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 19 results. Next

A183063 Number of even divisors of n.

Original entry on oeis.org

0, 1, 0, 2, 0, 2, 0, 3, 0, 2, 0, 4, 0, 2, 0, 4, 0, 3, 0, 4, 0, 2, 0, 6, 0, 2, 0, 4, 0, 4, 0, 5, 0, 2, 0, 6, 0, 2, 0, 6, 0, 4, 0, 4, 0, 2, 0, 8, 0, 3, 0, 4, 0, 4, 0, 6, 0, 2, 0, 8, 0, 2, 0, 6, 0, 4, 0, 4, 0, 4, 0, 9, 0, 2, 0, 4, 0, 4, 0, 8, 0, 2, 0, 8, 0, 2
Offset: 1

Views

Author

Jaroslav Krizek, Dec 22 2010

Keywords

Comments

Number of divisors of n that are divisible by 2. More generally, it appears that the sequence formed by starting with an initial set of k-1 zeros followed by the members of A000005, with k-1 zeros between every one of them, can be defined as "the number of divisors of n that are divisible by k", (k >= 1). For example if k = 1 we have A000005 by definition, if k = 2 we have this sequence. Note that if k >= 3 the sequences are not included in the OEIS because the usual OEIS policy is not to include sequences like this where alternate terms are zero; this is an exception. - Omar E. Pol, Oct 18 2011
Number of zeros in n-th row of triangle A247795. - Reinhard Zumkeller, Sep 28 2014
a(n) is also the number of partitions of n into equal parts, minus the number of partitions of n into consecutive parts. - Omar E. Pol, May 04 2017
a(n) is also the number of partitions of n into an even number of equal parts. - Omar E. Pol, May 14 2017

Examples

			For n = 12, set of even divisors is {2, 4, 6, 12}, so a(12) = 4.
On the other hand, there are six partitions of 12 into equal parts: [12], [6, 6], [4, 4, 4], [3, 3, 3, 3], [2, 2, 2, 2, 2, 2] and [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]. And there are two partitions of 12 into consecutive parts: [12] and [5, 4, 3], so a(12) = 6 - 2 = 4, equaling the number of even divisors of 12. - _Omar E. Pol_, May 04 2017
		

Crossrefs

Programs

  • Haskell
    a183063 = sum . map (1 -) . a247795_row
    -- Reinhard Zumkeller, Sep 28 2014, Jan 15 2013, Jan 10 2012
    
  • Magma
    [IsOdd(n) select 0 else #[d:d in Divisors(n)|IsEven(d)]:n in [1..100]]; // Marius A. Burtea, Dec 16 2019
    
  • Maple
    A183063 := proc(n)
        if type(n,'even') then
            numtheory[tau](n/2) ;
        else
            0;
        end if;
    end proc: # R. J. Mathar, Jun 18 2015
  • Mathematica
    Table[Length[Select[Divisors[n], EvenQ]], {n, 90}] (* Alonso del Arte, Jan 10 2012 *)
    a[n_] := (e = IntegerExponent[n, 2]) * DivisorSigma[0, n / 2^e]; Array[a, 100] (* Amiram Eldar, Jul 06 2022 *)
  • PARI
    a(n)=if(n%2,0,numdiv(n/2)) \\ Charles R Greathouse IV, Jul 29 2011
    
  • Python
    from sympy import divisor_count
    def A183063(n): return divisor_count(n>>(m:=(~n&n-1).bit_length()))*m # Chai Wah Wu, Jul 16 2022
  • Sage
    def A183063(n): return len([1 for d in divisors(n) if is_even(d)])
    [A183063(n) for n in (1..80)]  # Peter Luschny, Feb 01 2012
    

Formula

a(n) = A000005(n) - A001227(n).
a(2n-1) = 0; a(2n) = A000005(n).
G.f.: Sum_{d>=1} x^(2*d)/(1 - x^(2*d)) and generally for the number of divisors that are divisible by k: Sum_{d>=1} x^(k*d)/(1 - x^(k*d)). - Geoffrey Critzer, Apr 15 2014
Dirichlet g.f.: zeta(s)^2/2^s and generally for the number of divisors that are divisible by k: zeta(s)^2/k^s. - Geoffrey Critzer, Mar 28 2015
From Ridouane Oudra, Sep 02 2019: (Start)
a(n) = Sum_{i=1..n} (floor(n/(2*i)) - floor((n-1)/(2*i))).
a(n) = 2*A000005(n) - A000005(2n). (End)
Conjecture: a(n) = lim_{x->n} f(Pi*x), where f(x) = sin(x)*Sum_{k>0} (cot(x/(2*k))/(2*k) - 1/x). - Velin Yanev, Dec 16 2019
a(n) = A000005(A000265(n))*A007814(n) - Chai Wah Wu, Jul 16 2022
Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma - 1/2 - log(2)/2)*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 27 2022
Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A000005(k) = 1-log(2) (A244009). - Amiram Eldar, Mar 01 2023

Extensions

Formula corrected by Charles R Greathouse IV, Jul 29 2011

A024168 a(n) = n! * (1 + Sum_{j=1..n} (-1)^j/j).

Original entry on oeis.org

1, 0, 1, 1, 10, 26, 276, 1212, 14736, 92304, 1285920, 10516320, 166112640, 1680462720, 29753498880, 359124192000, 7053661440000, 98989454592000, 2137497610752000, 34210080898560000, 805846718380032000, 14489879077804032000, 369868281883398144000
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of permutations of n letters all cycles of which have length <= n/2, a quantity which arises in the solution to the One Hundred Prisoners problem. - Jim Ferry (jferry(AT)alum.mit.edu), Mar 29 2007

Crossrefs

A075829(n) = a(n-1)/gcd(a(n-1), a(n)).

Programs

  • Maple
    a := n -> n!*((-1)^n*LerchPhi(-1, 1, n + 1) + 1 - log(2));
    seq(simplify(a(n)), n=0..21); # Peter Luschny, Dec 27 2018
  • Mathematica
    f[k_] := (k + 1) (-1)^(k + 1)
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 22}]    (* A024168 signed *)
    (* Clark Kimberling, Dec 30 2011 *)
  • PARI
    x='x+O('x^33); concat([0],Vec(serlaplace((x-log(1+x))/(1-x)))) \\ Joerg Arndt, Dec 27 2018

Formula

From Michael Somos, Oct 29 2002: (Start)
E.g.f.: (log(x+1)-1)/(x-1).
a(n) = a(n-1)+a(n-2)*(n-1)^2, n>=2. (End)
a(0) = 1, a(n) = a(n-1)*n + (-1)^n*(n-1)!. - Daniel Suteu, Feb 06 2017
a(n) = n!*((-1)^n*LerchPhi(-1, 1, n+1) + 1 - log(2)). - Peter Luschny, Dec 27 2018
Limit_{n->oo} a(n)/n! = 1 - log(2) = A244009. - Alois P. Heinz, Jul 08 2022

Extensions

More terms from Michael Somos, Oct 29 2002
a(0)=1 prepended and edited by Alois P. Heinz, Sep 24 2023

A114112 a(1)=1, a(2)=2; thereafter a(n) = n+1 if n odd, n-1 if n even.

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14, 13, 16, 15, 18, 17, 20, 19, 22, 21, 24, 23, 26, 25, 28, 27, 30, 29, 32, 31, 34, 33, 36, 35, 38, 37, 40, 39, 42, 41, 44, 43, 46, 45, 48, 47, 50, 49, 52, 51, 54, 53, 56, 55, 58, 57, 60, 59, 62, 61, 64, 63, 66, 65, 68, 67, 70, 69, 72, 71
Offset: 1

Views

Author

Leroy Quet, Nov 13 2005

Keywords

Comments

a(1)=1; for n>1, a(n) is the smallest positive integer not occurring earlier in the sequence such that a(n) does not divide Sum_{k=1..n-1} a(k). - Leroy Quet, Nov 13 2005 (This was the original definition. A simple induction argument shows that this is the same as the present definition. - N. J. A. Sloane, Mar 12 2018)
Define b(1)=2; for n>1, b(n) is the smallest number not yet in the sequence which shares a prime factor with the sum of all preceding terms. Then a simple induction argument shows that the b(n) sequence is the same as the present sequence with the first term omitted. - David James Sycamore, Feb 26 2018
Here are the details of the two induction arguments (Start)
For a(n), let A(n) = a(1)+...+a(n). The claim is that for n>2 a(n)=n+1 if n odd, n-1 if n even.
The induction hypotheses are: for i
For b(n), the argument is very similar, except that the missing numbers when looking for b(n) are slightly different, and (setting B(n) = b(1)+...b(n)), we have B(2i)=(i+1)(2i+1), B(2i+1)=(i+2)(2i+1). - N. J. A. Sloane, Mar 14 2018
When sequence a(n) is increasing, then the Cesàro means sequence c(n) = (a(1)+...+a(n))/n is also increasing, but the converse is false. This sequence is a such an example where c(n) is increasing, while a(n) is not increasing (Arnaudiès et al.). See proof in A354008. - Bernard Schott, May 11 2022

References

  • J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Crossrefs

All of A014681, A103889, A113981, A114112, A114285 are essentially the same sequence. - N. J. A. Sloane, Mar 12 2018
Cf. A114113 (partial sums).
See A084265 for the partial sums of the b(n) sequence.
About Cesàro mean theorem: A033999, A141310, A237420, A354008.
Cf. A244009.

Programs

  • Mathematica
    a[1] = 1; a[n_] := a[n] = Block[{k = 1, s, t = Table[ a[i], {i, n - 1}]}, s = Plus @@ t; While[ Position[t, k] != {} || Mod[s, k] == 0, k++ ]; k]; Array[a, 72] (* Robert G. Wilson v, Nov 18 2005 *)
  • PARI
    a(n) = if (n<=2, n, if (n%2, n+1, n-1)); \\ Michel Marcus, May 16 2022
    
  • Python
    def A114112(n): return n + (0 if n <= 2 else -1+2*(n%2)) # Chai Wah Wu, May 24 2022

Formula

G.f.: x*(x^4-2*x^3+x^2+x+1)/((1-x)*(1-x^2)). - N. J. A. Sloane, Mar 12 2018
The g.f. for the b(n) sequence is x*(x^3-3*x^2+2*x+2)/((1-x)*(1-x^2)). - Conjectured (correctly) by Colin Barker, Mar 04 2018
E.g.f.: 1 - x + x^2/2 + (x - 1)*cosh(x) + (x + 1)*sinh(x). - Stefano Spezia, Sep 02 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - log(2) (A244009). - Amiram Eldar, Jun 29 2025

Extensions

More terms from Robert G. Wilson v, Nov 18 2005
Entry edited (with simpler definition) by N. J. A. Sloane, Mar 12 2018

A035328 a(n) = n*(2*n-1)*(2*n+1).

Original entry on oeis.org

0, 3, 30, 105, 252, 495, 858, 1365, 2040, 2907, 3990, 5313, 6900, 8775, 10962, 13485, 16368, 19635, 23310, 27417, 31980, 37023, 42570, 48645, 55272, 62475, 70278, 78705, 87780, 97527, 107970, 119133, 131040, 143715, 157182, 171465, 186588
Offset: 0

Keywords

Comments

Bisection of A027480. For n>1, gives area of triangle two of whose cevians bound three smaller triangles with areas n-1, n, n+1 contiguously. - Lekraj Beedassy, Dec 21 2006

References

  • Eric Harold Neville, Jacobian Elliptic Functions, 2nd ed., 1951, p. 38.
  • Konrad Knopp, Theory and Application of Infinite Series, Dover, p. 269

Programs

  • Magma
    [n*(2*n-1)*(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jun 07 2011
    
  • Mathematica
    Table[n(2n-1)(2n+1),{n,0,40}] (* Harvey P. Dale, Jan 11 2014 *)
  • PARI
    vector(100,n,(n-1)*(2*n-1)*(2*n-3)) \\ Derek Orr, Jan 29 2015

Formula

a(n) = 3*A000447(n) = 3*A000292(2*n-1).
Sum_{n>=1} 1/a(n) = 2*log(2) - 1. - Benoit Cloitre, Apr 05 2002
a(n) = A204558(2*n) / (2*n). - Reinhard Zumkeller, Jan 18 2012
G.f.: 3*x*(1 + 6*x + x^2)/(1 - x)^4. - Colin Barker, Mar 27 2012
Product_{n>=1} 4*n^3/a(n) = Pi/2. - Daniel Suteu, Feb 05 2017
a(n) = Sum_{i=0..2*n} A046092(n-1)+i = Sum_{i=2*n+1..4*n-1} A046092(n-1)+i for n>0. Example: for n = 5, A046092(4) = 40 and a(5) = 40 + 41 + 42 + ... + 49 + 50 = 51 + 52 + 53 + ... + 58 + 59 = 495. - Bruno Berselli, Oct 26 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - log(2) (A244009). - Amiram Eldar, Jan 30 2021
E.g.f.: exp(x)*x*(3 + 12*x + 4*x^2). - Stefano Spezia, Sep 03 2023

Extensions

More terms from Benoit Cloitre, Apr 05 2002

A188859 Decimal expansion of 2 - log(4).

Original entry on oeis.org

6, 1, 3, 7, 0, 5, 6, 3, 8, 8, 8, 0, 1, 0, 9, 3, 8, 1, 1, 6, 5, 5, 3, 5, 7, 5, 7, 0, 8, 3, 6, 4, 6, 8, 6, 3, 8, 4, 8, 9, 9, 9, 7, 3, 1, 2, 7, 9, 4, 8, 9, 4, 9, 1, 7, 5, 8, 6, 3, 9, 9, 8, 1, 0, 1, 3, 2, 1, 2, 7, 5, 6, 0, 6, 0, 6, 1, 0, 5, 6, 8, 7, 8, 8, 2, 7, 3, 3, 4, 6, 0, 0, 7, 1, 6, 2, 6, 2, 4, 9, 1, 5, 9, 9, 7
Offset: 0

Author

Keywords

Comments

Limit as n increases without bound of the probability that n mod m is less than m/2, with m chosen uniformly at random from 1..n. (As usual, 0 <= n mod m < m.)

Examples

			0.61370563888010938116553575708364686384899973127949...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[2 - Log[4], 10, 120][[1]]
  • PARI
    vecextract(eval(Vec(Str(2-log(4)))),"3..")

Formula

From Amiram Eldar, Aug 15 2020: (Start)
Equals Sum_{k>=1} 1/(2*k^2 + k).
Equals -Integral_{x=0..1} log(1-x^2) dx. (End)
Equals Sum_{k>=1} A023416(k)/(k*(k+1)) (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021
Equals 1/(1 + 2/(3 + 1^2/(4 + 3^2/(5 + 2^2/(6 + 4^2/(7 + 3^2/(8 + 5^2/(9 + 4^2/(10 + 6^2/(11 + ... + (n-1)^2/((2*n) + (n+1)^2/((2*n+1) + ... )))))))))))). Cf. A016639. - Peter Bala, Mar 04 2024
Equals 1/2 + Sum_{k>=1} 1/(k*(4*k^2-1)^2). - Sean A. Irvine, Apr 06 2025

A033931 a(n) = lcm(n,n+1,n+2).

Original entry on oeis.org

6, 12, 60, 60, 210, 168, 504, 360, 990, 660, 1716, 1092, 2730, 1680, 4080, 2448, 5814, 3420, 7980, 4620, 10626, 6072, 13800, 7800, 17550, 9828, 21924, 12180, 26970, 14880, 32736, 17952, 39270, 21420, 46620, 25308, 54834, 29640, 63960, 34440, 74046, 39732, 85140
Offset: 1

Author

Keywords

Comments

Also denominator of h(n+2) - h(n-1), where h(n) is the n-th harmonic number Sum_{k=1..n} 1/k, the numerator is A188386. - Reinhard Zumkeller, Jul 04 2012

Programs

  • Haskell
    a033931 n = lcm n (lcm (n + 1) (n + 2))  -- Reinhard Zumkeller, Jul 04 2012
    
  • Magma
    [Numerator((n^3-n)/(n^2+1)): n in [2..50]]; // Vincenzo Librandi, Aug 19 2014
    
  • Maple
    a:= n-> ilcm($n..n+2):
    seq(a(n), n=1..50);  # Alois P. Heinz, Jul 18 2025
  • Mathematica
    LCM@@@Partition[Range[50],3,1] (* or *) LinearRecurrence[{0,4,0,-6,0,4,0,-1},{6,12,60,60,210,168,504,360},50] (* Harvey P. Dale, Jun 29 2019 *)
  • PARI
    a(n) = lcm(n^2+n,n+2) \\ Charles R Greathouse IV, Sep 30 2016

Formula

a(n) = n*(n+1)*(n+2)*[3-(-1)^n]/4.
From Reinhard Zumkeller, Jul 04 2012: (Start)
a(n) = 6 * A067046(n).
A007947(a(n)) = A078637(n). (End)
From Amiram Eldar, Sep 29 2022: (Start)
Sum_{n>=1} 1/a(n) = 1 - log(2) (A244009).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2) - 2. (End)

A100381 a(n) = 2^n*binomial(n,2).

Original entry on oeis.org

0, 0, 4, 24, 96, 320, 960, 2688, 7168, 18432, 46080, 112640, 270336, 638976, 1490944, 3440640, 7864320, 17825792, 40108032, 89653248, 199229440, 440401920, 968884224, 2122317824, 4630511616, 10066329600, 21810380800, 47110422528
Offset: 0

Author

Jorge Coveiro, Dec 30 2004

Keywords

Comments

From Enrique Navarrete, Jun 13 2023: (Start)
a(n) is the number of ways to partition the set [n]={1,2,...,n} into two sets S,T and select 2 elements in total (from either S or T or both).
Example. For n=4, sample partitions are given (where S(i),T(j) means i elements are selected from S, j elements are selected from T):
S={ }, T={1,2,3,4}: partition [4] in 1 way, S(0),T(2) (6 ways);
S={1}, T={2,3,4}: partition [4] in 4 such ways, S(1),T(1) or S(0),T(2) (24 ways);
S={1,2}, T={3,4}: partition [4], in such 6 ways, S(1),T(1) or S(0),T(2) or S(2),T(0) (36 ways);
S={1,2,3}, T={4}: partition [4] in 4 such ways, S(1),T(1) or S(2),T(0) (24 ways);
S={1,2,3,4}, T={ }: partition [4] in 1 way, S(2),T(0) (6 ways). (End)

References

  • Jolley, Summation of Series, Dover (1961), eq (214) page 40.

Crossrefs

Programs

  • Maple
    seq(2^n*binomial(n,2),n=0..20);
  • Mathematica
    Range[0,20]! CoefficientList[Series[2x^2 Exp[2x],{x,0,20}],x]
    Table[2^n Binomial[n,2],{n,0,30}] (* or *) LinearRecurrence[{6,-12,8},{0,0,4},30] (* Harvey P. Dale, Aug 15 2020 *)
  • PARI
    a(n)=binomial(n,2)<Charles R Greathouse IV, Oct 16 2015

Formula

Sum_{n>=2} 1/a(n) = 1 - log(2) = 0.3068528.... - Graeme McRae, Jul 28 2006
a(n) = Sum_{k=0..n} k*2^k = 2*A001815(n). - Zerinvary Lajos, Oct 09 2006
E.g.f.: 2*x^2*exp(2x).
a(n) = 4*A001788(n-1). - Johannes W. Meijer, Jun 27 2009
Sum_{j=1..k} (j+2)/a(j+1) = 1 - 1/((k+1)*2^k). [Jolley]
G.f.: -4*x^2 / (2*x-1)^3. - R. J. Mathar, Oct 05 2011
Sum_{n>=2} (-1)^n/a(n) = 3*log(3/2) - 1. - Amiram Eldar, Jul 20 2020
From Peter Bala Mar 04 2024: (Start)
Sum_{k = 2..n+2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/(3*n + 4)))).
Sum_{k = 2..n+2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/(n + 4)))).
Letting n -> oo in the above gives the continued fraction representations
1 - log(2) = Sum_{k >= 2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/((3*n + 4) - ... )))) (an equivalent continued fraction for 1 - log(2) was conjectured by the Ramanujan machine) and
3*log(3/2) - 1 = Sum_{k >= 2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/((n + 4) + ... )))). (End)

A245238 Decimal expansion of the Dickman function evaluated at 1/4.

Original entry on oeis.org

0, 0, 4, 9, 1, 0, 9, 2, 5, 6, 4, 7, 7, 6, 0, 8, 3, 2, 3, 5, 2, 7, 3, 9, 1, 5, 0, 9, 2, 3, 6, 1, 5, 1, 8, 6, 0, 3, 2, 4, 8, 4, 2, 9, 7, 4, 1, 7, 6, 9, 2, 9, 4, 5, 9, 7, 7, 9, 6, 1, 6, 5, 7, 5, 2, 8, 0, 3, 0, 6, 3, 1
Offset: 0

Author

Keywords

Comments

Density of the fourth-root-smooth numbers.

Examples

			F(1/4) = 0.00491092564776083235273915092361518603248429741769294597796...
		

References

  • Karl Dickman, On the frequency of numbers containing prime factors of a certain relative magnitude, Arkiv för Matematik, Astronomi och Fysik 22A 10 (1930), pp. 1-14.

Crossrefs

F(1/2) = A244009, F(1/3) = A175475.

Programs

  • Mathematica
    RealDigits[1-Log[4]+PolyLog[2, 1/4]+2*Log[2]^2-Pi^2/12-PolyLog[3, 1/4]-PolyLog[2, 1/4]*Log[2]-2/3*Log[2]^3+13*Zeta[3]/24,10,100,-1][[1]] (* Vaclav Kotesovec, Jul 15 2014 *)
  • PARI
    1-log(4)+polylog(2,1/4)+2*log(2)^2-Pi^2/12-polylog(3,1/4)-polylog(2,1/4)*log(2)-2/3*log(2)^3+13*zeta(3)/24

A277559 Decimal expansion of the first derivative of the infinite power tower function x^x^x... at x = sqrt(2).

Original entry on oeis.org

9, 2, 1, 7, 5, 3, 6, 7, 0, 0, 1, 9, 2, 3, 1, 5, 4, 4, 7, 0, 5, 1, 3, 1, 3, 6, 3, 2, 6, 5, 2, 4, 7, 9, 1, 9, 6, 0, 8, 2, 3, 9, 7, 9, 9, 6, 0, 3, 7, 9, 5, 4, 2, 9, 0, 3, 1, 1, 2, 0, 8, 4, 1, 2, 7, 3, 3, 3, 2, 2, 5, 3, 6, 7, 3, 5, 0, 3, 0, 2, 9, 0, 7, 5, 7, 4, 5, 7, 5, 1, 5, 2, 2, 5, 4, 3, 0, 7, 9, 3, 2, 4, 2, 0, 2
Offset: 1

Author

Alois P. Heinz, Oct 19 2016

Keywords

Comments

It is known that sqrt(2)^sqrt(2)^sqrt(2)^... = 2.

Examples

			9.21753670019231544705131363265247919608239799603795429...
		

Programs

  • Magma
    Sqrt(8)/(1-Log(2)); // G. C. Greubel, Jul 27 2018
  • Mathematica
    RealDigits[Sqrt[8]/(1-Log[2]), 10, 100][[1]] (* G. C. Greubel, Jul 27 2018 *)
  • PARI
    sqrt(8)/(1-log(2)) \\ Michel Marcus, Oct 20 2016
    

Formula

Equals 2^(3/2)/(1-log(2)).
Equals A010466/A244009. - Michel Marcus, Oct 20 2016

A346152 a(n) is the least prime divisor p_j of n such that if n = Product_{i=1..k} p_i^e_i and p_1 < p_2 < ... < p_k, then Product_{i=1..j-1} p_i^e_i <= sqrt(n) < Product_{i=j..k} p_i^e_i. a(1) = 1.

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 2, 13, 7, 5, 2, 17, 3, 19, 5, 7, 11, 23, 2, 5, 13, 3, 7, 29, 3, 31, 2, 11, 17, 7, 3, 37, 19, 13, 2, 41, 7, 43, 11, 3, 23, 47, 2, 7, 5, 17, 13, 53, 3, 11, 2, 19, 29, 59, 3, 61, 31, 3, 2, 13, 11, 67, 17, 23, 5, 71, 3, 73, 37, 5
Offset: 1

Author

Amiram Eldar, Jul 07 2021

Keywords

Comments

First differs from A088387 at n = 30.
First differs from A197861 at n = 24.
Erdős (1982) proved that for any 0 <= alpha <= 1, the asymptotic density g(alpha) of numbers k with a(k) < k^alpha exists, and that it is continuous and strictly increasing between g(0) = 0 and g(1) = 1.
In the case of alpha = 1/2, the sequence is A063539 \ {1} whose asymptotic density is g(1/2) = 1 - log(2) (A244009).

Examples

			a(4) = 2 since 1 <= sqrt(4) < 2^2.
a(6) = 3 since 2 <= sqrt(6) < 2*3.
a(30) = 3 since 2 <= sqrt(30) < 2*3*5.
		

Programs

  • Mathematica
    a[1] = 1; a[n_] := Module[{fct = FactorInteger[n], prods, ind}, prods = Rest @ FoldList[Times, 1, Power @@@ fct]; ind = FirstPosition[prods^2, _?(# > n &)][[1]]; fct[[ind, 1]]]; Array[a, 100]

Formula

a(n) <= A006530(n).
a(p^e) = p for prime p and e>=1.
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