A025527 -id:A025527 - cp's OEIS frontend

A250270 Products of terms of A003418.

1, 2, 4, 6, 8, 12, 16, 24, 32, 36, 48, 60, 64, 72, 96, 120, 128, 144, 192, 216, 240, 256, 288, 360, 384, 420, 432, 480, 512, 576, 720, 768, 840, 864, 960, 1024, 1152, 1296, 1440, 1536, 1680, 1728, 1920, 2048, 2160, 2304, 2520, 2592, 2880, 3072, 3360, 3456

Offset: 1

Matthew Vandermast, Dec 16 2014

nonn

Includes all factorials and Jordan-Polya numbers, since n! = Product_{i = 1..n} A003418(floor(n/i)) for positive n.

			720 = 2*6*60 = 12*60. Since 2, 6, 12 and 60 are all terms of A003418, 720 is a term of this sequence.

Michel Marcus, Table of n, a(n) for n = 1..5000

Range of values of A253139. Subsequences include A000142, A001013, A001813, A025527, A064350, A166338, A250569.

Subsequence of A025487.

f(n) = lcm(vector(n, i, i)); \\ A003418
mul(x,y) = x*y;
lista(nn) = {my(v = vector(nn, k, f(k)), lim = f(nn+1), ok = 0, nv); while (!ok,  nv = select(x->(xMichel Marcus, May 09 2021

A055377 a(n) = largest prime <= n/2.

Original entry on oeis.org

2, 2, 3, 3, 3, 3, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 7, 11, 11, 11, 11, 13, 13, 13, 13, 13, 13, 13, 13, 17, 17, 17, 17, 19, 19, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 23, 29, 29, 29, 29, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 37

Offset: 4

Labos Elemer, Jun 22 2000; David W. Wilson, Jun 10 2005

nonn

Also largest prime factor of any composite <= n. E.g., a(15) = 7 since 7 is the largest prime factor of {4,6,8,9,10,12,14,15}, the composites <= 15.
Also largest prime dividing A025527(n) = n!/lcm[1,...,n]. [Comment from Ray Chandler, Apr 26 2007: Primes > n/2 don't appear as factors of A025527(n) since they appear once in n! and again in the denominator lcm[1,...,n]. Primes <= n/2 appear more times in the numerator than the denominator so they appear in the fraction.]
a(n) is the largest prime factor whose exponent in the factorization of n! is greater than 1. - Michel Marcus, Nov 11 2018

			n = 10, n! = 3628800, lcm[1,...,10] = 2520, A025527(10) = 1440 = 32*9*5 so a(7) = 5 (offset = 3).

Michael De Vlieger, Table of n, a(n) for n = 4..10000

Cf. A000040, A000142, A003418, A006530, A007917, A025527, A060265, A079952, A079953.

Table[Prime@ PrimePi[n/2], {n, 4, 78}] (* Michael De Vlieger, Sep 21 2017 *)

a(n) = precprime(n/2); \\ Michel Marcus, Sep 20 2017

a(n) = Max(gpf((n+2) mod k): 1 < k < (n+2) and k not prime), with gpf=A006530 (greatest prime factor). - Reinhard Zumkeller, Mar 27 2004

Where defined, that is for n > 2, a(A000040(n)) = A000040(A079952(n)). - Peter Munn, Sep 18 2017

More terms from James Sellers, Jul 04 2000

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, May 14 2007

A081530 a(n) = running sum of the first n harmonic numbers, multiplied by the LCM of 1..n.

Original entry on oeis.org

1, 5, 26, 77, 522, 669, 5772, 13827, 48610, 55991, 699612, 785633, 11359222, 12530955, 13726712, 29889983, 550271934, 593094837, 12094689300, 12932216325, 13780828710, 14640022575, 356714770680, 376932115005, 1986818142426

Offset: 1

Amarnath Murthy, Mar 27 2003

nonn

Consider triangle in A081525. Write terms in k-th row with denominator = LCM of terms in that row. Sequence gives sum of numerators of terms in n-th row.

			(1), 2*(1 + 3/2), 6*(1 + 3/2 + 11/6), 12*(1 + 3/2 + 11/6 + 25/12).

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A081525, A081526, A081527, A081528, A081529.

Cf. A001705, A025527.

H:=n->add(1/i,i=1..n):seq((n+1)*ilcm(seq(j,j=1..n))*(H(n+1)-1),n=1..30); # C. Ronaldo

Table[Sum[HarmonicNumber[k], {k, n}] LCM @@ Range[n], {n, 36}] (* Wouter Meeussen *)

a(n) = lcm(1..n)*(n+1)*(H(n+1)-1), where H(n) is the n-th harmonic number. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 19 2004

Equal to A001705(n) / A025527(n). - Martin Fuller, Jan 03 2006

More terms from Wouter Meeussen, Apr 13 2003

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

Clark Kimberling

nonn

			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

Cf. A003418, A002944, A025527.

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

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

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

A205958 a(0) = 1 and a(n) = A180000(n)*a(floor(n/2))^2 for n > 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 3, 48, 16, 40, 40, 270, 270, 945, 63, 129024, 129024, 64512, 64512, 2016000, 96000, 528000, 528000, 144342000, 28868400, 187644600, 20849400, 1787836050, 1787836050, 59594535, 59594535, 3999321544458240, 121191561953280, 1030128276602880

Offset: 0

Peter Luschny, Feb 04 2012

nonn

lcm(1,2,..,n) = (n!*a(n)) / ((n/2)!*a(n/2))^2.

lcm(1,2,..,n)*a(n) is a divisor of n! and n!/(lcm(1,2,..,n)*a(n)) is a square.

Cf. A180000, A025527, A205959.

def A205958(n) :
    if n == 0 : return 1
    return A180000(n)*A205958(n//2)^2

A244656 Least product of consecutive positive integers which is divisible by each of 1, 2, ..., n.

Original entry on oeis.org

2, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 55440, 55440, 360360, 360360, 360360, 2162160, 85765680, 85765680, 33522128640, 33522128640, 33522128640, 33522128640, 19275223968000, 19275223968000, 19275223968000

Offset: 1

Rick L. Shepherd, Jul 03 2014, Sep 14 2014

nonn

For n > 1, clearly a(n) is bounded below by lcm(1,2,...,n) and bounded above by n!. Further, a(n) is a positive multiple of lcm(1,2,...,n). Any product of two or more consecutive positive integers may be expressed as m!/k!, where 0 <= k <= m-2. For this sequence, the m corresponding to a(n) may or may not be a multiple of n. Whenever a(n) can be expressed as the product of exactly two consecutive integers, it is a term of A002378. See the a-file link for further comments.

			a(7) = 20*21 = 21!/19! = 420 because 420 is divisible by 1, 2, 3, 4, 5, 6, and 7, and no positive integer less than 420 is divisible by each of these. Here, 420 = lcm(1,2,3,4,5,6,7). 420 is an oblong (or promic) number (A002378).
a(11) = 7*8*9*10*11 = 11!/6! = 55440. Here, 27720 = lcm(1,2,3,4,5,6,7,8,9,10,11), but 27720 cannot be represented as a product of consecutive positive integers.
a(31) = 6081487775*6081487776 = 36984493563555938400, also a promic number.

Rick L. Shepherd, Table of n, a(n) for n = 1..36
Rick L. Shepherd, Sample program output and calculation notes

Cf. A003418, A000142, A025527, A002378.

{a(n) =
local(L, M, i, k = 0, s = 0, ret = 0, d, divs2,
   st, pr, prt = 1); /* Use prt = 0 to suppress printing. */
if(n < 1, return, if(n < 3, ret = 2,
L = lcm(vector(n, i, i));
M = n!/L;
while(k < M,
   k++;
   s += L; d = divisors(s); divs2 = #d \ 2;
   st = 2; pr = d[st];
   i = 0;
   while(st + i <= divs2,
      if(d[st + i + 1] == d[st + i] + 1,
         pr *= d[st + i + 1]; i++;
         if(pr == s,
            if(prt,
               print1("k*L = ", k, "*", L, " = ",
                 s, " = ", d[st], "*");
               if(d[st + i] > d[st] + 2, print1("...*"),
                 if(d[st + i] == d[st] + 2,
                   print1(d[st] + 1, "*")));
               print(d[st + i], " = ", d[st + i], "!/",
                 d[st] - 1, "!"));
            ret = s; break(2),
            if(pr > s, st++; pr = d[st]; i = 0)),
         if(pr < s, st += i + 1, st++); pr = d[st]; i = 0)))));
return(ret)}

A059955 a(n) = floor( prime(n)!/lcm(1..prime(n)) ) modulo prime(n).

Original entry on oeis.org

1, 2, 5, 10, 3, 10, 4, 3, 28, 17, 18, 30, 20, 41, 42, 14, 19, 30, 37, 63, 50, 7, 12, 83, 30, 91, 19, 69, 91, 97, 56, 22, 80, 39, 137, 44, 9, 154, 19, 37, 141, 141, 168, 126, 183, 200, 205, 136, 55, 95, 204, 126, 213, 230, 68, 63, 158, 202, 162, 102, 182, 104, 38, 165

Offset: 2

Lekraj Beedassy, Mar 13 2001

nonn

a(n) gives also the smallest coefficient for which the multiple M of lcm(1 through p(n)-1) satisfies p(n) divides M + 1. This computes the solution of the puzzle requiring the smallest number such that grouping in 2's, 3's, etc. up to the n-th prime,all leave a remainder of one except the last which leaves no remainder.

			a(2)=1 because prime(2)=3 and floor(3!/lcm(1,2,3)) mod 3 = 1 mod 3 = 1;
a(3)=2 because prime(3)=5 and floor(5!/lcm(1,2,3,4,5)) mod 5 = 2 mod 5 = 2;
a(4)=5 because prime(4)=7 and floor(7!/lcm(1,2,3,4,5,6,7)) mod 7 = 12 mod 7 = 5;
a(7)=10 because prime(7)=17 and floor(17!/lcm(1,2,...,17)) mod 17 = 29030400 mod 17 = 10.

Cf. A003418, A025527, A099796.

[Floor( Factorial(p)/Lcm([1..p]) ) mod p: p in PrimesInInterval(3,400)]; // Bruno Berselli, Feb 08 2015

for n from 2 to 150 do printf(`%d,`, floor(ithprime(n)!/ilcm(i $ i=1..ithprime(n))) mod ithprime(n) ); od:

More terms from James Sellers, Mar 15 2001

A175455 a(n) = H(n) * (lcm(1,2,...,n))^2, where H(n) = harmonic numbers (1/1 + 1/2 + ... + 1/n).

Original entry on oeis.org

1, 6, 66, 300, 8220, 8820, 457380, 1917720, 17965080, 18600120, 2320468920, 2384502120, 412970037480, 422245703880, 430902992520, 1756076802480, 516336630329520, 524676485052720, 192260441419366320, 194970060218934000, 197550649551855600, 200013939369644400

Offset: 1

Jaroslav Krizek, May 17 2010

nonn

			For n = 3, a(3) = (1/1 + 1/2 + 1/3) * (1*2*3)^2 = (11/6) * 36 = 66.

Andrew Howroyd, Table of n, a(n) for n = 1..200

Cf. A000142, A001008, A002805, A003418, A025527, A025529.

a(n)={sum(k=1, n, 1/k)*lcm([1..n])^2} \\ Andrew Howroyd, Jan 08 2020

a(n) = (A001008(n) / A002805(n)) * (A003418(n))^2.
a(n) = A000142(n) * A025529(n) / A025527(n) = A025529(n) * A003418(n).
a(n) = (1/1 + 1/2 + ... + 1/n) * (lcm(1,2,...,n))^2.

Terms a(13) and beyond from Andrew Howroyd, Jan 08 2020

A216915 T(n, k) = Product{1<=j<=n, gcd(j,k)=1 | j} / lcm{1<=j<=n, gcd(j,k)=1 | j} for n >= 0, k >= 1, square array read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 12, 1, 2, 1, 1, 1, 1, 12, 1, 2, 1, 1, 1, 1, 1, 48, 1, 2, 1, 2, 1, 1, 1, 1, 144, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1440, 3, 8, 1, 12, 1, 2, 1, 1, 1, 1, 1440, 3, 8, 1, 12, 1, 2, 1, 1, 1, 1, 1, 17280, 3, 80

Offset: 1

Peter Luschny, Oct 02 2012

T(n,k) = Product(R(n,k))/lcm(R(n,k)) where R(n,k) is the set of all integers up to n that are relatively prime to k.

T(n,k) = A216919(n,k)/A216917(n,k).

			   k |n=0  1  2  3  4  5  6  7  8   9   10
  ---+------------------------------------
   1 |  1  1  1  1  2  2 12 12 48 144 1440
   2 |  1  1  1  1  1  1  1  1  1   3    3
   3 |  1  1  1  1  2  2  2  2  8   8   80
   4 |  1  1  1  1  1  1  1  1  1   3    3
   5 |  1  1  1  1  2  2 12 12 48 144  144
   6 |  1  1  1  1  1  1  1  1  1   1    1
   7 |  1  1  1  1  2  2 12 12 48 144 1440
   8 |  1  1  1  1  1  1  1  1  1   3    3
   9 |  1  1  1  1  2  2  2  2  8   8   80
  10 |  1  1  1  1  1  1  1  1  1   3    3
  11 |  1  1  1  1  2  2 12 12 48 144 1440
  12 |  1  1  1  1  1  1  1  1  1   1    1
  13 |  1  1  1  1  2  2 12 12 48 144 1440

def A216915(n, k):
    def R(n, k): return [j for j in (1..n) if gcd(j, k) == 1]
    return mul(R(n,k))/lcm(R(n, k))
for k in (1..13): [A216915(n, k) for n in (0..10)]

For n > 0:
A(n,1) = A025527(n);
A(4,n) = A000034(n);
A(n,n) = A128247(n).

A093627 a(n) = lcm(1,2,3,...,n) * (sum of Farey fractions of order n).

Original entry on oeis.org

1, 3, 15, 42, 330, 390, 3990, 9660, 36540, 41580, 595980, 651420, 10630620, 11711700, 13153140, 29189160, 594233640, 630990360, 14083949880, 15015120120, 16411875480, 17575838280, 463140798120, 484557713640, 2690500012200

Offset: 1

Robert G. Wilson v and Clark Kimberling, Apr 06 2004

nonn

Eric Weisstein's World of Mathematics, Farey Sequence.

Cf. A025527, A093593.

Farey[n_] := Union[ Flatten[ Join[{0}, Table[a/b, {b, n}, {a, b}]]]]; Table[LCM @@ Range[n] Plus @@ Farey[n], {n, 26}]

a(n) = A093593(n)/A025527(n).

cp's OEIS Frontend

A250270 Products of terms of A003418.

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PARI

A055377 a(n) = largest prime <= n/2.

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Mathematica

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A081530 a(n) = running sum of the first n harmonic numbers, multiplied by the LCM of 1..n.

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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).

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A205958 a(0) = 1 and a(n) = A180000(n)*a(floor(n/2))^2 for n > 0.

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Sage

A244656 Least product of consecutive positive integers which is divisible by each of 1, 2, ..., n.

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A059955 a(n) = floor( prime(n)!/lcm(1..prime(n)) ) modulo prime(n).

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Magma

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A175455 a(n) = H(n) * (lcm(1,2,...,n))^2, where H(n) = harmonic numbers (1/1 + 1/2 + ... + 1/n).

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