Aliquot cycles and generalizations
Some records of Amicable Number
1. Amicable Number coprime to 30
I found the first example of this type of Amicable Number :
21 Kohmoto 1997 193D
7^4*11*13^2*17*19*23^2*29*31*37*53*59^2*61*71*79*97*139*193*211*283*439*467*643*659*877*1259*1459^2*1667*1699*2801*2917*3541*74587*366983*2201897*19126727*714991703*22108315965121*8689461455946208667857*89*100329964009286143948575662850542265921787709
7^4*11*13^2*17*19*23^2*29*31*37*53*59^2*61*71*79*97*139*193*211*283*439*467*643*659*877*1259*1459^2*1667*1699*2801*2917*3541*74587*366983*2201897*19126727*714991703*22108315965121*8689461455946208667857*9029696760835752955371809656548803932960893899
Eric Weisstein's MathWorld
Jan Pedersen's various lists and statistics for amicable pairs.
2. Odd type(2,1) Amicable Number
Record :
21 Kohmoto 1999 224D
5*7^2*11^4*13^3*17*19^4*43*79*151*167*179*191*239*773^2*911*971*1609*2053*2339*2671*3221*4201*4637*21031*100823*598303*1008229*68559571*3950417818889*1426954029512242718954035759901*1153609640874638003808774525507672237049*131*28420327112587581861832969210407013231939163
5*7^2*11^4*13^3*17*19^4*43*79*151*167*179*191*239*773^2*911*971*1609*2053*2339*2671*3221*4201*4637*21031*100823*598303*1008229*68559571*3950417818889*1426954029512242718954035759901*1153609640874638003808774525507672237049*3751483178861560805761951935773725746615969647
3. Exponential Amicable Number
Record :
E2 Kohmoto 1999 Pedersen 1999
3^2*5^2*7^2*11^2*223^2*1709^2*6709^2*17880047^2*45716542082243281^2*2898754812701092802998976497201^2*6090040568037149797177908276107366402366002553215630764396750799*334490068174622922740599249948504186249982867689670475718202956403594633020341444340722916757717494303515174781666762751*334490068174622922740599249948504186249982867689670475730351400965289518849517115914879948957761883812267394072503101391^2
3^2*5^2*7^2*11^2*223*1709*6709*17880047*45716542082243281*2898754812701092802998976497201*6090040568037149797177908276107366402366002553215630764396750799^2*334490068174622922740599249948504186249982867689670475718202956403594633020341444340722916757717494303515174781666762751^2*334490068174622922740599249948504186249982867689670475730351400965289518849517115914879948957761883812267394072503101391
4.Amicable Quadruple
Record :
2 ^ (n-1) * M(n) * 5 ^ 9 * 7 ^ 2 * 11 ^ 4 * 17 ^ 2 * 19 * 29 ^ 2 * 67
* 71 ^ 2 * 109 * 131 * 139 * 179 * 307 * 431 * 521 * 653 *
1019 * 1279 * 2557 * 3221 * 5113 * 5171 * 6949 *
[ 173 * 1933058921 * 149 * 103540742849 ]
[ 173 * 1933058921 * 15531111427499 ]
[ 336352252427 * 149 * 103540742849 ]
[ 336352252427 * 15531111427499 ]
502773191775265222765583948694790874226250760534483493351148819807844330258789326474132715563910971453566180338254762609427137706831601465327827988327624239612616422975162247101757678239790219281860603094956768570673376648722519323886069644008339910500678304687500=2^2*3^2*5^9*7^4*11*13*17^2*19*29*41*43*47*79*157*163*223*433*601*1201*1303*1399*2053*2719*5167*13187*16787*52747*98543*2848337*500739672615943*7010355416623201*16506961423173486727453*10109028245165675006759491729*53*9163813886186194062277465733355041*228493422236252646229906698810905701952497928588074391826432954923
512259478412534377912104400556956794642748699075565891620859485411803428388235904156327495106845080879150954656413067981695149046958062327829974666090661755061950759427059051380484075775357409938284527759528193106202843727966022755931701332051097954213561695312500=2^2*3^2*5^9*7^4*11*13*17^2*19*29*41*43*47*79*157*163*223*433*601*1201*1303*1399*2053*2719*5167*13187*16787*52747*98543*2848337*500739672615943*7010355416623201*16506961423173486727453*10109028245165675006759491729*494845949854054479362983149603295787*228493422236252646229906698809925173866676005823410702992961841963
Generalized Aliquot Cycles
1.[ Rational Amicable Pair ]
Two integers which satisfy the following equations :
sigma(a)=sigma(b)= polynomial(a,b)/polynomial(a,b)
and the following conditions are needed.
c1 all the degrees of terms of numerator of right fraction are same.
c2 all the degrees of terms of denominator of right fraction are same.
c3 (degree of numerator) - (degree of denominator) = 1
The reason of conditions.
If a=b and c1,c2,c3 are satisfied then the equation becomes :
sigma(a)=(m*a^r)/(n*a^s) , r-s=1
so,
sigma(a)=m/n*a
If m/n is an integer then it is the definition of multiple perfect number.
It means that equation of RA has solutions at least for the case such that a=b and m/n is an integer.
1. Rational AP No.1
sigma(a)=sigma(b)=(a+b)^n/(a^(n-1)+b^(n-1))
n=1
sigma(a)=sigma(b)=(a+b)/2
there is no example for not a=b.
because, if a<b then sigma(b)<(b+b)/2=b , but for all n, 1<n -> n<sigma(n)
so, obvious one : a=b=1 exists.
n=2
sigma(a)=sigma(b)=(a+b)^2/(a+b)=a+b
it is AP
n=3
sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2)
ex.1 Sloane A038362, A038363
2^(m-1)*M(m)*3*5^2*13*31*139*277*3877*[11*19, 239 ]
where M(m) means Mersenne Prime. m=3 or 7<=m, a*[b, c] means a*b,a*c
17 digits to 4197936 digits
ex.2-1
2^(m-1)*M(m)*3*7*11^2*17^2*19^2*23*127*307*359*3739*22433*68209*[83*1931, 162287 ]
m=5 or 7<m
38 digits to 4197954 digits
ex.2-2 transformed from ex.2-1 : 19^2*127 -> 19^4*151*911
2^(m-1)*M(m)*3*7*11^2*17^2*19^4*23*151*307*359*911*3739*22433*68209*[83*1931, 162287 ]
5<=m
44 digits to 4197960 digits
ex.3-1
2^11*3^7*13*17*19^2*23*41*127*227*271*541*2269*124429*[29*569, 17099 ]
37 digits
ex.3-2 transformed from ex.3-1
2^11*3^7*13*17*19^4*23*41*151*227*271*541*911*2269*124429*[29*569, 17099 ]
43 digits
2.Rational AP No.2
sigma(a)=sigma(b)=1/k*(a+b)^n/(a^(n-1)+b^(n-1))
ex.1
k=2, n=4
2^(m-1)*M(m)*3*5*7*23^2*59*79*137*547*2477*158527*173428537*8671426849
*[83*1931, 162287 ]
m=5 or 7<m
48 digits to 4197965 digits
ex.2
k=6, n=6
2^(m-1)*M(m)*3^10*11*13*17*23^3*43*53*59*79*89*103*107*229*823*1031*
801*1831*3851*4271*19751*9322471*[83*1931, 162287]
2<m
71 digits to 4197988 digits
3.Ratoinal AP No.3
sigma(a)=sigma(b)=3/2*(a+b)^3/(a^2+a*b+b^2)
ex.
2^8*3^2*13*17*41*53*73^2*1801*11971*[5*11, 71]
4.Rational AP No.4
sigma(a)=sigma(b)=k*a*b/(a+b)
equivalently,
1/sigma(a)=1/sigma(b)=1/(k*a)+1/(k*b)
So, it is also called a k-multiple reciprocal AP.
ex.
k=6
2^9 *11*13*[3*7, 31 ]
k=9
2^19*5^3*7^2*13*31*41*367*[19*23, 3^3*11 ]
k=8
2^9*7^2*31*97*[3^3*11, 3*5*19]
2^22*3^6*19*31*61*137*151*197*547*1093*178481*[5*47, 11*23]
the name is analogy of k-multiple AP :
sigma(a)=sigma(b)=k*a+k*b
2. [Irrational Amicacble Number]
Two numbers a,b which satisfy the following equation :
sigma(a)=sigma(b)=m*(a^2+b^2-n^2*GCD(a,b)^2)^(1/2)
m=2,n=7
2^8*7*37*73*[ 3*11, 47 ]
m=3,n=29
2^7*3^2*7*13*[5*11, 71 ]
m=3,n=43
2^8*3^2*5*7*73*[17*19, 359 ]
where k*[a, b] means k*a, k*b
[Unitary phi Amicable Number]
Unitary phi or Unitary totient function is defined as follows :
If n=Product p(i)^r(i) then
uphi(n)=Product (p(i)^r(i)-1)
ex. uphi(12)=(4-1)*(3-1)=6
All terms of uphi(n) are unitary divisors of n, it is not the sum of
divisors but a difference of divisors.
And if we replace sigma in the equation of amicable number with uphi,
then we obtain another equation which defines unitary phi Amicable Number.
uphi(a)=uphi(b)=a-b
the signature of b must be negative, if not, no example exists.
ex.
(2,1), (60,36), (195,147), ...
A067741 on Sloane's table.
[Irrational Unitary phi Amimcable Number]
Two numbers a,b such that :
uphi(a)=uphi(b)=m*(a^2-b^2)^(1/2)
ex.
m= 1 2^2*[3^2*5, 3*17 ] , 2^3*7*[3^2*5, 3*17 ] ,
2^3*13*[3^3*5, 3*53 ]
I think only finite examples exist. These three are all.
m= 2 A046709, A046710 on Sloane's table
2^4*3^2*[5^2*11^2, 5*11*73 ]
2^4*3^2*[5^2*7, 5*37 ]
2^4*3^3*13*[5^2*7, 5*37 ]
2^5*3^2*31*[5^2*7, 5*37 ],
2^5*3^3*13*31*[5^2*7, 5*37 ],
2^8*3^2*17*[5^2*7, 5*37 ],
2^8*3^3*13*17*[5^2*7, 5*37 ],
2^11*3^2*11^2*23*89*[5^2*7, 5*37 ],
2^11*3^3*11^2*13*23*89*[5^2*7, 5*37 ],
2^16*3^2*17*257*[5^2*7, 5*37 ],
2^16*3^3*13*17*257*[5^2*7, 5*37 ],
2^17*3^2*17*257*131071*[5^2*7, 5*37 ],
2^17*3^3*13*17*257*13071*[5^2*7, 5*37 ],
2^32*3^2*17*257*65537*[5^2*7, 5*37 ],
2^32*3^3*13*17*257*65537*[5^2*7, 5*37 ]
[Theorem]
The following formulas give Irrational Unitary phi Amicable Number
for the case m=2 :
2^(2^(n+1))*3^2*(Product F(i), i=2 to n)*[5^2*7, 5*37 ] , n=1,2,3,4
2^(2^(n+1))*3^3*13*(Product F(i), i=2 to n)*[5^2*7, 5*37 ] ,
n=1,2,3,4
where, if n=1 then the (Product F(i), i=2 to n)=1
and, F(i) means Fermat Prime
m=3 (3^2*7, 5*13)
m=4 2^4*5^2*[3^4*11, 3^2*101]
[ Another type of I.U.P.A.N. ]
two numbers a,b such that :
uphi(a)=uphi(b)=m*((a^2-b^2)^(1/2)+a-b)
m=1 2^4*[3^2*5, 3*17 ] 2^5*31*[3^2*5, 3*17 ]
m=2 2^6*3^4*[5^2*7, 5*37 ], 2^9*3^4*73*[5^2*7, 5*37 ]
3.[ op-perfect number ]
opsigma function is defined as follows.
If x=p1^r1*p2^r2*p3^r3* ---
then opsigma(x)=(1+p1+p1^3+p1^5+ --- +p1^r1)*
(1+p2+p2^3+p2^5+ --- +p2^r2)*
(1+p3+p3^3+p3^5+ --- +p3^r3)*
---
If ri is even it must be
(1+pi+pi^3+pi^5+ --- +pi^(ri-1))
It is sum of op-divisors of x. If all powers of each prime factors
of a divisor of x are odd, then I call it op-divisor.op is for odd power. ex. opsigma(2^5*3^3)=(1+2+2^3+2^5)*(1+3+3^3)
The following sequence is solutions of the equation that
opsigma(x)=2*x
2*3, 2^3*3*11, 2^5*3*11*43, 2^7*3^3*5*19*31, 2^7*3^5*5*19*23*137,
2^9*3^3*5*19*31*683, 2^9*3^5*5*19*23*137*683,
2^11*3^3*5*19*31*683*2731, 2^11*3^5*5*19*23*137*683*2731,
2^13*3^3*7*11*31*83*331, 2^13*3^5*7*11*23*83*137*331,
2^15*3^3*7*11*31*83*331*43691, 2^15*3^5*7*!1*23*83*137*331*43691,
2^17*3^3*7*11*31*83*331*43691*174763,
2^17*3^5*7*11*23*83*137*331*43691*174763
6, 264, 45408, 10177920, 9310826880, 27806077440, 25437179036160, 303753589954560, 277875743791011840, 14504815632384,
13269098919960576, 2534919599177957376, 2318960803647990104064, 1772040615644549459607552, 1621074187711734778226147328
op-perfect numbers have an interesting character.
If x is opp and x=2^k*y, gcd(2^k,y)=1, (2^(k+4)+1)/3 is prime,
then 4*x*(2^(k+4)+1)/3 is also opp.
If Bateman and Shefridge and Wagstaff's conjecture is correct,then
it is written as follows.
If x is opp and x=2^k*y, gcd(2^k,y)=1, 2^(k+4)-1 is prime,
then 4*x*(2^(k+4)+1)/3 is also opp.
But it is not necessary, because the case of k=7, 2^11-1 is not
prime but (2^11+1)/3=683 is prime.
4.[ (-1)sigma sequence ]
I have studied Polynomial divisor function which is defined as follows :
If n=Product p(i)^r(i)
then p-sigma(n)=Product
(k(0)+k(1)*p(i)+k(2)*p(i)^2+...+k(r(i))*p(i)^r(i))
where coefficients k(j) are integer.
Ex.
ordinary sigma ... all k(j)=1
(-1)sigma ... k(0)=-1 and other k(j)=1
If we iterate (-1)sigma, another aliquot sequence is obtained :
a(1)=(-1)sigma(a(0)), a(2)=(-1)sigma(a(1)), ...
Three cases are observed :
1. it goes to infinite.
2. it becomes cyclic.
3. 1, 1, 1, ...
[Further generalized Perfect Number and Amicable Number and Sociable
Number ]
1.Definition.
If x=Product p(i)^r(i),
then (-1)sigma(x)=Product (-1+Sum p(i)^s(i), s(i)=1 to r(i))
Ex.
(-1)sigma(24)=(-1+2+4+8)*(-1+3)=1-2-4-8-3+6+12+24
all terms are divisors of 24, but it is not the sum of divisors.
It is a difference of divisors.
2. (-1)sigma sociable number. A049057, A049058, A049059, A051152 on
Sloane's table
(-1)sigma sequence is defined as follows :
a(n)=(-1)sigma(a(n-1))
If a(k+1)=a(1) then these k-tuple numbers are called a (-1)sigma
sociable number of order k.
k=1
2^2*5, 2^3*3*13, 2^4*3*7*29, 2^5*3*5*61, 2^10*3*5*17*409,
2^9*3*5*17*1021, 2^7*3*5*11^2*13*23*131
k=2
2^2 - 5,
2^3*3^3 - 2*13*19,
2^5*3*5^2*7 - 2^2*3*29*61
k=3
2^2*13 - 2^2*3*5 - 2^3*5
2^2*3*13 - 2^3*3*5 - 2^3*13
2^2*3*5*13 - 2^5*3*5 - 2^3*61
2^5*3*13 - 2^3*3*61 - 2^3*3*5*13
2^2*3*5*13*17 - 2^9*3*5 - 2^3*1021
2^9*3*13 - 2^3*3*1021 - 2^3*3*5*13*17
2^3*3*5*17*61 - 2^9*3*5*13 - 2^5*3*1021
2^9*3*5*61 - 2^5*3*5*1021 - 2^5*3*5*17*61
k=4
2^2*3*5*29*61 - 2^7*3*7*5^2 - 2^2*3*11*23*29 - 2^5*5^2*7*11
k=8
2^3*5*7*29 - 2^5*3*7*13 - 2^4*3^2*61 - 2^2*3*5*11*29 - 2^6*5^2*7 -
2*3*5^3*29 - 2^4*7^2*11 - 2*5^2*11*29
k=8
2^6*3*5^2 - 2*5^3*29 - 2^3*7^2*11 - 2*5^2*11*13 - 2^3*3*5*29 -
2^5*7*13 - 2^3*3^2*61 - 2^2*3*5*11*13
3.Theorem
[ Theorem S3 ] .
If 2^(2^(n+1)+2)-3 is prime then
2^3*13*(Product F(i) i=0 to n) --- 2^3*3*(2^(2^(n+1)+2)-3) ---
2^(2^(n+1)+1)*3*13
are (-1)sigma sociable number of order 3. F(i) means Fermat Prime.
The theorem gives only 2 examples for n=1, 2
for F(1)
a,b,c= 2^3*13*3*5,
2^3*3*61,
2^5*3*13
for F(2)
a,b,c= 2^3*13*3*5*17
2^3*3*1021
2^9*3*13
for F(0), it becomes a=b=c
for F(3),F(4), 2^18-3 and 2^34-3 are not prime.
[ Proof ]
(-1)sigma(2^3*13*(Product F(i) i=0 to n)) = 13*2^2*3*2^(2^(n+1)-1)
= 2^(2^(n+1)+1)*3*13
(-1)sigma(2^(2^(n+1)+1)*3*13) = (2^(2^(n+1)+2)-3)*2*2^2*3 =
2^3*3*(2^(2^(n+1)+2)-3)
(-1)sigma(2^3*3*(2^(2^(n+1)+2)-3) = 13*2*2^2*(2^(2^(n+1))-1) =
2^3*13*(2^(2^n)-1)*(2^(2^n)+1) = 2^3*13*(Product F(i) i=0 to n))
4. Perfect Number A034094
(-1)sigma(x) = m*x , 2<=m
m=2
2^3*3^3*13*19, 2^4*3^3*7*19*29, 2^5*3^3*5*19*61,
2^5*3^2*5^2*7*11*29*61, 2^9*3^3*5*17*19*1021,
2^7*3^3*5*11^2*13*19*23*131, 2^10*3^2*5^2*7*11*17*29*409,
2^9*3^2*5^2*7*11*17*29*1021,
m=3
2^6*3^3*5^4*7*19^2*41*379,
2^24*3^3*5^5*7^2*11*17*19*29*61*233*239*467*479*70051
5.Super Perfect Number A051153
(-1)sigma((-1)sigma(a))=2*a
13*19, 2^3*3^2*5*29, 2^4*3^3*7*29,
2^5*3^3*5^2*7, 2^2*3^3*29*61, 2^5*3^3*5*61,
2^7*3*5*11*19*29*1021
6.Amicable Number A051594, A051595
(-1)sigma(a)=(-1)sigma(b)=a+b
2^5*3^3*7*19*61*[ 11*29, 281 ]
2^10*3^3*7*17*19*409*[ 11*29, 281 ]
2^9*3^3*7*17*19*1021*[ 11*29, 281 ]
2^8*3^4*7*17*127*509*[ 5*19, 73 ]
where a*[b, c] means a*b,a*c
5. [ 1/m-sociable number ]
The sequence which is defined as follows is called 1/m-sigma sequence :
a(n)=1/m*sigma(a(n-1))
Multiple Perfect Number is a fixed point of this mapping.
Because,
if a(n)=a(n-1) then
m*a(n)=sigma(a(n))
If the sequence becomes a cyclic sequence, then it is called a
1/m-sociable number of order k. k is number of the members.
[1/2-sociable number of order 2]
2^(m-1)*M(n), 2^(n-1)*M(m)
where M(n) and M(m) are Mersenne Prime, m and n are different.
Proof
1/2*sigma(2^(m-1)*M(n))
= 1/2*(2^m-1)*2^n
= 2^(n-1)* M(m)
1/2*sigma(2(n-1)*M(m))
= 2^(m-1)*M(n)
ex. 12,14 48,62
a sporadic solution
2^5*5, 3^3*7
[1/4-sigma sociable number of order k]
a(n)=1/4*sigma(a(n-1)), a(k+1)=a(1)
k=12
2 ^ 3 * 3 ^ 2 * 5 ^ 2* 7 * 13 * 19 * 31 * 37 ,
2 ^ 10 * 3 * 5 ^ 2* 7 * 13 * 19 * 31 ,
2 ^ 11 * 5 * 7 * 23 * 31 * 89 ,
2 ^ 11 * 3 ^ 6 * 5 ^ 2* 7 * 13 ,
2 ^ 2 * 3 ^ 2 * 5 * 7 ^ 2 * 13 * 31 * 1093 ,
2 ^ 6 * 3 ^ 2 * 7 ^2 * 13 * 19 * 547 ,
2 ^ 3 * 3 * 5 * 7 * 13 * 19 *127 * 137 ,
2 ^ 15 * 3 ^ 3* 5 ^ 2 * 7 * 23 ,
2 ^ 7* 3 ^ 2 * 5 ^ 2 * 17 * 31 * 257 ,
2 ^ 5* 3 ^ 4 * 5 * 13 * 17 * 31 * 43 ,
2 ^ 8* 3 ^ 5 * 7 ^2 * 11 ^ 3 ,
2 ^ 3 * 3 ^ 2 * 7 ^ 2* 13 * 19 * 61 * 73
k=2
2^7*3^3*7*13*19*73,
2^8*3*5^3*7*17*37
k=2
2^9*3^2*5*7*13,
2^3*3^2*7*11*13*31
k=25
2 ^ 15 * 5 * 7 * 13 * 31 * 83 * 181
2 ^ 11 * 3 ^ 3 * 5 * 7 ^ 3 * 13 * 17 * 257
2 ^ 9 * 3 ^ 6 * 5 ^ 4 * 7 ^ 2 * 13 * 43
2 * 3 ^ 2 * 7 * 11 ^ 3* 19 * 31 * 71 * 1093
2 ^ 15 * 3 ^ 4 * 5 * 13 * 61 * 547
2 ^ 3 * 3 ^ 2 * 5 * 7 * 11 ^ 2 * 17 * 31 * 137* 257
2 ^ 10 * 3 ^ 6 * 5 * 7 * 13 * 19 * 23 * 43
2 ^ 10 * 3 ^ 2 * 5 * 7 * 11 * 23 * 89 * 1093
2 ^ 9 *3 ^ 5 * 5 * 13 * 23 * 89 * 547
2 ^ 8 * 3 ^ 5 * 5 * 7 ^ 2 * 11 * 13 * 31 * 137
2 ^ 10 * 3 ^ 4 * 7 ^ 3 * 13 * 19 * 23 * 73
2 ^ 9 * 3 * 5 ^ 3 * 7 * 11 ^ 2 * 23 * 37* 89
2 ^ 10 * 3 ^ 5 * 5 * 7 * 11 * 13 * 19 ^ 2 * 31
2 ^ 12 * 3 ^ 3 * 7 ^ 2 * 13 * 23 * 89 * 127
2 ^ 13 * 3 ^ 4 * 5 ^ 2 * 7 * 19 * 8191
2 ^ 16 * 3 * 5 * 11 ^ 2 * 31 * 43 * 127
2 ^ 15 * 3 * 7 * 11 * 19 * 131071
2 ^ 24* 3 ^ 2 * 5 ^ 2 * 17 * 257
3 ^ 3 * 13 * 31 ^ 2 * 43 * 601* 1801
2 ^ 6 * 3 * 5 * 7 ^ 2 * 11 * 17 * 43 * 53 * 331
2 ^ 9 * 3 ^ 8 * 11 * 19 * 83 * 127
2 ^ 11 * 3 ^ 3 * 5 * 7 * 11 * 13 * 31 * 757
2 ^ 14 * 3 ^ 4 * 5 ^ 2 * 7 ^ 2 * 13 * 379
2 * 3 * 5 * 7 ^ 2 * 11 ^ 2 * 19 ^ 2 * 31 ^ 2 * 151
2 ^ 4 * 3 ^ 5 * 7 * 19 ^ 3 * 127 * 331
[1/3-sigma sociable number of order 2]
2^8*5*11*31*37,
2^9*3*7*19*73
6. [fixed points of a divisor function]
fixed points of a function such that :
It n=Product p(i)^r(i) then
f(n)=Product (-2+p(i)^(r(i)-1)+p(i)^r(i))
ex.f(200)=f(2^3*5^2)=(-2+2^2+2^3)*(-2+5+5^2)=280
numbers satisfy :
f(x)=x
2^2, 2^3*5, 2^4*5*11, 2^5*5*11*23, 2^6*5*11*23*47, 2^10*3^2*5*7*13*29*59
7. [(1+e)sigma perfect number ]
(1+e)sigma(x)=2*x
If n=Product p(i)^r(i) then (1+e)sigma(n)=Product (1+Sum p(i)^s(i), s(i) divides r(i))
2*3, 2^2*7, 2^3*3*11, 2^4*3*23, 2^5*3*5*7, 2^6*3*5*79, 2^8*3^2*7*13*31, 2^10*3*5*7*19*1063
8. [(+2)-sigma perfect number ]
(+2)sigma(n)=2*n
if n=Product p(i)^r(i) then
(+2)sigma(n)=Product (2+Sum p(i)^s(i), s(i)=1 to r(i))
ex. (+2)sigma(6)=(2+2)*(2+3)=20
even examples 2^n.
odd examlples 3^2*7, 3^4*7*61, 3^6*7*61*547
9. [g-reduced Amicable pair ]
Two numbers a,b such that :
sigma(a)=sigma(b)=a+b+GCD(a,b)
3*5*[2*19, 59], 3*7*[2^2*11, 83], 3^2*13*[2*5, 17], 3*7^2*[2^2*41, 293], 3^4*11*[2*17, 53]
a comment : No (even,even)pair or (odd,odd,)pair is known.
10. [ 3-infinitary Perfect Number ]
3-i-sigma(a)=2*a
3-i-sigma(a) means sum of 3-i-divisor of a.
If n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its trinary expansion everywhere that the corresponding r(i) has a digit b, then d is a 3-i-divisor of n.
ex. 2^3*3 is a 3-i-divisor of 2^5*3^2, because 2^3*3 = 2^10*3^1 and 2^5*3^2 = 2^12*3^2 in trinary expanded power. All corresponding digits satisfy the condition. 1<=1, 0<=2, 1<=2.
2*3, 2^2*7, 2^4*3^3*7, 2^3*3^2*7*13, 2^9*3^4*5*7*19, 2^6*3*5*19*37*73, 2^10*3^6*5*19^2*127*379*757
Theorems for generating Amicable Numbers.
1. a method to make a type 2,1 AP from a type 2,1 AP.
[Theorem 21]
If k*a*b, k*c is a type 2,1 AP.
and (a+1)*(2*b+1)-1, (a+1)^2*(2*b+1)-1, 2*b+1 are primes,
then k*(2*b+1)*a*((a+1)*(2*b+1)-1), k*(2*b+1)*((a+1)^2*(2*b+1)-1) is also a type 2,1 AP.
ex.
2^2*5*11, 2^2*71 is type 2,1
137, 827, 23 are primes,
so, 2^2*23*5*137, 2^2*23*827 is a type 2,1
comment :
I found the method when I was considering about an interesting pair of double APs which I named a "brothers". They are as follows :
double AP is two numbers such that,
sigma(a)=sigma(b)=2*a+2*b
ex.1
2^2*3^2*5*11*13*19*59*71*227*283*383*907*1531*12979*119723*493201*97633865794514039*1757409584301252701*23872651793148216690383*1145887286071114401138383* [660031076776961895055708607]
[23*27501294865706745627321191]
146 digits
ex.2
2^2*3^2*5*11*13*19*59*71*227*283*383*907*1531*12979*119723*493201*97633865794514039*1757409584301252701*23872651793148216690383* [13750647432853372813660607]
[23*572943643035557200569191]
121 digits
They were strange, because almost all of the factors of the two examples are same.
I studied about the structure of them.
It was difficult to find the formula, but at last I succeeded.
It is possible to apply this theorem for the following cases.
1.You should calculate both cases of k*a*b, k*c and k*b*a, k*c for each type 2,1 APs.
I mean for example, 2^2*5*11, 2^2*71 is also written as 2^2*11*5, 2^2*71, so,
(11+1)*(2*5+1)-1, (11+1)^2*(2*5+1)-1 should be tested with my method.
2.You should calculate the reverse of my method which is as follows.
[Theorem 21R]
If k*a*b, k*c is a type 2,1 AP, and (b+1)/(a+1), (b-a)/(2*a+2), a*b0+a+b0 are primes, and gcd(k,k0)=k0,
where k0=(b+1)/(a+1), b0=(b-a)/(2*a+2)
then k/k0*a*b0, k/k0*( a*b0+a+b0) is a type 2,1 AP.
ex.
2^2*23*5*137, 2^2*23*827 is a type 2,1 and
(137+1)/(5+1), (137-5)/(2*5+2), 5*11+5+11 are primes
so, 2^2*23/23*5*11, 2^2*23/23*71 is a type 2,1.
ex. of no good
my 160 digits' type 2,1 satisfies regrettably only the first and fourth conditions.
3*5^4*17*19^3*43*71*103*113*181*607*773*1213*1747*2239*13913*27647*206209*4839559850381*267598622366967013*330751897245571228067*93599618317321044970596118139*[89*8423965648558894047353650632509]
[758156908370300464261828556925899]
(8423965648558894047353650632509+1)/(89+1)
=93599618317321044970596118139 is prime.
2. For the generation of type (3,2) numbers use the following theorem:
Let (au,ap) be an amicable pair with gcd(a,p)=1 and p prime.
Put C := (p+u)(p+1). If C = D1.D2 and if r, s, and q given as
r := p +D1
s := p + D2
q := u + r + s
are all prime, then (auq,ars) is an amicable pair.
If (au,ap) has type (n,1) then (auq,ars) has type (n+1,2).
comment :
This is a known theorem which Jan Pedersen told me.
3. Theorems which generate e-Amicable Number form ordinal AN or Unitary AN.
[Theorem 1.]
If k*a,k*b is an Amicable Pair and k=gcd(k*a,k*b) and the members are square free, then
k^2*a^2*b,k^2*b^2*a is an e-AP.
where e-AP is defined as follows,
esigma(a)=esigma(b)=a+b
if x=Product p(i)^r(i) then
esigma(x)=Product Sum p(i)^s(i), s(i) divides r(i)
[Proof]
gcd(k,a)=1, gcd(k,b)=1, gcd(a,b)=1 , so
esigma(k^2*a^2*b)=esigma(k^2)*esigma(a^2)*esigma(b)
=(Product u(i)`2+u(i))*(Product v(i)^2+v(i))*b
=k*(Product u(i)+1)*a*(Product v(i)+1)*b
=k*sigma(k)*a*sigma(a)*b
=k*a*b*sigma(k*a)
=k*a*b*(k*a+k*b)
=k^2*a^2*b+k^2*b^2*a
equally,
esigma(k^2*b^2*a)
=k^2*a^2*b+k^2*b^2*a
where k=Product u(i), a=Product v(i)
ex. From 187 digits UAP on the table, we get the following e-aAP.
3^2*5^2*7^2*11^2*223^2*1709^2*6709^2*17880047^2*45716542082243281^2*2898754812701092802998976497201^2*334490068174622922740599249948504186249982867689670475730351400965289518849517115914879948957761883812267394072503101391^2*6090040568037149797177908276107366402366002553215630764396750799*334490068174622922740599249948504186249982867689670475718202956403594633020341444340722916757717494303515174781666762751
3^2*5^2*7^2*11^2*6090040568037149797177908276107366402366002553215630764396750799^2*334490068174622922740599249948504186249982867689670475718202956403594633020341444340722916757717494303515174781666762751^2*223*1709*6709*17880047*45716542082243281*2898754812701092802998976497201*334490068174622922740599249948504186249982867689670475730351400965289518849517115914879948957761883812267394072503101391
557 digits
[Theorem 2.]
If k*a,k*b is an Unitary Amicable Pair and gcd(k,a)=1,gcd(k,b)=1,
gcd(a,b)=1, and r(i)<=2 for all i, and a,b are square free,
then k*(Product p(i))*a^2*b, k*(Product p(i))*b^2*a is an e-AP.
where k=Product p(i)^r(i)
[Proof]
esigma(k*(Product p(i))*a^2*b)
=esigma((Product p(i)^(r(i)+1))*(Product u(i)^2)*b)
r(i)+1 is 2 or 3, and b is square free so,
=(Product p(i)^(r(i)+1)+p(i))*(Product u(i)^2+u(i))*b
=usigma(k)*(Product p(i))*usigma(a)*a*b
=usigma(k*a)*(Product p(i))*a*b
=(k*a+k*b)*(Product p(i))*a*b
=k*(Product p(i))*a^2*b+ k*(Product p(i))*b^2*a
equally,
esigma(k*(Product p(i))*a^2*b)
=k*(Product p(i))*a^2*b+ k*(Product p(i))*b^2*a
where a=Product u(i)
[Theorem 3.]
If k*a,k*b is an Unitary Amicable Pair and gcd(k,a)=1,gcd(k,b)=1,
gcd(a,b)=1, and r(i)+1 are primes for all i, and a,b are square free,
then k*(Product p(i))*a^2*b, k*(Product p(i))*b^2*a is an e-AP.
where k=Product p(i)^r(i)
[Proof]
esigma(k*(Product p(i))*a^2*b)
=esigma((Product p(i)^(r(i)+1))*(Product u(i)^2)*b)
r(i)+1 are primes , and b is square free so,
=(Product p(i)^(r(i)+1)+p(i))*(Product u(i)^2+u(i))*b
=usigma(k)*(Product p(i))*usigma(a)*a*b
=usigma(k*a)*(Product p(i))*a*b
=(k*a+k*b)*(Product p(i))*a*b
=k*(Product p(i))*a^2*b+ k*(Product p(i))*b^2*a
equally,
esigma(k*(Product p(i))*a^2*b)
=k*(Product p(i))*a^2*b+ k*(Product p(i))*b^2*a
where a=Product u(i)
How to calculate Rational Amicable Number.
The explanation is a little part of algorithm.
A RAN is defined as follows,
sigma(a)=sigma(b)=(a+b)^3/(a^2+b^2) ...(1)
[ a method to solve this equation. ]
If two numbers are k*x and k*y, and gcd(k,x)=1, gcd(k,y)=1,
then,
sigma(k*x)=sigma(k*y)
sigma(x)=sigma(y)
so, at first, we find two numbers x,y which satisfy this equation.
and,
(k*x+k*y)^3/((k*x)^2+(k*y)^2)
=k*(x+y)^3/(x^2+y^2) ...(2)
so, secondly, we calculate (x+y)^3/(x^2+y^2)
and we denote ((x+y)^3/(x^2+y^2))/sigma(x) s.
the equation (1) becomes
sigma(k)=k*(x+y)^3/(x^2+y^2)/sigma(x)
=s*k ...(3)
if s is an integer, k is multiple perfect number, but
generally s is a rational number, so (3) defines
Rational Perfect Number.
it is solvable when all of the prime factors of numerator of s
are "good" which is defined as follows,
if u is a factor of (p^r-1)/(p-1), p is prime,
and both p,r are small,
then u is "good".
it takes many pages to explain the concept of "good"
and thirdly, we use "sprout" algorithm to solve (3).
I am sorry, the paper about it has 50 pages. I will write it some day.
For example,
53*1999 and 10799 are satisfy the condition, but the numerator
of s is 2*23*4651, and 4651 is not good.
Because the smallest power of p(i) which makes p(i)^r-1 to have 4651 as a factor are,
2^1550, 3^4650, 5^2325, 7^2325, 11^4651, 13^15, 17^1550,
19^2325, 23^2325, 29^930, 31^1550, ... etc.
In the case of this RAN, "good" means smaller than
2^32, 3^16, 5^8, 7^4, 11^2, 13^1, ...
So, it is absoutely not good.
it mean , there is no "sprout" for the "seed" (53*1999,107999).
under construction