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 35 results. Next

A025537 a(n) = (1/s(1) + 1/s(2) + ... + 1/s(n+1)) * LCM{1, 2, ..., n}, where s(k) = LCM{1,2,...,k}/k = A002944(k).

Original entry on oeis.org

1, 2, 5, 17, 35, 181, 182, 1278, 2559, 7687, 7688, 84580, 84581, 1099567, 1099582, 1099590, 2199181, 37386095, 37386096, 710335844, 710335865, 710335887, 710335888, 16337725448, 16337725453, 81688627291, 81688627300, 245065881928, 245065881929
Offset: 0

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Author

Keywords

Examples

			n=4: LCM{1,2,3,4} = 12, so a(4) = 12*(1/1 + 1/1 + 1/2 + 1/3 + 1/12) = 12*35/12 = 35. - _N. J. A. Sloane_, Sep 04 2019
		

Crossrefs

Programs

  • PARI
    s(n) = lcm([1..n])/n; \\ A002944
    a(n) = lcm([1..n])*sum(k=1, n+1, 1/s(k)); \\ Michel Marcus, Sep 04 2019

Formula

a(n) = A003418(n) * Sum_{k=1..n+1} 1/A002944(k). - Sean A. Irvine, Sep 04 2019

Extensions

Name improved by Sean A. Irvine, Sep 04 2019 and N. J. A. Sloane, Sep 04 2019

A081529 Duplicate of A002944.

Original entry on oeis.org

1, 1, 2, 3, 12, 10, 60, 105, 280, 252, 2520, 2310, 27720, 25740, 24024, 45045, 720720, 680680, 12252240, 11639628, 11085360, 10581480, 232792560, 223092870, 1070845776, 1029659400, 2974571600, 2868336900
Offset: 1

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Author

Keywords

Crossrefs

Cf. A003418.

Formula

Equals A003418(n) / n - Martin Fuller, Jan 03 2006

Extensions

More terms from Martin Fuller, Jan 03 2006

A092593 a(n) is the smallest number k > 1 for which A001142(k)/A002944(k+1)^n is an integer.

Original entry on oeis.org

2, 3, 9, 9, 15, 15, 38, 45, 45, 45, 61, 61, 225, 225, 225, 225, 225, 225, 225, 225, 225, 225, 635, 635, 1545, 1545, 1545, 1545, 2137, 2137, 2137, 2137, 2137, 2137, 2137, 2137, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660
Offset: 1

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Author

Labos Elemer, Mar 10 2004

Keywords

Comments

a(62) > 12500. - Robert Israel, Jan 24 2019

Examples

			n=4, A001142(9) = 1*9*36*...*9*1 = 11759522374656,
A002944(10) = lcm(1,2,...,10)/10=252 and A001142(9) = 2916*(252^4) = 11759522374656,
so a(4)=9, the smallest relevant number.
		

Crossrefs

Programs

  • Maple
    A001142:= proc(n) option remember; procname(n-1)*n^(n-1)/(n-1)! end proc:
    A001142(0):= 1:
    A002944:= proc(n) option remember; ilcm(n,procname(n-1)*(n-1))/n end proc:
    A002944(1):= 1:
    f:= proc(n) option remember; local k;
    for k from procname(n-1) do
       if type(A001142(k)/A002944(k+1)^n, integer) then return k fi
    od
    end proc:
    f(1):= 2:
    map(f, [$1..61]); # Robert Israel, Jan 23 2019
  • Mathematica
    Table[fla=1;Do[s1=Apply[Times, Table[Binomial[n, j], {j, 0, n}]]; s2=Apply[LCM, Table[Binomial[n, j], {j, 0, n}]]; If[IntegerQ[s1/(s2^k)]&&!Equal[n, 1]&&Equal[fla, 1], Print[{n, k}];fla=0], {n, 1, 230}], {k, 1, 25}]

Extensions

Corrected and extended by Robert Israel, Jan 23 2019

A025538 a(n) = (1/s(1) - 1/s(2) + ... + d/s(n+1)) * LCM{1, 2, ..., n}, where d = (-1)^n, s = A002944, i.e., s(k) = LCM of row k of Pascal's triangle.

Original entry on oeis.org

1, 0, 1, 1, 3, 9, 10, 66, 135, 395, 396, 4344, 4345, 56471, 56486, 56478, 112957, 1920251, 1920252, 36484768, 36484789, 36484767, 36484768, 839149640, 839149645, 4195748199, 4195748208, 12587244596, 12587244597, 365030093283, 365030093284
Offset: 0

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Author

Keywords

Formula

a(n) = A003418(n) * Sum_{k=1..n+1} (-1)^(k+1)/A002944(k). - Sean A. Irvine, Sep 04 2019

Extensions

Title improved by Sean A. Irvine, Sep 04 2019

A056611 Quotient: squarefree kernel of A002944(n) divided by that of A001405.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 3, 1, 1, 5, 5, 5, 5, 7, 7, 7, 7, 21, 21, 15, 5, 55, 165, 33, 33, 143, 143, 1001, 1001, 1001, 1001, 91, 91, 221, 221, 221, 221, 323, 323, 323, 323, 2261, 2261, 24871, 24871, 572033, 572033, 81719, 81719, 24035, 24035, 312455, 312455, 85215
Offset: 1

Views

Author

Labos Elemer, Aug 07 2000

Keywords

Examples

			n=14,LCM[1,14,....,14,1]=24024,its sqf kernel is 6006; binomial[14,7]=3432,its sqf kernel is 858.a(14)=6006/858=7
		

Crossrefs

Formula

A092592 a(n) = A001142(n)/A002944(n), i.e., the product of C(n,j) binomial coefficients (for j=0..n) is divided by the least common multiple of them.

Original entry on oeis.org

1, 1, 3, 8, 250, 2700, 252105, 39337984, 46664771328, 12859560000000, 100271725559775000, 155537541908508672000, 8147102459616398435002560, 1112629444624748831692923019008, 198709463856994388404947123486328125
Offset: 1

Views

Author

Labos Elemer, Mar 10 2004

Keywords

Crossrefs

Programs

  • PARI
    a(n) = my(v = vector(n+1, j, binomial(n, j-1))); prod(k=1, #v, v[k])/ lcm(v); \\ Michel Marcus, Jul 29 2017

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

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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

A276781 a(n) = 1+n-(nearest power of prime <= n); for n > 1, a(n) = minimal b such that the numbers binomial(n,k) for b <= k <= n-b have a common divisor greater than 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 1, 2
Offset: 1

Views

Author

N. J. A. Sloane, Sep 29 2016, following a suggestion from Eric Desbiaux

Keywords

Comments

The definition in the video has "b < k < n-b" rather than "b <= k <= n-b", but that appears to be a typographical error.
From Antti Karttunen, Jan 21 2020: (Start)
a(n) = 1 if n is a power of prime (term of A000961), otherwise a(n) is one more than the distance to the nearest preceding prime power.
For n > 1, a(n) indicates the maximum region on the row n of Pascal's triangle (A007318) such that binomial terms C(n,a(n)) .. C(n,n-a(n)) all share a common prime factor. Because for all prime powers, p^k, the binomial terms C(p^k,1) .. C(p^k,p^k-1) have p as their prime factor, we have a(A000961(n)) = 1 for all n, while for each successive n that is not a prime power, the region of shared prime factor shrinks one step more towards the center of the triangle. From this follows that this is the ordinal transform of A025528 (equally, of A065515, or of A003418(n) from n >= 1 onward), equivalent to the simple definition given above.
(End)

Examples

			Row 6 of Pascal's triangle is 1,6,15,20,15,6,1 and [15,20,15] have a common divisor of 5. Since 15 = binomial(6,2), a(6)=2.
		

Crossrefs

Cf. A007318, A010055, A276782 (positions of records), A000961 (positions of ones), A024619 (positions of terms > 1).

Programs

  • Maple
    mygcd:=proc(lis) local i,g,m;
    m:=nops(lis); g:=lis[1];
    for i from 2 to m do g:=gcd(g,lis[i]); od:
    g; end;
    f:=proc(n) local b,lis; global mygcd;
    for b from floor(n/2) by -1 to 1 do
    lis:=[seq(binomial(n,i),i=b..n-b)];
    if mygcd(lis)=1 then break; fi; od:
    b+1;
    end;
    [seq(f(n),n=2..120)];
  • Mathematica
    Table[b = 1; While[GCD @@ Map[Binomial[n, #] &, Range[b, n - b]] == 1, b++]; b, {n, 92}] (* Michael De Vlieger, Oct 03 2016 *)
  • PARI
    A276781(n) = if(1==n,1,forstep(k=n,1,-1,if(isprimepower(k),return(1+n-k)))); \\ Antti Karttunen, Jan 21 2020
    
  • Python
    from sympy import factorint
    def A276781(n): return 1+n-next(filter(lambda m:len(factorint(m))<=1, range(n,0,-1))) # Chai Wah Wu, Oct 25 2024

Formula

If A010055(n) == 1, a(n) = 1, otherwise a(n) = 1 + a(n-1). - Antti Karttunen, Jan 21 2020

Extensions

Term a(1) = 1 prepended and alternative simpler definition added to the name by Antti Karttunen, Jan 20 2020

A025527 a(n) = n!/lcm{1,2,...,n} = (n-1)!/lcm{C(n-1,0), C(n-1,1), ..., C(n-1,n-1)}.

Original entry on oeis.org

1, 1, 1, 2, 2, 12, 12, 48, 144, 1440, 1440, 17280, 17280, 241920, 3628800, 29030400, 29030400, 522547200, 522547200, 10450944000, 219469824000, 4828336128000, 4828336128000, 115880067072000, 579400335360000, 15064408719360000
Offset: 1

Views

Author

Clark Kimberling, Dec 11 1999

Keywords

Comments

a(n) = a(n-1) iff n is prime. Thus a(1)=a(2)=a(3)=1 is the only triple in this sequence. - Franz Vrabec, Sep 10 2005
a(k) = a(k+1) for k in A006093. - Lekraj Beedassy, Aug 03 2006
Partial products of A048671. - Peter Luschny, Sep 09 2009

Examples

			a(5) = 2 as 5!/lcm(1..5) = 120/60 = 2.
		

Crossrefs

Programs

Formula

a(n) = A000142(n)/A003418(n) = A000254(n)/A025529(n). - Franz Vrabec, Sep 13 2005
log a(n) = n log n - 2n + O(n/log^4 n). (The error term can be improved. On the Riemann Hypothesis it is O(n^k) for any k > 1/2.) - Charles R Greathouse IV, Oct 16 2012
a(n) = A205957(n), 1 <= n <= 11. - Daniel Forgues, Apr 22 2014
Conjecture: a(A006093(n)) = phi(A000142(A006093(n))) / phi(A003418(A006093(n))), where phi is the Euler totient function. - Fred Daniel Kline, Jun 03 2017

A048671 a(n) is the least common multiple of the proper divisors of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 6, 1, 4, 3, 10, 1, 12, 1, 14, 15, 8, 1, 18, 1, 20, 21, 22, 1, 24, 5, 26, 9, 28, 1, 30, 1, 16, 33, 34, 35, 36, 1, 38, 39, 40, 1, 42, 1, 44, 45, 46, 1, 48, 7, 50, 51, 52, 1, 54, 55, 56, 57, 58, 1, 60, 1, 62, 63, 32, 65, 66, 1, 68, 69, 70, 1, 72, 1, 74, 75, 76, 77, 78, 1
Offset: 1

Views

Author

Keywords

Comments

A proper divisor d of n is a divisor of n such that 1 <= d < n.
Previous name was: a(n) = q(n)/q(n-1), where q(n) = n!/A003418(n).

Examples

			8!/lcm(8) = 48 = 40320/840 while 7!/lcm(7) = 5040/420 = 12 so a(8) = 48/12 = 4.
a(5) = 1 = lcm(1,2,3,4,5)/lcm(1,5,10,10,5,1).
		

Crossrefs

Cf. A182936 gives the dual (greatest common divisor).

Programs

Formula

a(n) = A025527(n)/A025527(n-1).
a(n) = (n*A003418(n-1))/A003418(n).
a(n) = A003418(n-1)/A002944(n). [corrected by Michel Marcus, May 18 2020]
From Henry Bottomley, May 19 2000: (Start)
a(n) = n/A014963(n) = lcm(A052126(n), A032742(n)).
a(n) = n if n not a prime power, a(n) = n/p if n = p^m (i.e., a(n) = 1 if n = p). (End)
From Vladeta Jovovic, Jul 04 2002: (Start)
a(n) = n*Product_{d | n} d^mu(d).
Product_{d | n} a(d) = A007956(n). (End)
a(n) = Product_{k=1..n-1} if(gcd(n, k) > 1, 1 - exp(2*pi*i*k/n), 1), where i = sqrt(-1). - Paul Barry, Apr 15 2005
From Peter Luschny, Jun 09 2011: (Start)
a(n) = Product_{k=1..n-1} if(gcd(k,n) > 1, 2*Pi/Gamma(k/n)^2, 1).
a(n) = Product_{k=1..n-1} if(gcd(k,n) > 1, 2*sin(Pi*k/n), 1). (End)

Extensions

New definition based on a comment of David Wasserman by Peter Luschny, Mar 23 2011
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