 440393924631
Offset: 1

Views

Author

William Rex Marshall, Dec 16 2010

Keywords

Comments

Numerators of the matching fractions are given in A182975.
Denominators of the matching fractions are given in A182976.
The initial (zeroth) term of A038567 is ignored.

Examples

			a(1)=1, a(2)=2 and a(3)=7 because the 1st, 2nd and 7th fractions match in the following two sequences:
1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5 ... (A020652/A038567)
1/2, 1/3, 1/4, 2/3, 1/5, 1/6, 2/5, 3/4 ... (A182972/A182973)
		

References

  • S. Cook, Problem 511: An Enumeration Problem, Journal of Recreational Mathematics, Vol. 9:2 (1976-77), 137. Solution by the Problem Editor, JRM, Vol. 10:2 (1977-78), 122-123.

Crossrefs

A182975 Numerators of fractions with the same position in A020652/A038567 and A182972/A182973.

Original entry on oeis.org

1, 1, 2, 9, 30, 59, 67, 97, 197, 513, 1729, 1922, 3239, 4646, 10734, 15592, 39474, 238623, 498579
Offset: 1

Views

Author

William Rex Marshall, Dec 16 2010

Keywords

Comments

The positions of the matching fractions are given in A182974.
The denominators of the matching fractions are given in A182976.
The initial (zeroth) term of A038567 is ignored.

Examples

			The matching fractions are 1/2, 1/3, 2/5, 9/23, 30/73, 59/143 ... (this is A182975/A182976).
		

References

  • S. Cook, Problem 511: An Enumeration Problem, Journal of Recreational Mathematics, Vol. 9:2 (1976-77), 137. Solution by the Problem Editor, JRM, Vol. 10:2 (1977-78), 122-123.

Crossrefs

A279067 Least prime q such that (r-q)/(q-p), where pA038566/A038567.

Original entry on oeis.org

5, 11, 29, 37, 6421, 367, 149, 14281, 251, 701, 521, 631, 84913, 127, 331, 75479, 787, 7057, 1949, 3407, 388621, 1847, 1277, 1087, 2879, 1399, 13859, 4621, 43391, 1657, 743507, 40213, 1151, 162209, 1973, 3491, 736577, 2579, 8039, 1264129, 14369, 43691, 4547, 4201, 8147, 29101
Offset: 1

Views

Author

Keywords

Comments

Almost a bisection of A275785 with only the term 5 being in both A279066 & A279067.
The union of A279066 & A279067 is A275785 with only 5 as a common term.
Records: 5, 11, 29, 37, 6421, 14281, 84913, 388621, 743507, 1264129, 1491377, 1613279, 15733451, 27196633, 106132883, 125747441, 304328911, 344278939, 756574061, 1166821769, 2691812749, ..., .
1/n = A179256(n).

Examples

			Row 1:        1/1                          5
Row 2:        1/2                         11
Row 3:     1/3  2/3                   29      37
Row 4:     1/4  3/4                 6421     367
Row 5: 1/5 2/5  3/5 4/5       149  14281     251
Row 6:     1/6  5/6                 521      631
Row 7: 1/7    ..    6/7  84913 127  331    75479   787 7057
Row 8: 1/8 3/8  5/8 7/8       1949 3407   388621  1847
etc.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{p = 2, q = 3, r = 5}, While[(r - q) != n(q - p), p = q; q = r; r = NextPrime@ r]; q]; Farey[n_] := Union@ Flatten@ Table[a/b, {b, n}, {a, 0, b}]; ff = Rest@ Reverse@ Sort[ Farey[25], Denominator[#2] < Denominator[#1] &]; f@# & /@ ff

A279066 Least prime q such that (q-p)/(r-q), where pA038566/A038567.

Original entry on oeis.org

5, 3, 31, 23, 8123, 89, 139, 7963, 337, 409, 199, 797, 45439, 113, 953, 88547, 293, 2633, 1933, 3643, 137029, 13381, 523, 2861, 1381, 1259, 7621, 7433, 156157, 3089, 546781, 30911, 1951, 294563, 1129, 3229, 285871, 10369, 14221, 3651341, 25819, 3967, 1669, 6173, 23473, 51383
Offset: 1

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Author

Keywords

Comments

Almost a bisection of A275785 with only the term 5 being in both A279066 & A279067.
The union of A279066 & A279067 is A275785 with only 5 as a common term.
1/n = A179210(n).
Records: 5, 31, 8123, 45439, 88547, 137029, 156157, 546781, 3651341, 11931613, 16613347, 54636251, 72510257, 102626747, 148379059, 290018137, 847428851, 1165527283, 8232085373, 32592174133, 40113962921, ..., .

Examples

			Row 1:        1/1                                       5
Row 2:        1/2                                       3
Row 3:     1/3  2/3                                 31      23
Row 4:     1/4  3/4                               8123      89
Row 5: 1/5 2/5  3/5 4/5                      139  7963     337    409
Row 6:     1/6  5/6                                199     797
Row 7:    1/7 .. 6/7                   45439 113   953   88547    293   2633
Row 8: 1/8 3/8  5/8 7/8                     1933  3643  137029  13381
etc.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{p = 2, q = 3, r = 5}, While[q != n(r - q) + p, p = q; q = r; r = NextPrime@ r]; q]; Farey[n_] := Union@ Flatten@ Table[a/b, {b, n}, {a, 0, b}]; ff = Rest@ Reverse@ Sort[ Farey[25], Denominator[#2] < Denominator[#1] &]; f@# & /@ ff

A341864 Least increasing sequence of primes a(n) == A020652(n) (mod A038567(n)).

Original entry on oeis.org

3, 7, 11, 13, 19, 31, 37, 43, 59, 61, 71, 113, 149, 157, 179, 229, 251, 257, 283, 293, 311, 379, 389, 409, 419, 421, 431, 461, 463, 467, 479, 617, 673, 751, 829, 863, 919, 953, 1009, 1021, 1033, 1069, 1097, 1123, 1151, 1171, 1237, 1277, 1291, 1409, 1423, 1489, 1607, 1621, 1973, 1987, 2027, 2087
Offset: 1

Views

Author

J. M. Bergot and Robert Israel, Feb 22 2021

Keywords

Comments

A020652/A038567 is an enumeration of the fractions < 1 (in lowest terms) arranged by increasing denominator and then increasing numerator.
a(n) is the least prime > a(n-1) congruent to A020652(n) (mod A038567(n)).

Examples

			a(5) = 19 == A020652(5) = 3 (mod A038567(5) = 4) and is the least prime > a(4) = 13 with this property.
		

Crossrefs

Programs

  • Maple
    N:= 100: # for a(1)..a(N)
    A:=Vector(N): A[1]:= 3: n:= 1:
    for d from 3 while n < N do
      for m from 1 to d-1 while n < N do
        if igcd(m,d)=1 then
          n:= n+1;
          for k from ceil((A[n-1]+1 - m)/d) do
            q:= d*k+m;
            if isprime(q) then A[n]:= q; break fi
          od
        fi
    od od:
    convert(A,list);

A343634 Number of the fraction which has (the digits of) n as repeating period in its decimal expansion, according to the canonical enumeration A038566/A038567.

Original entry on oeis.org

23, 24, 3, 25, 26, 4, 27, 28, 1, 2951, 23, 327, 2952, 2953, 328, 2954, 2955, 34, 2956, 2957, 329, 24, 2958, 330, 2959, 2960, 35, 2961, 2962, 331, 2963, 2964, 3, 2965, 2966, 36, 2967, 2968, 332, 2969, 2970, 333, 2971, 25, 37, 2972, 2973, 334, 2974, 2975, 335, 2976, 2977, 38, 26
Offset: 1

Views

Author

M. F. Hasler, Oct 18 2021

Keywords

Comments

The fraction f = n/(10^L-1), where L is the number of decimal digits of n, has the infinite decimal expansion 0.{n}{n}{n}... (with special cases n = 9, 99 etc., where f = 0.999... = 1). This sequence lists the index of this fraction, for given n, corresponding to the "canonical enumeration of positive fractions <= 1", i.e., m such that f = A038566(m) / A038567(m-1).
Inspired by Angelini's blog post, which however takes a different approach to encoding fractions (in a more "decimal" way) and to deal with n's made of digits 9 or with repetitions, such as 11, 111, or 1010, etc.

Examples

			a(n = 1) = 23 because A038566(23)/A038567(22) = 1/9, the unique fraction (in lowest terms) whose decimal expansion is 0.111..., i.e., period = (1), repeated.
a(n = 2) = 24 because A038566(24)/A038567(23) = 2/9, the unique fraction (in lowest terms) whose decimal expansion is 0.222..., i.e., period = (2), repeated.
a(n = 3) = 3 because A038566(3)/A038567(2) = 1/3, the unique fraction (in lowest terms) whose decimal expansion is 0.333..., i.e., period = (3), repeated.
a(n = 9) = 1 because A038566(1)/A038567(0) = 1/1, the unique fraction (in lowest terms) equal to 0.999..., i.e., period = (9), repeated.
a(n = 10) = 2951 because A038566(2951)/A038567(2950) = 10/99, the unique fraction (in lowest terms) whose decimal expansion is 0.1010..., i.e., period = (10), repeated.
a(11) = a(1) because the unique fraction that has decimal expansion 0.1111..., i.e., period (11) repeated, is 1/9, the same as for 0.111..., i.e., period (1), repeated.
		

Crossrefs

Programs

  • PARI
    apply( {A343634(n, d=10^(logint(n,10)+1)-1, g=gcd(n,d)) = A002088(d\g-1) - sum(k=1, n\=g, gcd(k, d) > 1) + n}, [1..55])

Formula

a(n) = m such that A038566(m)/A038567(m-1) = n/A002283(A055642(n)), where A002283(1, 2, 3, ...) = (9, 99, 999, ...) and A055642(n) = number of digits of n.

A003418 Least common multiple (or LCM) of {1, 2, ..., n} for n >= 1, a(0) = 1.

Original entry on oeis.org

1, 1, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 12252240, 232792560, 232792560, 232792560, 232792560, 5354228880, 5354228880, 26771144400, 26771144400, 80313433200, 80313433200, 2329089562800, 2329089562800
Offset: 0

Views

Author

Roland Anderson (roland.anderson(AT)swipnet.se)

Keywords

Comments

The minimal exponent of the symmetric group S_n, i.e., the least positive integer for which x^a(n)=1 for all x in S_n. - Franz Vrabec, Dec 28 2008
Product over all primes of highest power of prime less than or equal to n. a(0) = 1 by convention.
Also smallest number whose set of divisors contains an n-term arithmetic progression. - Reinhard Zumkeller, Dec 09 2002
An assertion equivalent to the Riemann hypothesis is: | log(a(n)) - n | < sqrt(n) * log(n)^2. - Lekraj Beedassy, Aug 27 2006. (This is wrong for n = 1 and n = 2. Should "for n large enough" be added? - Georgi Guninski, Oct 22 2011)
Corollary 3 of Farhi gives a proof that a(n) >= 2^(n-1). - Jonathan Vos Post, Jun 15 2009
Appears to be row products of the triangle T(n,k) = b(A010766) where b = A130087/A130086. - Mats Granvik, Jul 08 2009
Greg Martin (see link) proved that "the product of the Gamma function sampled over the set of all rational numbers in the open interval (0,1) whose denominator in lowest terms is at most n" equals (2*Pi)^(1/2)*a(n)^(-1/2). - Jonathan Vos Post, Jul 28 2009
a(n) = lcm(A188666(n), A188666(n)+1, ..., n). - Reinhard Zumkeller, Apr 25 2011
a(n+1) is the smallest integer such that all polynomials a(n+1)*(1^i + 2^i + ... + m^i) in m, for i=0,1,...,n, are polynomials with integer coefficients. - Vladimir Shevelev, Dec 23 2011
It appears that A020500(n) = a(n)/a(n-1). - Asher Auel, corrected by Bill McEachen, Apr 05 2024
n-th distinct value = A051451(n). - Matthew Vandermast, Nov 27 2009
a(n+1) = least common multiple of n-th row in A213999. - Reinhard Zumkeller, Jul 03 2012
For n > 2, (n-1) = Sum_{k=2..n} exp(a(n)*2*i*Pi/k). - Eric Desbiaux, Sep 13 2012
First column minus second column of A027446. - Eric Desbiaux, Mar 29 2013
For n > 0, a(n) is the smallest number k such that n is the n-th divisor of k. - Michel Lagneau, Apr 24 2014
Slowest growing integer > 0 in Z converging to 0 in Z^ when considered as profinite integer. - Herbert Eberle, May 01 2016
What is the largest number of consecutive terms that are all equal? I found 112 equal terms from a(370261) to a(370372). - Dmitry Kamenetsky, May 05 2019
Answer: there exist arbitrarily long sequences of consecutive terms with the same value; also, the maximal run of consecutive terms with different values is 5 from a(1) to a(5) (see link Roger B. Eggleton). - Bernard Schott, Aug 07 2019
Related to the inequality (54) in Ramanujan's paper about highly composite numbers A002182, also used in A199337: a(A329570(m))^2 is a (not minimal) bound above which all highly composite numbers are divisible by m, according to the right part of that inequality. - M. F. Hasler, Jan 04 2020
For n > 2, a(n) is of the form 2^e_1 * p_2^e_2 * ... * p_m^e_m, where e_m = 1 and e = floor(log_2(p_m)) <= e_1. Therefore, 2^e * p_m^e_m is a primitive Zumkeler number (A180332). Therefore, 2^e_1 * p_m^e_m is a Zumkeller number (A083207). Therefore, for n > 2, a(n) = 2^e_1 * p_m^e_m * r, where r is relatively prime to 2*p_m, is a Zumkeller number (see my proof at A002182 for details). - Ivan N. Ianakiev, May 10 2020
For n > 1, 2|(a(n)+2) ... n|(a(n)+n), so a(n)+2 .. a(n)+n are all composite and (part of) a prime gap of at least n. (Compare n!+2 .. n!+n). - Stephen E. Witham, Oct 09 2021

Examples

			LCM of {1,2,3,4,5,6} = 60. The primes up to 6 are 2, 3 and 5. floor(log(6)/log(2)) = 2 so the exponent of 2 is 2.
floor(log(6)/log(3)) = 1 so the exponent of 3 is 1.
floor(log(6)/log(5)) = 1 so the exponent of 5 is 1. Therefore, a(6) = 2^2 * 3^1 * 5^1 = 60. - _David A. Corneth_, Jun 02 2017
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 365.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row products of A133233.
Cf. A025528 (number of prime factors of a(n) with multiplicity).
Cf. A275120 (lengths of runs of consecutive equal terms), A276781 (ordinal transform from term a(1)=1 onward).

Programs

  • Haskell
    a003418 = foldl lcm 1 . enumFromTo 2
    -- Reinhard Zumkeller, Apr 04 2012, Apr 25 2011
    
  • Magma
    [1] cat [Exponent(SymmetricGroup(n)) : n in [1..28]]; // Arkadiusz Wesolowski, Sep 10 2013
    
  • Magma
    [Lcm([1..n]): n in [0..30]]; // Bruno Berselli, Feb 06 2015
    
  • Maple
    A003418 := n-> lcm(seq(i,i=1..n));
    HalfFarey := proc(n) local a,b,c,d,k,s; a := 0; b := 1; c := 1; d := n; s := NULL; do k := iquo(n + b, d); a, b, c, d := c, d, k*c - a, k*d - b; if 2*a > b then break fi; s := s,(a/b); od: [s] end: LCM := proc(n) local i; (1/2)*mul(2*sin(Pi*i),i=HalfFarey(n))^2 end: # Peter Luschny
    # next Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, ilcm(n, a(n-1))) end:
    seq(a(n), n=0..33);  # Alois P. Heinz, Jun 10 2021
  • Mathematica
    Table[LCM @@ Range[n], {n, 1, 40}] (* Stefan Steinerberger, Apr 01 2006 *)
    FoldList[ LCM, 1, Range@ 28]
    A003418[0] := 1; A003418[1] := 1; A003418[n_] := A003418[n] = LCM[n,A003418[n-1]]; (* Enrique Pérez Herrero, Jan 08 2011 *)
    Table[Product[Prime[i]^Floor[Log[Prime[i], n]], {i, PrimePi[n]}], {n, 0, 28}] (* Wei Zhou, Jun 25 2011 *)
    Table[Product[Cyclotomic[n, 1], {n, 2, m}], {m, 0, 28}] (* Fred Daniel Kline, May 22 2014 *)
    a1[n_] := 1/12 (Pi^2+3(-1)^n (PolyGamma[1,1+n/2] - PolyGamma[1,(1+n)/2])) // Simplify
    a[n_] := Denominator[Sqrt[a1[n]]];
    Table[If[IntegerQ[a[n]], a[n], a[n]*(a[n])[[2]]], {n, 0, 28}] (* Gerry Martens, Apr 07 2018 [Corrected by Vaclav Kotesovec, Jul 16 2021] *)
  • PARI
    a(n)=local(t); t=n>=0; forprime(p=2,n,t*=p^(log(n)\log(p))); t
    
  • PARI
    a(n)=if(n<1,n==0,1/content(vector(n,k,1/k)))
    
  • PARI
    a(n)=my(v=primes(primepi(n)),k=sqrtint(n),L=log(n+.5));prod(i=1,#v,if(v[i]>k,v[i],v[i]^(L\log(v[i])))) \\ Charles R Greathouse IV, Dec 21 2011
    
  • PARI
    a(n)=lcm(vector(n,i,i)) \\ Bill Allombert, Apr 18 2012 [via Charles R Greathouse IV]
    
  • PARI
    n=1; lim=100; i=1; j=1; until(n==lim, a=lcm(j,i+1); i++; j=a; n++; print(n" "a);); \\ Mike Winkler, Sep 07 2013
    
  • Python
    from functools import reduce
    from operator import mul
    from sympy import sieve
    def integerlog(n,b): # find largest integer k>=0 such that b^k <= n
        kmin, kmax = 0,1
        while b**kmax <= n:
            kmax *= 2
        while True:
            kmid = (kmax+kmin)//2
            if b**kmid > n:
                kmax = kmid
            else:
                kmin = kmid
            if kmax-kmin <= 1:
                break
        return kmin
    def A003418(n):
        return reduce(mul,(p**integerlog(n,p) for p in sieve.primerange(1,n+1)),1) # Chai Wah Wu, Mar 13 2021
    
  • Python
    # generates initial segment of sequence
    from math import gcd
    from itertools import accumulate
    def lcm(a, b): return a * b // gcd(a, b)
    def aupton(nn): return [1] + list(accumulate(range(1, nn+1), lcm))
    print(aupton(30)) # Michael S. Branicky, Jun 10 2021
  • Sage
    [lcm(range(1,n)) for n in range(1, 30)] # Zerinvary Lajos, Jun 06 2009
    
  • Scheme
    (define (A003418 n) (let loop ((n n) (m 1)) (if (zero? n) m (loop (- n 1) (lcm m n))))) ;; Antti Karttunen, Jan 03 2018
    

Formula

The prime number theorem implies that lcm(1,2,...,n) = exp(n(1+o(1))) as n -> infinity. In other words, log(lcm(1,2,...,n))/n -> 1 as n -> infinity. - Jonathan Sondow, Jan 17 2005
a(n) = Product (p^(floor(log n/log p))), where p runs through primes not exceeding n (i.e., primes 2 through A007917(n)). - Lekraj Beedassy, Jul 27 2004
Greg Martin showed that a(n) = lcm(1,2,3,...,n) = Product_{i = Farey(n), 0 < i < 1} 2*Pi/Gamma(i)^2. This can be rewritten (for n > 1) as a(n) = (1/2)*(Product_{i = Farey(n), 0 < i <= 1/2} 2*sin(i*Pi))^2. - Peter Luschny, Aug 08 2009
Recursive formula useful for computations: a(0)=1; a(1)=1; a(n)=lcm(n,a(n-1)). - Enrique Pérez Herrero, Jan 08 2011
From Enrique Pérez Herrero, Jun 01 2011: (Start)
a(n)/a(n-1) = A014963(n).
if n is a prime power p^k then a(n)=a(p^k)=p*a(n-1), otherwise a(n)=a(n-1).
a(n) = Product_{k=2..n} (1 + (A007947(k)-1)*floor(1/A001221(k))), for n > 1. (End)
a(n) = A079542(n+1, 2) for n > 1.
a(n) = exp(Sum_{k=1..n} Sum_{d|k} moebius(d)*log(k/d)). - Peter Luschny, Sep 01 2012
a(n) = A025529(n) - A027457(n). - Eric Desbiaux, Mar 14 2013
a(n) = exp(Psi(n)) = 2 * Product_{k=2..A002088(n)} (1 - exp(2*Pi*i * A038566(k+1) / A038567(k))), where i is the imaginary unit, and Psi the second Chebyshev's function. - Eric Desbiaux, Aug 13 2014
a(n) = A064446(n)*A038610(n). - Anthony Browne, Jun 16 2016
a(n) = A000142(n) / A025527(n) = A000793(n) * A225558(n). - Antti Karttunen, Jun 02 2017
log(a(n)) = Sum_{k>=1} (A309229(n, k)/k - 1/k). - Mats Granvik, Aug 10 2019
From Petros Hadjicostas, Jul 24 2020: (Start)
Nair (1982) proved that 2^n <= a(n) <= 4^n for n >= 9. See also Farhi (2009). Nair also proved that
a(n) = lcm(m*binomial(n,m): 1 <= m <= n) and
a(n) = gcd(a(m)*binomial(n,m): n/2 <= m <= n). (End)
Sum_{n>=1} 1/a(n) = A064859. - Bernard Schott, Aug 24 2020
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