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.

A214075 Triangle read by rows: T(n,k) = floor(A213998(n,k) / A213999(n,k)), 0<=k<=n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 2, 0, 1, 4, 7, 6, 2, 0, 1, 5, 12, 14, 8, 2, 0, 1, 6, 17, 26, 22, 11, 2, 0, 1, 7, 24, 44, 49, 34, 13, 2, 0, 1, 8, 31, 68, 93, 83, 47, 16, 2, 0, 1, 9, 40, 100, 162, 177, 131, 64, 19, 2, 0, 1, 10, 49, 140, 263, 340, 309
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 03 2012

Keywords

Comments

T(n,0) = 1;
T(n,1) = n - 1 for n > 0, cf. A001477;
T(n,2) = A074148(n-2) for n > 2;
T(n,n-2) = A022819(n) for n > 1;
T(n,n-1) = A055980(n) for n > 0;
T(n,n) = A000007(n).

Examples

			Start of triangle preceded by triangle "A213998/A213999":
. 0:                      1                                 1
. 1:                   1   1/2                             1  0
. 2:               1    3/2   1/3                         1  1  0
. 3:            1   5/2    11/6   1/4                    1 2  1  0
. 4:        1   7/2   13/3    25/12   1/5               1 3  4  2 0
. 5:     1   9/2   47/6   77/12   137/60   1/6         1 4  7  6 2 0
. 6:  1  11/2   37/3   57/4   87/10   49/20   1/7,    1 5 12 14 8 2 0.
		

Programs

  • Haskell
    a214075 n k = a214075_tabl !! n !! k
    a214075_row n = a214075_tabl !! n
    a214075_tabl = zipWith (zipWith div) a213998_tabl a213999_tabl

A001008 a(n) = numerator of harmonic number H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 3, 11, 25, 137, 49, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 55835135, 18858053, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 315404588903, 9227046511387
Offset: 1

Views

Author

Keywords

Comments

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
By Wolstenholme's theorem, p^2 divides a(p-1) for all primes p > 3.
From Alexander Adamchuk, Dec 11 2006: (Start)
p divides a(p^2-1) for all primes p > 3.
p divides a((p-1)/2) for primes p in A001220.
p divides a((p+1)/2) or a((p-3)/2) for primes p in A125854.
a(n) is prime for n in A056903. Corresponding primes are given by A067657. (End)
a(n+1) is the numerator of the polynomial A[1, n](1) where the polynomial A[genus 1, level n](m) is defined to be Sum_{d = 1..n - 1} m^(n - d)/d. (See the Mathematica procedure generating A[1, n](m) below.) - Artur Jasinski, Oct 16 2008
Better solutions to the card stacking problem have been found by M. Paterson and U. Zwick (see link). - Hugo Pfoertner, Jan 01 2012
a(n) = A213999(n, n-1). - Reinhard Zumkeller, Jul 03 2012
a(n) coincides with A175441(n) if and only if n is not from the sequence A256102. The quotient a(n) / A175441(n) for n in A256102 is given as corresponding entry of A256103. - Wolfdieter Lang, Apr 23 2015
For a very short proof that the Harmonic series diverges, see the Goldmakher link. - N. J. A. Sloane, Nov 09 2015
All terms are odd while corresponding denominators (A002805) are all even for n > 1 (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

Examples

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ].
Coincidences with A175441: the first 19 entries coincide because 20 is the first entry of A256102. Indeed, a(20)/A175441(20) = 55835135 / 11167027 = 5 = A256103(1). - _Wolfdieter Lang_, Apr 23 2015
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
  • H. W. Gould, Combinatorial Identities, Morgantown Printing and Binding Co., 1972, # 1.45, page 6, #3.122, page 36.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, page 347.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A145609-A145640. - Artur Jasinski, Oct 16 2008
Cf. A003506. - Paul Curtz, Nov 30 2013
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358.
Cf. A195505.

Programs

  • GAP
    List([1..30],n->NumeratorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018
  • Haskell
    import Data.Ratio ((%), numerator)
    a001008 = numerator . sum . map (1 %) . enumFromTo 1
    a001008_list = map numerator $ scanl1 (+) $ map (1 %) [1..]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Numerator(HarmonicNumber(n)): n in [1..30]]; // Bruno Berselli, Feb 17 2016
    
  • Maple
    A001008 := proc(n)
        add(1/k,k=1..n) ;
        numer(%) ;
    end proc:
    seq( A001008(n),n=1..40) ; # Zerinvary Lajos, Mar 28 2007; R. J. Mathar, Dec 02 2016
  • Mathematica
    Table[Numerator[HarmonicNumber[n]], {n, 30}]
    (* Procedure generating A[1,n](m) (see Comments section) *) m =1; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, k], {r, 1, 20}]; aa (* Artur Jasinski, Oct 16 2008 *)
    Numerator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
    Numerator[Table[((n - 1)/2)*HypergeometricPFQ[{1, 1, 2 - n}, {2, 3}, 1] + 1, {n, 1, 29}]] (* Artur Jasinski, Jan 08 2021 *)
  • PARI
    A001008(n) = numerator(sum(i=1,n,1/i)) \\ Michael B. Porter, Dec 08 2009
    
  • PARI
    H1008=List(1); A001008(n)={for(k=#H1008,n-1,listput(H1008,H1008[k]+1/(k+1))); numerator(H1008[n])} \\ about 100x faster for n=1..1500. - M. F. Hasler, Jul 03 2019
    
  • Python
    from sympy import Integer
    [sum(1/Integer(i) for i in range(1, n + 1)).numerator() for n in range(1, 31)]  # Indranil Ghosh, Mar 23 2017
    
  • Sage
    def harmonic(a, b):  # See the F. Johansson link.
        if b - a == 1:
            return 1, a
        m = (a+b)//2
        p, q = harmonic(a,m)
        r, s = harmonic(m,b)
        return p*s+q*r, q*s
    def A001008(n): H = harmonic(1,n+1); return numerator(H[0]/H[1])
    [A001008(n) for n in (1..29)] # Peter Luschny, Sep 01 2012
    

Formula

H(n) ~ log n + gamma + O(1/n). [See Hardy and Wright, Th. 422.]
log n + gamma - 1/n < H(n) < log n + gamma + 1/n [follows easily from Hardy and Wright, Th. 422]. - David Applegate and N. J. A. Sloane, Oct 14 2008
G.f. for H(n): log(1-x)/(x-1). - Benoit Cloitre, Jun 15 2003
H(n) = sqrt(Sum_{i = 1..n} Sum_{j = 1..n} 1/(i*j)). - Alexander Adamchuk, Oct 24 2004
a(n) is the numerator of Gamma/n + Psi(1 + n)/n = Gamma + Psi(n), where Psi is the digamma function. - Artur Jasinski, Nov 02 2008
H(n) = 3/2 + 2*Sum_{k = 0..n-3} binomial(k+2, 2)/((n-2-k)*(n-1)*n), n > 1. - Gary Detlefs, Aug 02 2011
H(n) = (-1)^(n-1)*(n+1)*n*Sum_{k = 0..n-1} k!*Stirling2(n-1, k) * Stirling1(n+k+1,n+1)/(n+k+1)!. - Vladimir Kruchinin, Feb 05 2013
H(n) = n*Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)/(k+1)^2. (Wenchang Chu) - Gary Detlefs, Apr 13 2013
H(n) = (1/2)*Sum_{k = 1..n} (-1)^(k-1)*binomial(n,k)*binomial(n+k, k)/k. (H. W. Gould) - Gary Detlefs, Apr 13 2013
E.g.f. for H(n) = a(n)/A002805(n): (gamma + log(x) - Ei(-x)) * exp(x), where gamma is the Euler-Mascheroni constant, and Ei(x) is the exponential integral. - Vladimir Reshetnikov, Apr 24 2013
H(n) = residue((psi(-s)+gamma)^2/2, {s, n}) where psi is the digamma function and gamma is the Euler-Mascheroni constant. - Jean-François Alcover, Feb 19 2014
H(n) = Sum_{m >= 1} n/(m^2 + n*m) = gamma + digamma(1+n), numerators and denominators. (see Mathworld link on Digamma). - Richard R. Forberg, Jan 18 2015
H(n) = (1/2) Sum_{j >= 1} Sum_{k = 1..n} ((1 - 2*k + 2*n)/((-1 + k + j*n)*(k + j*n))) + log(n) + 1/(2*n). - Dimitri Papadopoulos, Jan 13 2016
H(n) = (n!)^2*Sum_{k = 1..n} 1/(k*(n-k)!*(n+k)!). - Vladimir Kruchinin, Mar 31 2016
a(n) = Stirling1(n+1, 2) / gcd(Stirling1(n+1, 2), n!) = A000254(n) / gcd(A000254(n), n!). - Max Alekseyev, Mar 01 2018
From Peter Bala, Jan 31 2019: (Start)
H(n) = 1 + (1 + 1/2)*(n-1)/(n+1) + (1/2 + 1/3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3 + 1/4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
H(n)/n = 1 + (1/2^2 - 1)*(n-1)/(n+1) + (1/3^2 - 1/2^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^2 - 1/3^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... .
For odd n >= 3, (1/2)*H((n-1)/2) = (n-1)/(n+1) + (1/2)*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ... . Cf. A195505. See the Bala link in A036970. (End)
H(n) = ((n-1)/2) * hypergeom([1,1,2-n], [2,3], 1) + 1. - Artur Jasinski, Jan 08 2021
Conjecture: for nonzero m, H(n) = (1/m)*Sum_{k = 1..n} ((-1)^(k+1)/k) * binomial(m*k,k)*binomial(n+(m-1)*k,n-k). The case m = 1 is well-known; the case m = 2 is given above by Detlefs (dated Apr 13 2013). - Peter Bala, Mar 04 2022
a(n) = the (reduced) numerator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024
H(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)/k (H. W. Gould). - Gary Detlefs, May 28 2024

Extensions

Edited by Max Alekseyev, Oct 21 2011
Changed title, deleting the incorrect name "Wolstenholme numbers" which conflicted with the definition of the latter in both Weisstein's World of Mathematics and in Wikipedia, as well as with OEIS A007406. - Stanislav Sykora, Mar 25 2016

A002805 Denominators of harmonic numbers H(n) = Sum_{i=1..n} 1/i.

Original entry on oeis.org

1, 2, 6, 12, 60, 20, 140, 280, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 4084080, 77597520, 15519504, 5173168, 5173168, 118982864, 356948592, 8923714800, 8923714800, 80313433200, 80313433200, 2329089562800, 2329089562800, 72201776446800
Offset: 1

Views

Author

Keywords

Comments

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
If n is not in {1, 2, 6} then a(n) has at least one prime factor other than 2 or 5. E.g., a(5) = 60 has a prime factor 3 and a(7) = 140 has a prime factor 7. This implies that every H(n) = A001008(n)/A002805(n), n not from {1, 2, 6}, has an infinite decimal representation. For a proof see the J. Havil reference. - Wolfdieter Lang, Jun 29 2007
a(n) = A213999(n,n-1). - Reinhard Zumkeller, Jul 03 2012
From Wolfdieter Lang, Apr 16 2015: (Start)
a(n)/A001008(n) = 1/H(n) is the solution of the following version of the classical cistern and pipes problem. A cistern is connected to n different pipes of water. For the k-th pipe it takes k time units (say, days) to fill the empty cistern, for k = 1, 2, ..., n. How long does it take for the n pipes together to fill the empty cistern? 1/H(n) gives the answer as a fraction of the time unit.
In general, if the k-th pipe needs d(k) days to fill the empty cistern then all pipes together need 1/Sum_{k=1..n} 1/d(k) = HM(d(1), ..., d(n))/n days, where HM denotes the harmonic mean HM. For the described problem, HM(1, 2, ..., n)/n = A102928(n)/(n*A175441(n)) = 1/H(n).
For a classical cistern and pipes problem see, e.g., the Hunger-Vogel reference (in Greek and German) given in A256101, problem 27, p. 29, where n = 3, and d(1), d(2) and d(3) are 6, 4 and 3 days. On p. 97 of this reference one finds remarks on the history of such problems (called in German 'Brunnenaufgabe'). (End)
From Wolfdieter Lang, Apr 17 2015: (Start)
An example of the above mentioned cistern and pipes problems appears in Chiu Chang Suan Shu (nine books on arithmetic) in book VI, problem 26. The numbers are there 1/2, 1, 5/2, 3 and 5 (days) and the result is 15/75 (day). See the reference (in German) on p. 68.
A historical account on such cistern problems is found in the Johannes Tropfke reference, given in A256101, section 4.2.1.2 Zisternenprobleme (Leistungsprobleme), pp. 578-579.
In Fibonacci's Liber Abaci such problems appear on p. 281 and p. 284 of L. E. Sigler's translation. (End)
All terms > 1 are even while corresponding numerators (A001008) are all odd (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

Examples

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ] = A001008/A002805.
		

References

  • Chiu Chang Suan Shu, Neun Bücher arithmetischer Technik, translated and commented by Kurt Vogel, Ostwalds Klassiker der exakten Wissenschaften, Band 4, Friedr. Vieweg & Sohn, Braunschweig, 1968, p. 68.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
  • J. Havil, Gamma, (in German), Springer, 2007, p. 35-6; Gamma: Exploring Euler's Constant, Princeton Univ. Press, 2003.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
  • L. E. Sigler, Fibonacci's Liber Abaci, Springer, 2003, pp. 281, 284.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A001008 (numerators), A075135, A025529, A203810, A203811, A203812.
Partial sums: A027612/A027611 = 1, 5/2, 13/3, 77/12, 87/10, 223/20,...
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358, Sum 1/n^2: A007406/A007407, Sum 1/n^3: A007408/A007409.

Programs

  • GAP
    List([1..30],n->DenominatorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018
    
  • Haskell
    import Data.Ratio ((%), denominator)
    a002805 = denominator . sum . map (1 %) . enumFromTo 1
    a002805_list = map denominator $ scanl1 (+) $ map (1 %) [1..]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Denominator(HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Apr 16 2015
    
  • Maple
    seq(denom(sum((2*k-1)/k, k=1..n), n=1..30); # Gary Detlefs, Jul 18 2011
    f:=n->denom(add(1/k, k=1..n)); # N. J. A. Sloane, Nov 15 2013
  • Mathematica
    Denominator[ Drop[ FoldList[ #1 + 1/#2 &, 0, Range[ 30 ] ], 1 ] ] (* Harvey P. Dale, Feb 09 2000 *)
    Table[Denominator[HarmonicNumber[n]], {n, 1, 40}] (* Stefan Steinerberger, Apr 20 2006 *)
    Denominator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)
  • PARI
    a(n)=denominator(sum(k=2,n,1/k)) \\ Charles R Greathouse IV, Feb 11 2011
    
  • Python
    from fractions import Fraction
    def a(n): return sum(Fraction(1, i) for i in range(1, n+1)).denominator
    print([a(n) for n in range(1, 30)]) # Michael S. Branicky, Dec 24 2021
  • Sage
    def harmonic(a, b): # See the F. Johansson link.
        if b - a == 1 : return 1, a
        m = (a+b)//2
        p, q = harmonic(a,m)
        r, s = harmonic(m,b)
        return p*s+q*r, q*s
    def A002805(n) : H = harmonic(1,n+1); return denominator(H[0]/H[1])
    [A002805(n) for n in (1..29)] # Peter Luschny, Sep 01 2012
    

Formula

a(n) = Denominator(Sum_{k=1..n} (2*k-1)/k). - Gary Detlefs, Jul 18 2011
a(n) = n! / gcd(Stirling1(n+1, 2), n!) = n! / gcd(A000254(n),n!). - Max Alekseyev, Mar 01 2018
a(n) = the (reduced) denominator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024

Extensions

Definition edited by Daniel Forgues, May 19 2010

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

A007395 Constant sequence: the all 2's sequence.

Original entry on oeis.org

2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 1

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Author

Keywords

Comments

Continued fraction for 1 + sqrt(2). - Philippe Deléham, Nov 14 2006
a(n) = A213999(n,1). - Reinhard Zumkeller, Jul 03 2012
The least witness function W(k) is defined for odd composite numbers k. The sequence W(k) does not have its own entry in the OEIS because W(k) = 2 for all k with 9 <= k < 2047; then W(2047)=3. Cf. A089105. - N. J. A. Sloane, Sep 17 2014
a(n) = A254858(n-1,1). - Reinhard Zumkeller, Feb 09 2015
a(n) = number of permutations of length n+2 having exactly one ascent such that the first element the permutation is 2. - Ran Pan, Apr 20 2015
With alternating signs, this is the sequence of determinants of the 3 X 3 matrices m with m(i,j) = Fibonacci(n+i+j-2)^2. - Michel Marcus, Dec 23 2015
For p = prime(n+2), a(n) = ord_p(H_(p-1)), where ord_p denotes the p-adic valuation and H_i = 1 + 1/2 + ... + 1/i is a harmonic sum, except for n = 1944 and n = 157504, where ord_p(H_(p-1)) = 3, and any other term of A088164 that may exist (see Conrad link). The sequence a(n) = ord_p(H_(p-1)) does not have its own entry in the OEIS. - Felix Fröhlich, Mar 16 2016
This sequence is the only infinite bounded sequence of positive integers such that a(n) = (a(n-1) + a(n-2)) / gcd(a(n-1), a(n-2)) for all n >= 2. - Bernard Schott, Dec 28 2018

References

  • Titu Andreescu and Dorin Andrica, Number Theory, Birkhäuser, 2009, from 1999 Russian Mathematical Olympiad, p. 347.
  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 6.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

G.f.: 2/(1-x), and e.g.f.: 2*e^x. - Mohammad K. Azarian, Dec 22 2008
a(n) = A000005(A000040(n)). - Omar E. Pol, Feb 28 2018
a(n) = A002061(n) - A165900(n). - Torlach Rush, Feb 21 2019

A027611 Denominator of n * n-th harmonic number.

Original entry on oeis.org

1, 1, 2, 3, 12, 10, 20, 35, 280, 252, 2520, 2310, 27720, 25740, 24024, 45045, 720720, 680680, 4084080, 3879876, 739024, 235144, 5173168, 14872858, 356948592, 343219800, 2974571600, 2868336900, 80313433200, 77636318760
Offset: 1

Views

Author

Glen Burch (gburch(AT)erols.com)

Keywords

Comments

This is very similar to A128438, which is a different sequence. They differ at n=6 (and nowhere else?). - N. J. A. Sloane, Nov 21 2008
Denominator of 1/n + 2/(n-1) + 3/(n-2) + ... + (n-1)/2 + n.
Denominator of Sum_{k=1..n} frac(n/k) where frac(x/y) denotes the fractional part of x/y. - Benoit Cloitre, Oct 03 2002
Denominator of Sum_{d=2..n-1, n mod d > 0} n/d. Numerator = A079076. - Reinhard Zumkeller, Dec 21 2002
a(n) is odd iff n is a power of 2. - Benoit Cloitre, Oct 03 2002
Indices where a(n) differs from A128438 are terms of A074791. - Gary Detlefs, Sep 03 2011

Crossrefs

Programs

  • Haskell
    import Data.Ratio ((%), denominator)
    a027611 n = denominator $ sum $ map (n %) [1..n]
    -- Reinhard Zumkeller, Jul 03 2012
    
  • Magma
    [Denominator(n*HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Feb 19 2014
    
  • Maple
    a := n -> denom(add((n-j)/j, j=1..n));
    seq(a(n), n = 1..30); # Peter Luschny, May 12 2023
  • Mathematica
    a[n_]:=Denominator[n*HarmonicNumber[n]]; Array[a,100] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011 *)
  • PARI
    a(n) = denominator(n*sum(k=1, n, 1/k)); \\ Michel Marcus, Feb 15 2015
    
  • Python
    from sympy import harmonic
    def A027611(n): return (n*harmonic(n)).q # Chai Wah Wu, Sep 26 2021
    
  • SageMath
    [denominator(n*harmonic_number(n)) for n in (1..40)] # G. C. Greubel, Aug 24 2022

Formula

From Vladeta Jovovic, Sep 02 2002: (Start)
a(n) = denominators of coefficients in expansion of -log(1-x)/(1-x)^2.
a(n) = denominators of (n+1)*(harmonic(n+1) - 1).
a(n) = denominators of (n+1)*(Psi(n+2) + Euler-gamma - 1). (End)
a(n) = numerator(h(n)/h(n-1)) - denominator(h(n)/h(n-1)), n > 1, where h(n) is the n-th harmonic number. - Gary Detlefs, Sep 03 2011
a(n) = A213999(n, n-2) for n > 1. - Reinhard Zumkeller, Jul 03 2012
a(n) = denominators of coefficients of e.g.f. -1 + exp(x)*(1 + Sum_{j >= 0} (-x)^(j+1)/(j * j!)). - G. C. Greubel, Aug 24 2022

Extensions

Entry revised by N. J. A. Sloane following a suggestion of Eric W. Weisstein, Jul 02 2004

A010704 Period 2: repeat (3,6).

Original entry on oeis.org

3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, 3
Offset: 0

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Author

Keywords

Comments

Continued fraction expansion of A176105. - R. J. Mathar, Mar 08 2012
Digital roots of A007283. - Bruno Berselli, Nov 22 2018
Decimal expansion of 4/11. - Franklin T. Adams-Watters, Nov 28 2018

Crossrefs

Programs

Formula

G.f. 3*(1 + 2*x)/((1 - x)*(1 + x)). - R. J. Mathar, Nov 21 2011
From Reinhard Zumkeller, Jul 03 2012: (Start)
a(n) = 3*A000034(n).
a(n) = A213999(n,2). (End)
a(n + 1) = 9 - a(n). - David A. Corneth, Nov 29 2018
a(n) = 2 + 2^(1 - (-1)^n). - Vincenzo Librandi, Feb 28 2020
a(n) = 3*(3-(-1)^n)/2. - Aaron J Grech, Aug 02 2024

A213998 Numerators of the triangle of fractions read by rows: pf(n,0) = 1, pf(n,n) = 1/(n+1) and pf(n+1,k) = pf(n,k) + pf(n,k-1) with 0 < k < n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 11, 1, 1, 7, 13, 25, 1, 1, 9, 47, 77, 137, 1, 1, 11, 37, 57, 87, 49, 1, 1, 13, 107, 319, 459, 223, 363, 1, 1, 15, 73, 533, 743, 341, 481, 761, 1, 1, 17, 191, 275, 1879, 2509, 3349, 4609, 7129, 1, 1, 19, 121, 1207, 1627, 2131, 2761, 3601, 4861, 7381, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 03 2012

Keywords

Comments

T(n,0) = 1;
T(n,1) = A005408(n-1) for n > 0;
T(n,2) = A188386(n-2) for n > 2;
T(n,n-3) = A124837(n-2) for n > 2;
T(n,n-2) = A027612(n-1) for n > 1;
T(n,n-1) = A001008(n) for n > 0;
T(n,n) = 1;
A214075(n,k) = floor(T(n,k) / A213999(n,k)).

Examples

			Start of triangle pf with corresponding triangles of numerators and denominators:
. 0:                            1
. 1:                         1    1/2
. 2:                     1     3/2    1/3
. 3:                  1    5/2    11/6    1/4
. 4:              1   7/2    13/3    25/12    1/5
. 5:           1    9/2   47/6    77/12   137/60   1/6
. 6:        1  11/2   37/3    57/4    87/10    49/20    1/7
. 7:     1  13/2  107/6  319/12  459/20   223/20  363/140   1/8
. 8:  1  15/2  73/3  533/12  743/15  341/10   481/35   761/280  1/9,
.
. 0:   numerators     1                          1    denominators
. 1:                1  1                        1  2       A213999
. 2:              1   3  1                     1 2  3
. 3:            1   5  11 1                   1 2 6  4
. 4:          1  7  13  25  1                1 2 3  12 5
. 5:        1  9  47  77 137  1             1 2 6 12  60 6
. 6:      1 11  37 57  87  49  1           1 2 3 4 10  20  7
. 7:    1 13 107 319 459 223 363 1        1 2 6 12 20 20 140 8
. 8:  1 15 73 533 743 341 481 761 1,     1 2 3 12 15 10 35 280 9.
		

Crossrefs

Cf. A005408, A188386 (columns).
Cf. A001008, A027612, A124837 (diagonals).
Cf. A213999 (denominators).

Programs

  • Haskell
    import Data.Ratio ((%), numerator, denominator, Ratio)
    a213998 n k = a213998_tabl !! n !! k
    a213998_row n = a213998_tabl !! n
    a213998_tabl = map (map numerator) $ iterate pf [1] where
       pf row = zipWith (+) ([0] ++ row) (row ++ [-1 % (x * (x + 1))])
                where x = denominator $ last row
  • Mathematica
    T[, 0] = 1; T[n, n_] := 1/(n + 1);
    T[n_, k_] := T[n, k] = T[n - 1, k] + T[n - 1, k - 1];
    Table[T[n, k] // Numerator, {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 10 2021 *)

A124838 Denominators of third-order harmonic numbers (defined by Conway and Guy, 1996).

Original entry on oeis.org

1, 2, 6, 4, 20, 10, 70, 56, 504, 420, 4620, 3960, 3432, 6006, 90090, 80080, 1361360, 408408, 369512, 67184, 470288, 1293292, 29745716, 27457584, 228813200, 212469400, 5736673800, 5354228880, 155272637520, 291136195350, 273491577450
Offset: 1

Views

Author

Jonathan Vos Post, Nov 10 2006

Keywords

Comments

Numerators are A124837. All fractions reduced. Thanks to Jonathan Sondow for verifying these calculations. He suggests that the equivalent definition in terms of first order harmonic numbers may be computationally simpler. We are happy with the description of A027612 Numerator of 1/n + 2/(n-1) + 3/(n-2) +...+ (n-1)/2 + n, but baffled by the description of A027611.

Examples

			a(1) = 1 = denominator of 1/1.
a(2) = 2 = denominator of 1/1 + 5/2 = 7/2.
a(3) = 6 = denominator of 7/2 + 13/3 = 47/6.
a(4) = 4 = denominator of 47/6 + 77/12 = 57/4.
a(5) = 20 = denominator of 57/4 + 87/10 = 549/20.
a(6) = 10 = denominator of 549/20 + 223/20 = 341/10
a(7) = 70 = denominator of 341/10 + 481/35 = 3349/70.
a(8) = 1260 = denominator of 3349/70 + 4609/280 = 88327/1260.
a(9) = 45 = denominator of 88327/1260 + 4861/252 = 3844/45.
a(10) = 504 = denominator of 3844/45 + 55991/2520 = 54251/504, or, untelescoping:
a(10) = 504 = denominator of 1/1 + 5/2 + 13/3 + 77/12 + 87/10 + 223/20 + 481/35 + 4609/252 + 4861/252 + 55991/2520 = 54251/504.
		

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, pp. 143 and 258-259, 1996.

Crossrefs

Programs

  • Haskell
    a124838 n = a213999 (n + 2) (n - 1) -- Reinhard Zumkeller, Jul 03 2012
  • Mathematica
    Table[Denominator[(n+2)!/2!/n!*Sum[1/k,{k,3,n+2}]],{n,1,40}] (* Alexander Adamchuk, Nov 11 2006 *)

Formula

A124837(n)/A124838(n) = Sum_{i=1..n} A027612(n)/A027611(n+1).
a(n) = denominator(Sum_{m=1..n} Sum_{L=1..m} Sum_{k=1..L} 1/k).
a(n) = denominator(((n+2)!/(2!*n!)) * Sum_{k=3..n+2} 1/k). - Alexander Adamchuk, Nov 11 2006
a(n) = A213999(n+2,n-1). - Reinhard Zumkeller, Jul 03 2012

Extensions

Corrected and extended by Alexander Adamchuk, Nov 11 2006

A347563 Binomial complement triangle, T(n,k) = LCM(1,...,n)/binomial(n,k) for 0 <= k <= n, a(0) = T(0,0) = 0, read by rows.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 6, 2, 2, 6, 12, 3, 2, 3, 12, 60, 12, 6, 6, 12, 60, 60, 10, 4, 3, 4, 10, 60, 420, 60, 20, 12, 12, 20, 60, 420, 840, 105, 30, 15, 12, 15, 30, 105, 840, 2520, 280, 70, 30, 20, 20, 30, 70, 280, 2520
Offset: 0

Views

Author

Gary Waters, Sep 06 2021

Keywords

Comments

The one's complement of each carry value, in base prime p, defined in Lucas's Theorem. Also works using Erdős's method (see formula below).
At row n of the triangle, the values are symmetrical with the largest values occurring at T(n,0) = T(n,n) = LCM(1,...,n). The smallest value(s) occur at k = n/2 when n is even, and at k = floor(n/2) and k = floor(n/2)+1 when n is odd. T(n,k) = T(n,n-k).
Conjecture: For all n, T(n,0) mod A213999(n-1,n-1) = 0, and T(n,k+1) mod A213999(n,k) = 0 for 0 <= k <= n-1 (computed and verified for rows = 0..2000).

Examples

			T(7,3) = 12. Triangle T(n,k) begins:
     0;
     1,   1;
     2,   1,  2;
     6,   2,  2,  6;
    12,   3,  2,  3, 12;
    60,  12,  6,  6, 12, 60;
    60,  10,  4,  3,  4, 10, 60;
   420,  60, 20, 12, 12, 20, 60, 420;
   840, 105, 30, 15, 12, 15, 30, 105, 840;
  2520, 280, 70, 30, 20, 20, 30,  70, 280, 2520;
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[(LCM@@Range(1,n))/Binomial[n, k], {n, 0, 11}, {k, 0, n}]]
  • PARI
    row(n) = vector(n+1, k, k--; lcm([1..n])/binomial(n,k)); \\ Michel Marcus, Sep 13 2021

Formula

T(n,k) = Product_{p<=n} p^u_p, where u_p = i_max - Sum_{i=1..i_max} v_p(i) = Sum_{i=1..i_max} NOT(v_p(i)), with v_p(i) = floor(n/p^i) - floor(k/p^i) - floor((n-k)/p^i) = {0, 1} and i_max = floor(log(n)/log(p)) (using Erdős's method).
Showing 1-10 of 10 results.