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.

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A008836 Liouville's function lambda(n) = (-1)^k, where k is number of primes dividing n (counted with multiplicity).

Original entry on oeis.org

1, -1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, 1, 1, -1, -1, -1, -1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, -1, -1, 1, -1, 1, -1, -1, 1, 1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1, -1, 1, 1, -1, 1, 1, 1, 1, 1, -1, 1, 1, -1, 1, 1, 1, 1, -1, -1, -1, 1, -1
Offset: 1

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Author

Keywords

Comments

Coons and Borwein: "We give a new proof of Fatou's theorem: if an algebraic function has a power series expansion with bounded integer coefficients, then it must be a rational function. This result is applied to show that for any non-trivial completely multiplicative function from N to {-1,1}, the series sum_{n=1..infinity} f(n)z^n is transcendental over {Z}[z]; in particular, sum_{n=1..infinity} lambda(n)z^n is transcendental, where lambda is Liouville's function. The transcendence of sum_{n=1..infinity} mu(n)z^n is also proved." - Jonathan Vos Post, Jun 11 2008
Coons proves that a(n) is not k-automatic for any k > 2. - Jonathan Vos Post, Oct 22 2008
The Riemann hypothesis is equivalent to the statement that for every fixed epsilon > 0, lim_{n -> infinity} (a(1) + a(2) + ... + a(n))/n^(1/2 + epsilon) = 0 (Borwein et al., theorem 1.2). - Arkadiusz Wesolowski, Oct 08 2013

Examples

			a(4) = 1 because since bigomega(4) = 2 (the prime divisor 2 is counted twice), then (-1)^2 = 1.
a(5) = -1 because 5 is prime and therefore bigomega(5) = 1 and (-1)^1 = -1.
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 37.
  • P. Borwein, S. Choi, B. Rooney and A. Weirathmueller, The Riemann Hypothesis: A Resource for the Aficionado and Virtuoso Alike, Springer, Berlin, 2008, pp. 1-11.
  • H. Gupta, On a table of values of L(n), Proceedings of the Indian Academy of Sciences. Section A, 12 (1940), 407-409.
  • H. Gupta, A table of values of Liouville's function L(n), Research Bulletin of East Panjab University, No. 3 (Feb. 1950), 45-55.
  • P. Ribenboim, Algebraic Numbers, p. 44.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 279.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.3.5 on page 99.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 112.

Crossrefs

Möbius transform of A010052.
Cf. A182448 (Dgf at s=2), A347328 (Dgf at s=3), A347329 (Dgf at s=4).

Programs

  • Haskell
    a008836 = (1 -) . (* 2) . a066829  -- Reinhard Zumkeller, Nov 19 2011
    
  • Maple
    A008836 := n -> (-1)^numtheory[bigomega](n); # Peter Luschny, Sep 15 2011
    with(numtheory): A008836 := proc(n) local i,it,s; it := ifactors(n): s := (-1)^add(it[2][i][2], i=1..nops(it[2])): RETURN(s) end:
  • Mathematica
    Table[LiouvilleLambda[n], {n, 100}] (* Enrique Pérez Herrero, Dec 28 2009 *)
    Table[If[OddQ[PrimeOmega[n]],-1,1],{n,110}] (* Harvey P. Dale, Sep 10 2014 *)
  • PARI
    {a(n) = if( n<1, 0, n=factor(n); (-1)^sum(i=1, matsize(n)[1], n[i,2]))}; /* Michael Somos, Jan 01 2006 */
    
  • PARI
    a(n)=(-1)^bigomega(n) \\ Charles R Greathouse IV, Jan 09 2013
    
  • Python
    from sympy import factorint
    def A008836(n): return -1 if sum(factorint(n).values()) % 2 else 1 # Chai Wah Wu, May 24 2022

Formula

Dirichlet g.f.: zeta(2s)/zeta(s); Dirichlet inverse of A008966.
Sum_{ d divides n } lambda(d) = 1 if n is a square, otherwise 0.
Completely multiplicative with a(p) = -1, p prime.
a(n) = (-1)^A001222(n) = (-1)^bigomega(n). - Jonathan Vos Post, Apr 16 2006
a(n) = A033999(A001222(n)). - Jaroslav Krizek, Sep 28 2009
Sum_{d|n} a(d) *(A000005(d))^2 = a(n) *Sum{d|n} A000005(d). - Vladimir Shevelev, May 22 2010
a(n) = 1 - 2*A066829(n). - Reinhard Zumkeller, Nov 19 2011
a(n) = i^(tau(n^2)-1) where tau(n) is A000005 and i is the imaginary unit. - Anthony Browne, May 11 2016
a(n) = A106400(A156552(n)). - Antti Karttunen, May 30 2017
Recurrence: a(1)=1, n > 1: a(n) = sign(1/2 - Sum_{dMats Granvik, Oct 11 2017
a(n) = Sum_{ d | n } A008683(d)*A010052(n/d). - Jinyuan Wang, Apr 20 2019
a(1) = 1; a(n) = -Sum_{d|n, d < n} mu(n/d)^2 * a(d). - Ilya Gutkovskiy, Mar 10 2021
a(n) = (-1)^A349905(n). - Antti Karttunen, Apr 26 2022
From Ridouane Oudra, Jun 02 2024: (Start)
a(n) = (-1)^A066829(n);
a(n) = (-1)^A063647(n);
a(n) = A101455(A048691(n));
a(n) = sin(tau(n^2)*Pi/2). (End)

A048691 a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.

Original entry on oeis.org

1, 3, 3, 5, 3, 9, 3, 7, 5, 9, 3, 15, 3, 9, 9, 9, 3, 15, 3, 15, 9, 9, 3, 21, 5, 9, 7, 15, 3, 27, 3, 11, 9, 9, 9, 25, 3, 9, 9, 21, 3, 27, 3, 15, 15, 9, 3, 27, 5, 15, 9, 15, 3, 21, 9, 21, 9, 9, 3, 45, 3, 9, 15, 13, 9, 27, 3, 15, 9, 27, 3, 35, 3, 9, 15, 15, 9, 27, 3, 27
Offset: 1

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Author

Keywords

Comments

Inverse Moebius transform of A034444: Sum_{d|n} 2^omega(d), where omega(n) = A001221(n) is the number of distinct primes dividing n.
Number of elements in the set {(x,y): x|n, y|n, gcd(x,y)=1}.
Number of elements in the set {(x,y): lcm(x,y)=n}.
Also gives total number of positive integral solutions (x,y), order being taken into account, to the optical or parallel resistor equation 1/x + 1/y = 1/n. Indeed, writing the latter as X*Y=N, with X=x-n, Y=y-n, N=n^2, the one-to-one correspondence between solutions (X, Y) and (x, y) is obvious, so that clearly, the solution pairs (x, y) are tau(N)=tau(n^2) in number. - Lekraj Beedassy, May 31 2002
Number of ordered pairs of positive integers (a,c) such that n^2 - ac = 0. Therefore number of quadratic equations of the form ax^2 + 2nx + c = 0 where a,n,c are positive integers and each equation has two equal (rational) roots, -n/a. (If a and c are positive integers, but, instead, the coefficient of x is odd, it is impossible for the equation to have equal roots.) - Rick L. Shepherd, Jun 19 2005
Problem A1 on the 21st Putnam competition in 1960 (see John Scholes link) asked for the number of pairs of positive integers (x,y) such that xy/(x+y) = n: the answer is a(n); for n = 4, the a(4) = 5 solutions (x,y) are (5,20), (6,12), (8,8), (12,6), (20,5). - Bernard Schott, Feb 12 2023
Numbers k such that a(k)/d(k) is an integer are in A217584 and the corresponding quotients are in A339055. - Bernard Schott, Feb 15 2023

References

  • A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:1938-1960 Soln. to Prob. 1 1960, p. 516, MAA, 1980.
  • Ross Honsberger, More Mathematical Morsels, Morsel 43, pp. 232-3, DMA No. 10 MAA, 1991.
  • Loren C. Larson, Problem-Solving Through Problems, Prob. 3.3.7, p. 102, Springer 1983.
  • Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Algebra, Prob. 9-9 pp. 143 Dover NY, 1988.
  • D. O. Shklarsky et al., The USSR Olympiad Problem Book, Soln. to Prob. 123, pp. 28, 217-8, Dover NY.
  • Wacław Sierpiński, Elementary Theory of Numbers, pp. 71-2, Elsevier, North Holland, 1988.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 91.
  • Charles W. Trigg, Mathematical Quickies, Question 194, pp. 53, 168, Dover, 1985.

Crossrefs

Partial sums give A061503.
For similar LCM sequences, see A070919, A070920, A070921.
For the earliest occurrence of 2n-1 see A016017.

Programs

Formula

a(n) = A000005(A000290(n)).
tau(n^2) = Sum_{d|n} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
Multiplicative with a(p^e) = 2e+1. - Vladeta Jovovic, Jul 23 2001
Also a(n) = Sum_{d|n} (tau(d)*moebius(n/d)^2), Dirichlet convolution of A000005 and A008966. - Benoit Cloitre, Sep 08 2002
a(n) = A055205(n) + A000005(n). - Reinhard Zumkeller, Dec 08 2009
Dirichlet g.f.: (zeta(s))^3/zeta(2s). - R. J. Mathar, Feb 11 2011
a(n) = Sum_{d|n} 2^omega(d). Inverse Mobius transform of A034444. - Enrique Pérez Herrero, Apr 14 2012
G.f.: Sum_{k>=1} 2^omega(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 10 2018
Sum_{k=1..n} a(k) ~ n*(6/Pi^2)*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Jan 26 2023

Extensions

Additional comments from Vladeta Jovovic, Apr 29 2001

A018892 Number of ways to write 1/n as a sum of exactly 2 unit fractions.

Original entry on oeis.org

1, 2, 2, 3, 2, 5, 2, 4, 3, 5, 2, 8, 2, 5, 5, 5, 2, 8, 2, 8, 5, 5, 2, 11, 3, 5, 4, 8, 2, 14, 2, 6, 5, 5, 5, 13, 2, 5, 5, 11, 2, 14, 2, 8, 8, 5, 2, 14, 3, 8, 5, 8, 2, 11, 5, 11, 5, 5, 2, 23, 2, 5, 8, 7, 5, 14, 2, 8, 5, 14, 2, 18, 2, 5, 8, 8, 5, 14, 2, 14, 5, 5, 2, 23, 5, 5, 5, 11, 2, 23, 5, 8, 5, 5, 5, 17, 2, 8, 8
Offset: 1

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Author

Keywords

Comments

Number of elements in the set {(x,y): x|n, y|n, x<=y, gcd(x,y)=1}. Number of divisors of n^2 less than or equal to n. - Vladeta Jovovic, May 03 2002
Equivalently, number of pairs (x,y) such that lcm(x,y)=n. - Benoit Cloitre, May 16 2002
Also, number of right triangles with an integer hypotenuse and height n. - Reinhard Zumkeller, Jul 10 2002
The triangles are to be considered as resting on their hypotenuse, with the height measured to the right angle. - Franklin T. Adams-Watters, Feb 19 2015
a(n) >= 2 for n>=2 because of the identities 1/n = 1/(2*n) + 1/(2*n) = 1/(n+1) + 1/(n*(n+1)). - Lekraj Beedassy, May 04 2004
a(n) is the number of divisors of n^2 that are <= n; e.g., a(12) counts these 8 divisors of 12: 1,2,3,4,6,8,9,12. - Clark Kimberling, Apr 21 2019

Examples

			n=1: 1/1 = 1/2 + 1/2.
n=2: 1/2 = 1/4 + 1/4 = 1/3 + 1/6.
n=3: 1/3 = 1/6 + 1/6 = 1/4 + 1/12.
		

References

  • K. S. Brown, Posting to netnews group sci.math, Aug 17 1996.
  • L. E. Dickson, History of The Theory of Numbers, Vol. 2 p. 690, Chelsea NY 1923.
  • A. M. and I. M. Yaglom, Challenging Mathematical Problems With Elementary Solutions, Vol. 1, Dover, N.Y., 1987, pp. 8 and 60, Problem 19.

Crossrefs

Programs

  • Haskell
    a018892 n = length [d | d <- [1..n], n^2 `mod` d == 0]
    -- Reinhard Zumkeller, Jan 08 2012
    
  • Mathematica
    f[j_, n_] := (Times @@ (j(Last /@ FactorInteger[n]) + 1) + j - 1)/j; Table[f[2, n], {n, 96}] (* Robert G. Wilson v, Aug 03 2005 *)
    a[n_] := (DivisorSigma[0, n^2] + 1)/2; Table[a[n], {n, 1, 99}](* Jean-François Alcover, Dec 19 2011, after Vladeta Jovovic *)
  • PARI
    A018892(n)=(numdiv(n^2)+1)/2 \\ M. F. Hasler, Dec 30 2007
    
  • PARI
    A018892s(n)=local(t=divisors(n^2));vector((#t+1)/2,i,[n+t[i],n+n^2/t[i]]) /* show solutions */ \\ M. F. Hasler, Dec 30 2007
    
  • PARI
    a(n)=sumdiv(n,d,sum(i=1,d,lcm(d,i)==n)) \\ Charles R Greathouse IV, Apr 08 2012
    
  • Python
    from math import prod
    from sympy import factorint
    def A018892(n): return prod((a<<1)+1 for a in factorint(n).values())+1>>1 # Chai Wah Wu, Aug 20 2023

Formula

If n = (p1^a1)(p2^a2)...(pt^at), a(n) = ((2*a1 + 1)(2*a2 + 1) ... (2*at + 1) + 1)/2.
a(n) = (tau(n^2)+1)/2. - Vladeta Jovovic, May 03 2002
a(n) = A063647(n)+1 = A046079(2*n)+1. - Lekraj Beedassy, Dec 01 2003
a(n) = Sum_{d|n} phi(2^omega(d)), where phi is A000010 and omega is A001221. - Enrique Pérez Herrero, Apr 13 2012
a(n) = A000005(n) + A089233(n). - James Spahlinger, Feb 16 2016
a(n) = n + Sum_{i=1..n} sign(n^2 mod -i). - Wesley Ivan Hurt, Apr 07 2021
a(n) = Sum_{d|n} mu(n/d)*A184389(d). - Ridouane Oudra, Feb 22 2022
Sum_{k=1..n} a(k) ~ (n/(2*zeta(2)))*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 + zeta(2) + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Oct 03 2024

Extensions

More terms from David W. Wilson, Sep 15 1996
First example corrected by Jason Orendorff (jason.orendorff(AT)gmail.com), Jan 02 2009
Incorrect Mathematica program deleted by N. J. A. Sloane, Jul 08 2009

A000977 Numbers that are divisible by at least three different primes.

Original entry on oeis.org

30, 42, 60, 66, 70, 78, 84, 90, 102, 105, 110, 114, 120, 126, 130, 132, 138, 140, 150, 154, 156, 165, 168, 170, 174, 180, 182, 186, 190, 195, 198, 204, 210, 220, 222, 228, 230, 231, 234, 238, 240, 246, 252, 255, 258, 260, 264, 266, 270, 273, 276, 280, 282, 285
Offset: 1

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Author

Keywords

Comments

a(n+1)-a(n) seems bounded and sequence appears to give n such that the number of integers of the form nk/(n+k) k>=1 is not equal to Sum_{ d | n} omega(d) (i.e., n such that A062799(n) is not equal to A063647(n)). - Benoit Cloitre, Aug 27 2002
The first differences are bounded: clearly a(n+1) - a(n) <= 30. - Charles R Greathouse IV, Dec 19 2011

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 844.

Crossrefs

Complement of A070915.

Programs

  • Haskell
    a000977 n = a000977_list !! (n-1)
    a000977_list = filter ((> 2) . a001221) [1..]
    -- Reinhard Zumkeller, May 03 2013
  • Maple
    A000977 := proc(n)
    if (nops(numtheory[factorset](n)) >= 3) then
       RETURN(n)
    fi: end:  seq(A000977(n), n=1..500); # Jani Melik, Feb 24 2011
  • Mathematica
    DeleteCases[Table[If[Count[PrimeQ[Divisors[i]], True] >= 3, i, 0], {i, 1, 274}], 0]
    Select[Range[300], PrimeNu[#] >= 3 &] (* Paolo Xausa, Mar 28 2024 *)
  • PARI
    is(n)=omega(n)>2 \\ Charles R Greathouse IV, Dec 19 2011
    

Formula

a(n) = n + O(n log log n / log n). - Charles R Greathouse IV, Dec 19 2011 A001221(a(n)) > 2. - Reinhard Zumkeller, May 03 2013
A033992 UNION A033993 UNION A051270 UNION A074969 UNION A176655 UNION ... - R. J. Mathar, Dec 05 2016

Extensions

More terms from Vit Planocka (planocka(AT)mistral.cz), Sep 17 2002

A127730 Triangle read by rows: row n consists of the positive integers m where m+n divides m*n.

Original entry on oeis.org

2, 6, 4, 12, 20, 3, 6, 12, 30, 42, 8, 24, 56, 18, 72, 10, 15, 40, 90, 110, 4, 6, 12, 24, 36, 60, 132, 156, 14, 35, 84, 182, 10, 30, 60, 210, 16, 48, 112, 240, 272, 9, 18, 36, 63, 90, 144, 306, 342, 5, 20, 30, 60, 80, 180, 380, 28, 42, 126, 420, 22, 99, 220, 462
Offset: 2

Views

Author

Leroy Quet, Jan 26 2007

Keywords

Comments

The maximum term of the n-th row, for n >= 2, is n*(n-1). The minimum term of row n is A063427(n). Row n contains A063647(n) terms (according to a comment by Benoit Cloitre). For p prime, row p^k has k terms. (Each term in row p^k is of the form p^k*(p^j -1), 1 <= j <= k.)

Examples

			Row 6 is (3,6,12,30) because 6+3 = 9 divides 6*3 = 18, 6+6 = 12 divides 6*6 = 36, 6+12 = 18 divides 6*12 = 72 and 6+30 = 36 divides 6*30 = 180.
		

Crossrefs

Programs

  • Maple
    for n from 2 to 20 do for m from 1 to n*(n-1) do if(m*n mod (m+n) = 0)then printf("%d, ",m): fi: od: od: # Nathaniel Johnston, Jun 22 2011
  • Mathematica
    f[n_] := Select[Range[n^2], Mod[n*#, n + # ] == 0 &];Table[f[n], {n, 2, 24}] // Flatten (* Ray Chandler, Feb 13 2007 *)
  • PARI
    arow(n)=local(d,m);d=divisors(n^2);vector(#d\2,k,m=d[ #d\2-k+1];n*(n-m)/m) \\ Franklin T. Adams-Watters, Aug 07 2009

Formula

Let d_n be the sequence of divisors of n^2 that are less than n, in reverse order. Then T(n,k) = n*(n-d_n(k))/d_n(k). - Franklin T. Adams-Watters, Aug 07 2009

Extensions

Extended by Ray Chandler, Feb 13 2007

A089233 Number of coprime pairs of divisors > 1 of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 2, 1, 1, 0, 3, 0, 1, 0, 2, 0, 6, 0, 0, 1, 1, 1, 4, 0, 1, 1, 3, 0, 6, 0, 2, 2, 1, 0, 4, 0, 2, 1, 2, 0, 3, 1, 3, 1, 1, 0, 11, 0, 1, 2, 0, 1, 6, 0, 2, 1, 6, 0, 6, 0, 1, 2, 2, 1, 6, 0, 4, 0, 1, 0, 11, 1, 1, 1, 3, 0, 11, 1, 2, 1, 1, 1, 5, 0, 2, 2, 4, 0, 6, 0, 3, 6
Offset: 1

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Author

Reinhard Zumkeller, Dec 11 2003

Keywords

Comments

Also the number of divisors of n^2 which do not divide n and which are less than n. See link for proof. - Andrew Weimholt, Dec 06 2009
a(A000961(n)) = 0; a(A006881(n)) = 1; a(A054753(n)) = 2; a(A065036(n)) = 3. - Robert G. Wilson v, Dec 16 2009
First occurrence of k beginning with 0: 1, 6, 12, 24, 36, 96, 30, 384, 144, 216, 288, 60, 432, 24576, 1152, 864, 120, 393216, 1728, 1572864, 180, 240, 18432, 25165824, 5184, 210, 480, 13824, 10368, 360, 15552, 960, 20736, 55296, 1179648, 31104, 900, ..., . Except for 1, each is divisible by 6. Also the first occurrence of k must occur at or before 6*2^(n-1). - Robert G. Wilson v, Dec 16 2009
a(3*2^n) = n; if x = 2^n, then a(x) = a(2*x); and if x is not a power of two, then a(x) = y then a(2*x) > y. - Robert G. Wilson v, Dec 16 2009
a(n) = 0 iff n is a prime power. - Franklin T. Adams-Watters, Aug 20 2013

Programs

  • Haskell
    a089233 n = sum $ [a063524 $ gcd u v | let ds = tail $ a027750_row n,
                                           u <- ds, v <- dropWhile (<= u) ds]
    -- Reinhard Zumkeller, Sep 04 2013
    
  • Magma
    [(NumberOfDivisors(n^2)-1)/2 - NumberOfDivisors(n)+1: n in [1..100]]; // Vincenzo Librandi, Dec 23 2018
  • Mathematica
    a[n_] := (DivisorSigma[0, n^2] - 1)/2 - DivisorSigma[0, n] + 1; Array[a, 104] (* Robert G. Wilson v, Dec 16 2009 *)
  • PARI
    a(n) = (numdiv(n^2)-1)/2 - numdiv(n) + 1; \\ Michel Marcus, Feb 17 2016
    

Formula

a(n) = #{(x,y): 1 < x < y, x|n, y|n and gcd(x, y) = 1}.
a(n) = A063647(n) - A000005(n) + 1.
a(n) = A018892(n) - A000005(n). - Franklin T. Adams-Watters, Aug 20 2013

A066446 Number of unordered divisor pairs of n.

Original entry on oeis.org

0, 1, 1, 3, 1, 6, 1, 6, 3, 6, 1, 15, 1, 6, 6, 10, 1, 15, 1, 15, 6, 6, 1, 28, 3, 6, 6, 15, 1, 28, 1, 15, 6, 6, 6, 36, 1, 6, 6, 28, 1, 28, 1, 15, 15, 6, 1, 45, 3, 15, 6, 15, 1, 28, 6, 28, 6, 6, 1, 66, 1, 6, 15, 21, 6, 28, 1, 15, 6, 28, 1, 66, 1, 6, 15, 15, 6, 28, 1, 45, 10, 6, 1, 66, 6, 6, 6, 28
Offset: 1

Views

Author

Robert G. Wilson v, Dec 28 2001

Keywords

Examples

			The divisors of 6 are 1, 2, 3 & 6. In unordered pairs they are {1, 2}, {1, 3}, {1, 6}, {2, 3}, {2, 6}, & {3, 6}. Since there are six pairs, a(6) = 6. Also d(6) = 4. 4*3/2 = 6.
		

Crossrefs

Programs

  • Haskell
    a066446 = a000217 . subtract 1 . a000005'
    -- Reinhard Zumkeller, Sep 08 2015
  • Maple
    with(numtheory): seq(tau(n)*(tau(n)-1)/2, n=1..60); # Ridouane Oudra, Apr 15 2023
  • Mathematica
    Table[ Binomial[ DivisorSigma[0, n], 2], {n, 1, 100}]
  • PARI
    { for (n=1, 1000, a=binomial(numdiv(n), 2); write("b066446.txt", n, " ", a) ) } \\ Harry J. Smith, Feb 15 2010
    

Formula

a(p) = 1 iff p is a prime.
Combinations of d(n), the number of divisors of n (A000005), taken two at a time. If the canonical factorization of n into prime powers is Product p^e(p) then d(n) = Product (e(p) + 1). Therefore a(n) = C(d(n), 2) = d(n)*{ d(n)-1 }/2 which is a triangular number (A000217).
a(n) = A184389(n) - A000005(n) = A035116(n) - A184389(n). - Reinhard Zumkeller, Sep 08 2015
a(n) = A000217(A000005(n)-1). - Antti Karttunen, Sep 21 2018
a(n) = Sum_{k|n, i|n, i < k} 1. - Wesley Ivan Hurt, Aug 20 2020
a(n) = Sum_{d|n} A063647(d). - Ridouane Oudra, Apr 15 2023

A126098 Where records occur in A018892.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 360, 420, 840, 1260, 1680, 2520, 4620, 7560, 9240, 13860, 18480, 27720, 55440, 83160, 110880, 120120, 180180, 240240, 360360, 720720, 1081080, 1441440, 1801800, 2042040, 2882880, 3063060, 4084080, 5405400, 6126120, 12252240, 18378360, 24504480
Offset: 1

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Author

N. J. A. Sloane, Mar 05 2007

Keywords

Comments

Remarkably similar to but ultimately different from A018894. - Jorg Brown and N. J. A. Sloane, Mar 06 2007
This sequence represents "where records occur" for a number of sequences in addition to A018892 including the following: A015995, A015996, A015999, A016001, A016002, A016003, A016005, A016006, A016007, A016008, A016009, A048691, A048785, A063647, A117677, A144943. - Ray Chandler, Dec 04 2008
Subsequence of A025487. - Ray Chandler, Sep 05 2008
Also record-setting elements of tau(n^2) (just as A002182 gives the record-setting elements of tau(n)). The point is that A018892 is (tau(n^2) + 1)/2. As tau(n^2) is odd, the record-setting elements of A018892 are also the record setting elements of tau(n^2). - Allen Tracht, Jan 20 2009

Crossrefs

Cf. A018892, A126097. Equals A117010(n) + 1.

Extensions

More terms from Jorg Brown (jorg(AT)google.com) and T. D. Noe, Mar 05 2007
a(27) corrected by hupo001(AT)gmail.com, Jan 10 2008

A343652 Number of maximal pairwise coprime sets of divisors of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 4, 1, 4, 1, 4, 2, 2, 1, 6, 2, 2, 3, 4, 1, 5, 1, 5, 2, 2, 2, 8, 1, 2, 2, 6, 1, 5, 1, 4, 4, 2, 1, 8, 2, 4, 2, 4, 1, 6, 2, 6, 2, 2, 1, 10, 1, 2, 4, 6, 2, 5, 1, 4, 2, 5, 1, 12, 1, 2, 4, 4, 2, 5, 1, 8, 4, 2, 1, 10, 2, 2
Offset: 1

Views

Author

Gus Wiseman, Apr 25 2021

Keywords

Comments

Also the number of maximal pairwise coprime sets of divisors > 1 of n. For example, the a(n) sets for n = 12, 30, 36, 60, 120 are:
{6} {30} {6} {30} {30}
{12} {2,15} {12} {60} {60}
{2,3} {3,10} {18} {2,15} {120}
{3,4} {5,6} {36} {3,10} {2,15}
{2,3,5} {2,3} {3,20} {3,10}
{2,9} {4,15} {3,20}
{3,4} {5,6} {3,40}
{4,9} {5,12} {4,15}
{2,3,5} {5,6}
{3,4,5} {5,12}
{5,24}
{8,15}
{2,3,5}
{3,4,5}
{3,5,8}

Examples

			The a(n) sets for n = 12, 30, 36, 60, 120:
  {1,6}    {1,30}     {1,6}    {1,30}     {1,30}
  {1,12}   {1,2,15}   {1,12}   {1,60}     {1,60}
  {1,2,3}  {1,3,10}   {1,18}   {1,2,15}   {1,120}
  {1,3,4}  {1,5,6}    {1,36}   {1,3,10}   {1,2,15}
           {1,2,3,5}  {1,2,3}  {1,3,20}   {1,3,10}
                      {1,2,9}  {1,4,15}   {1,3,20}
                      {1,3,4}  {1,5,6}    {1,3,40}
                      {1,4,9}  {1,5,12}   {1,4,15}
                               {1,2,3,5}  {1,5,6}
                               {1,3,4,5}  {1,5,12}
                                          {1,5,24}
                                          {1,8,15}
                                          {1,2,3,5}
                                          {1,3,4,5}
                                          {1,3,5,8}
		

Crossrefs

The case of pairs is A063647.
The case of triples is A066620.
The non-maximal version counting empty sets and singletons is A225520.
The non-maximal version with no 1's is A343653.
The non-maximal version is A343655.
The version for subsets of {1..n} is A343659.
The case without 1's or singletons is A343660.
A018892 counts pairwise coprime unordered pairs of divisors.
A048691 counts pairwise coprime ordered pairs of divisors.
A048785 counts pairwise coprime ordered triples of divisors.
A084422, A187106, A276187, and A320426 count pairwise coprime sets.
A100565 counts pairwise coprime unordered triples of divisors.
A305713 counts pairwise coprime non-singleton strict partitions.
A324837 counts minimal subsets of {1...n} with least common multiple n.
A325683 counts maximal Golomb rulers.
A326077 counts maximal pairwise indivisible sets.

Programs

  • Mathematica
    fasmax[y_]:=Complement[y,Union@@Most@*Subsets/@y];
    Table[Length[fasmax[Select[Subsets[Divisors[n]],CoprimeQ@@#&]]],{n,100}]

Formula

a(n) = A343660(n) + A005361(n).

A063428 a(n) is the smallest positive integer of the form n*k/(n+k).

Original entry on oeis.org

1, 2, 2, 4, 2, 6, 4, 6, 5, 10, 3, 12, 7, 6, 8, 16, 6, 18, 4, 12, 11, 22, 6, 20, 13, 18, 12, 28, 5, 30, 16, 22, 17, 10, 9, 36, 19, 26, 8, 40, 6, 42, 22, 18, 23, 46, 12, 42, 25, 34, 26, 52, 18, 30, 7, 38, 29, 58, 10, 60, 31, 14, 32, 40, 22, 66, 34, 46, 20, 70, 8, 72, 37, 30, 38, 28, 26
Offset: 2

Views

Author

Henry Bottomley, Jul 19 2001

Keywords

Comments

Or, smallest b such that 1/n + 1/c = 1/b has integer solutions.
Largest b is (n-1) since 1/n + 1/(n(n-1)) = 1/(n-1).
a(n) = smallest k such that k*n/(k-n) is an integer. - Derek Orr, May 29 2014

Examples

			a(6) = 2 because 6*3/(6+3) = 2 is the smallest integer of the form 6*k/(6+k).
a(10) = 5 since 1/10 + 1/10 = 1/5, 1/10 + 1/15 = 1/6, 1/10 + 1/40 = 1/8, 1/10 + 1/90 = 1/9 and so the first sum provides the value.
		

Crossrefs

Programs

  • Mathematica
    spi[n_]:=Module[{k=1},While[!IntegerQ[(n*k)/(n+k)],k++];(n*k)/(n+k)]; Array[ spi,80,2] (* Harvey P. Dale, May 05 2022 *)
  • PARI
    a(n)={my(k=1); if(n>1, while (n*k%(n + k), k++); n*k/(n + k))} \\ Harry J. Smith, Aug 20 2009

Formula

a(n) = n*A063427(n)/(n + A063427(n)) = 2n - A063649(n).
If n is prime a(n) = n - 1. - Benoit Cloitre, Dec 31 2001

Extensions

New description from Benoit Cloitre, Dec 31 2001
Entry revised by N. J. A. Sloane, Feb 13 2007
Definition revised by Franklin T. Adams-Watters, Aug 07 2009
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