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 24 results. Next

A136682 Numbers k such that A119682(k) is prime.

Original entry on oeis.org

2, 3, 5, 8, 23, 41, 47, 48, 49, 95, 125, 203, 209, 284, 323, 395, 504, 553, 655, 781, 954, 1022, 1474, 1797, 1869, 1923, 1934, 1968, 2043, 2678, 3413, 3439, 4032, 4142, 4540, 4895, 5018, 5110, 5194, 5357, 6591, 11504, 11949, 14084, 20365
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

A119682(k) = Numerator of Sum_{j=1..k} (-1)^(j+1)/j^2.

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ (-1)^(k+1)*1/k^2, {k,1,n} ] ]; If[ PrimeQ[f], Print[ {n,f} ] ], {n,1,125} ]

Extensions

a(12)-a(17) from Alexander Adamchuk, Apr 28 2008
a(18)-a(31) from Amiram Eldar, Mar 14 2019
a(32)-a(45) from Robert Price, Apr 14 2019

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

A120296 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^4.

Original entry on oeis.org

1, 15, 1231, 19615, 12280111, 4090037, 9824498837, 157151464517, 38193952437631, 7637983935923, 111835788321880643, 111830093529238643, 3194097388508809394723, 3194009594644356242723, 15970381078317764649391
Offset: 1

Views

Author

Alexander Adamchuk, Jul 10 2006

Keywords

Comments

p divides a(p-1) for prime p > 2 - similar to Wolstenholme's theorem for A007406(n) (= numerator of Sum_{k=1..n} 1/k^2) and for A007410(n) (= numerator of Sum_{k=1..n} 1/k^4).
Lim_{n -> infinity} a(n)/A334585(n) = A267315 = (7/8)*A013662. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 15/16, 1231/1296, 19615/20736, 12280111/12960000, 4090037/4320000, 9824498837/10372320000, ... = A120296/A334585. - _Petros Hadjicostas_, May 06 2020
		

Crossrefs

Cf. A007406, A007410, A013662, A119682, A267315, A334585 (denominators).

Programs

  • Mathematica
    Numerator[Table[Sum[(-1)^(k+1)/k^4,{k,1,n}],{n,1,20}]]
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^4)); \\ Michel Marcus, May 07 2020

Formula

a(n) = numerator(Sum_{k=1..n} (-1)^(k+1)/k^4).

Extensions

Name edited by Petros Hadjicostas, May 07 2020

A136675 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^3.

Original entry on oeis.org

1, 7, 197, 1549, 195353, 194353, 66879079, 533875007, 14436577189, 14420574181, 19209787242911, 19197460851911, 42198121495296467, 6025866788581781, 6027847576222613, 48209723660000029, 236907853607882606477
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

a(n) is prime for n in A136683.
Lim_{n -> infinity} a(n)/A334582(n) = A197070. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 7/8, 197/216, 1549/1728, 195353/216000, 194353/216000, 66879079/74088000, 533875007/592704000, ... = a(n)/A334582(n). - _Petros Hadjicostas_, May 06 2020
		

Crossrefs

Programs

  • Maple
    map(numer,ListTools:-PartialSums([seq((-1)^(k+1)/k^3, k=1..100)])); # Robert Israel, Nov 09 2023
  • Mathematica
    (* Program #1 *) Table[Numerator[Sum[(-1)^(k+1)/k^3, {k,1,n}]], {n,1,50}]
    (* Program #2 *) Numerator[Accumulate[Table[(-1)^(k+1) 1/k^3, {k,50}]]] (* Harvey P. Dale, Feb 12 2013 *)
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^3)); \\ Michel Marcus, May 07 2020

A136677 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^6.

Original entry on oeis.org

1, 63, 45991, 2942695, 45982595359, 5109066151, 601081707598999, 38469080386820311, 252396118308232060471, 252395862211967012407, 447134922152359540530757327, 447134770212444455649757327, 2158234586764514215343657417779543, 308319185132349039219686748825649
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

p divides a(p-1) for prime p > 2. a(n) is prime for n = {19, 47, 164, ...} = A136686.
Lim_{n -> infinity} a(n)/A334605(n) = A275703 = (31/32)*A013664. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 63/64, 45991/46656, 2942695/2985984, 45982595359/46656000000, 5109066151/5184000000, ... = a(n)/A334605(n). - _Petros Hadjicostas_, May 07 2020
		

Crossrefs

Programs

  • Mathematica
    Table[ Numerator[ Sum[ (-1)^(k+1)/k^6, {k,1,n} ] ], {n,1,30} ]
    Accumulate[Table[(-1)^(k+1)/k^6,{k,20}]]//Numerator (* Harvey P. Dale, Aug 21 2023 *)

A136676 Numerator of Sum_{k=1..n} (-1)^(k+1)/k^5.

Original entry on oeis.org

1, 31, 7565, 241837, 755989457, 755889457, 12705011703799, 406547611705943, 98792790681344149, 98791774426324117, 15910615688635928566967, 15910549913780913466967, 5907492176026179821253778331
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

a(n) is prime for n in A136685.
Lim_{n -> infinity} a(n)/A334604(n) = A267316 = (15/16)*A013663. - Petros Hadjicostas, May 07 2020

Examples

			The first few fractions are 1, 31/32, 7565/7776, 241837/248832, 755989457/777600000, 755889457/777600000, ... = a(n)/A334604(n). - _Petros Hadjicostas_, May 07 2020
		

Crossrefs

Programs

  • Mathematica
    Table[ Numerator[ Sum[ (-1)^(k+1)/k^5, {k,1,n} ] ], {n,1,30} ]
  • PARI
    a(n) = numerator(sum(k=1, n, (-1)^(k+1)/k^5)); \\ Michel Marcus, May 07 2020

A136681 Numbers k such that A058313(k) is prime.

Original entry on oeis.org

3, 4, 5, 6, 9, 10, 13, 16, 17, 18, 37, 43, 58, 121, 124, 126, 137, 203, 247, 283, 285, 286, 289, 317, 424, 508, 751, 790, 937, 958, 1066, 1097, 1151, 1166, 1194, 1199, 1235, 1414, 1418, 1460, 1498, 1573, 2090, 2122, 2691, 2718, 3030, 3426, 3600, 3653, 3737
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

A058313(k) = Numerator of Sum_{j=1..k} (-1)^(j+1)/j.

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ (-1)^(k+1)*1/k, {k,1,n} ] ]; If[ PrimeQ[f], Print[ {n,f} ] ], {n,1,317} ]
  • PARI
    isok(n) = isprime(numerator(sum(k=1, n, (-1)^(k+1)/k))); \\ Michel Marcus, Mar 14 2019

Extensions

a(25)-a(30) from James R. Buddenhagen, Sep 22 2015
a(31)-a(51) from Amiram Eldar, Mar 14 2019

A136683 Numbers k such that A136675(k) is prime.

Original entry on oeis.org

2, 3, 4, 5, 6, 9, 20, 21, 29, 119, 132, 151, 351, 434, 457, 462, 572, 611, 930, 1107, 1157, 1452, 1515, 2838, 3997, 5346, 6463, 6725, 7664, 10234, 14168, 14299
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

A136675(k) = Numerator of Sum_{j=1..k} (-1)^(j+1)/j^3.

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ (-1)^(k+1)*1/k^3, {k,1,n} ] ]; If[ PrimeQ[f], Print[ {n,f} ] ], {n,1,151} ]
    Flatten[Position[Numerator[Accumulate[Table[(-1)^(k+1) 1/k^3,{k,3000}]]],?PrimeQ] ] (* _Harvey P. Dale, Feb 12 2013 *)
  • PARI
    isok(n) = ispseudoprime(numerator(sum(k=1, n, (-1)^(k+1) / k^3))); \\ Daniel Suteu, Mar 15 2019

Extensions

More terms from Harvey P. Dale, Feb 12 2013
a(25)-a(28) from Amiram Eldar, Mar 15 2019
a(29)-a(32) from Robert Price, Apr 22 2019

A136684 Numbers k such that A120296(k) is prime.

Original entry on oeis.org

3, 5, 8, 11, 20, 38, 61, 65, 71, 80, 83, 93, 233, 704, 1649, 2909, 3417, 3634, 9371
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

A120296(k) = Numerator of Sum_{j=1..k} (-1)^(j+1)/j^4.

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ (-1)^(k+1)*1/k^4, {k,1,n} ] ]; If[ PrimeQ[f], Print[ {n,f} ] ], {n,1,100} ]
    Select[Range[1000],PrimeQ[Numerator[Sum[(-1)^(k+1) 1/k^4,{k,#}]]]&] (* Harvey P. Dale, Aug 28 2012 *)

Extensions

More terms from Harvey P. Dale, Aug 28 2012
a(15)-a(19) from Robert Price, Apr 23 2019

A136685 Numbers k such that A136676(k) is prime.

Original entry on oeis.org

2, 19, 51, 78, 84, 130, 294, 910, 2223, 2642, 3261, 4312, 4973, 7846, 9439
Offset: 1

Views

Author

Alexander Adamchuk, Jan 16 2008

Keywords

Comments

A136676(k) = Numerator of Sum_{j=1..k} (-1)^(j+1)/j^5.

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ (-1)^(k+1)*1/k^5, {k,1,n} ] ]; If[ PrimeQ[f], Print[ {n,f} ] ], {n,1,130} ]

Extensions

a(7)-a(8) from Amiram Eldar, Mar 14 2019
a(9)-a(15) from Robert Price, Apr 16 2019
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