Search:a228036 -id:a228036
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A020487
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Antiharmonic numbers: numbers k such that sigma_1(k) divides sigma_2(k).
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+10
30
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1, 4, 9, 16, 20, 25, 36, 49, 50, 64, 81, 100, 117, 121, 144, 169, 180, 196, 200, 225, 242, 256, 289, 324, 325, 361, 400, 441, 450, 468, 484, 500, 529, 576, 578, 605, 625, 650, 676, 729, 784, 800, 841, 900, 961, 968, 980, 1024, 1025, 1058, 1089, 1156, 1225, 1280, 1296
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OFFSET
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1,2
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COMMENTS
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Numbers k such that antiharmonic mean of divisors of k is an integer. Antiharmonic mean of divisors of number m = Product (p_i^e_i) isA001157(m)/A000203(m) = Product ((p_i^(e_i+1)+1)/(p_i+1)). So a(n) = k, for some n, ifA001157(k)/A000203(k) is an integer. -Jaroslav Krizek,Mar 09 2009
Squares are antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) -... + 1 is an integer. The nonsquare antiharmonic numbers areA227771.They include the primitive antiharmonic numbersA228023,except for its first term. -Jonathan Sondow,Aug 02 2013
The term "antiharmonic" is also known as "contraharmonic". -Pahikkala Jussi,Dec 11 2013
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LINKS
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Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru,The Biharmonic mean,Mathematical Reports, Vol. 18(68), No. 4 (2016), pp. 483-495,preprint,arXiv:1601.03081 [math.NT], 2016.
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EXAMPLE
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a(3) = 9 = 3^2; antiharmonic mean of divisors of 9 is (3^(2+1) + 1)/(3 + 1) = 7; 7 is integer. -Jaroslav Krizek,Mar 09 2009
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MATHEMATICA
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Select[Range[2000], Divisible[DivisorSigma[2, #], DivisorSigma[1, #]]&] (*Jean-François Alcover,Nov 14 2017 *)
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PROG
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(Magma) [n: n in [1..1300] | IsZero(DivisorSigma(2, n) mod DivisorSigma(1, n))]; //Bruno Berselli,Apr 10 2013
(Haskell)
a020487 n = a020487_list!! (n-1)
a020487_list = filter (\x -> a001157 x `mod` a000203 x == 0) [1..]
(Python)
from sympy import divisor_sigma
def ok(n): return divisor_sigma(n, 2)%divisor_sigma(n, 1) == 0
(Python) # faster for producing initial segment of sequence
from math import prod
from sympy import factorint
def ok(n):
f = factorint(n)
sigma1 = prod((p**( e+1)-1)//(p-1) for p, e in f.items())
sigma2 = prod((p**(2*e+2)-1)//(p**2-1) for p, e in f.items())
return sigma2%sigma1 == 0
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A227771
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Antiharmonic numbers that are not squares.
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+10
7
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20, 50, 117, 180, 200, 242, 325, 450, 468, 500, 578, 605, 650, 800, 968, 980, 1025, 1058, 1280, 1300, 1445, 1476, 1620, 1682, 1700, 1800, 1872, 2178, 2312, 2340, 2420, 2450, 2600, 2645, 2925, 3200, 3362, 3380, 3757, 3872, 4050, 4100, 4205, 4232, 4352, 4418, 4500, 4693, 5200
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OFFSET
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1,1
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COMMENTS
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Given prime factorization m = product (p_i^e_i), the antiharmonic (or contraharmonic) mean of the divisors of m is sigma_2(m)/sigma_1(m) = product (p_i^(e_i+1)+1)/(p_i+1). If this is an integer, then m is called antiharmonic.
All squares are trivially antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) -... + 1 is an integer. Sequence gives the nontrivial antiharmonic numbers.
The antiharmonic means of their divisors areA227986.
Sequence is infinite, since if n is in the sequence and gcd(n, k) = 1 then nk^2 is also in the sequence. -Charles R Greathouse IV,Aug 02 2013
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REFERENCES
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R. Guy, Unsolved Problems in Number Theory, B2 (see harmonic number).
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LINKS
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FORMULA
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EXAMPLE
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sigma_2(20)/sigma_1(20) = (1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2)/(1 + 2 + 4 + 5 + 10 + 20) = 546/42 = 13 is an integer, 20 is not a square, and no smaller number has these properties, so a(1) = 20.
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PROG
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(PARI) is(n)=if(issquare(n), return(0)); my(f=factor(n)); denominator(prod(i=1, #f~, (f[i, 1]^(f[i, 2]+1)+1)/(f[i, 1]+1)))==1 \\Charles R Greathouse IV,Aug 02 2013
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A228023
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Primitive antiharmonic numbers.
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+10
6
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1, 20, 50, 117, 200, 242, 325, 500, 578, 605, 650, 800, 968, 1025, 1058, 1280, 1445, 1476, 1682, 1700, 2312, 2340, 2600, 2645, 3200, 3362, 3757, 3872, 4205, 4232, 4352, 4418, 4693, 5618, 6728, 6962, 7514, 8228, 8405, 8833, 9248, 9425, 9472, 10082, 10400, 11045, 11849, 12493
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OFFSET
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1,2
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COMMENTS
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Antiharmonic numbers (A020487) which are not the product of an antiharmonic number and a relatively prime square > 1. Apart from the first term, a subsequence ofA227771(antiharmonic numbers that are not squares).
Is this sequence infinite? It seems that 4n^2 <= a(n) <= 8n^2 for n > 1, and that a(n) ~ 6n^2 as n -> infinity--seeA228036for motivation.
The antiharmonic mean of the divisors of a(n) isA228024(n).
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LINKS
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EXAMPLE
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200 = 2^3 * 5^2 is antiharmonic (since sigma_2(200)/sigma(200) = 119 is an integer) but 2^3 is not antiharmonic, so 200 is in this sequence.
180 = 2^2 * 3^2 * 5 is antiharmonic but 180/3^2 = 20 is also antiharmonic, so 180 is not in the sequence.
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PROG
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(PARI) isf(f)=denominator(prod(i=1, #f~, (f[i, 1]^(f[i, 2]+1)+1)/(f[i, 1]+1)))==1
nosmaller(f, startAt)=for(i=startAt, #f~, if(f[i, 2]%2==0&&f[i, 2], return(nosmaller(f, i+1)&&!(f[i, 2]=0)&&!isf(f)&&nosmaller(f, i+1)))); 1
is(n)=my(f); isf(f=factor(n))&&nosmaller(f, 1)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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