[math-fun] A sequence concerning the abundancy index (long)
I apologize for the blurriness of the following prose. There are no difficult concepts here, but I haven't been able to express it in English very well. Read carefully; I assure you there is a reward here. This came from my amateur fossickings in the world of multiply-perfect numbers. For n a positive integer, let S(n) be the sum of the divisors of n; I'm using S because I can't easily type a sigma. The abundancy index of n, T(n), is the rational number S(n)/n. Classical perfect numbers have T(n) = 2; the more general multiply-perfect numbers have integer T(n). All of this so far is standard. Suppose T(n) = U(n)/V(n), with the fraction expressed in lowest terms. Factor U(n) and V(n) into prime powers. Of these prime powers, one of them is the largest. Sometimes it is from the numerator, perhaps from the denominator. (The sequence of n where the largest prime power in T(n) is found the denominator, in this way, is 5, 8, 11, 13, 14, 17, 19 ..., which is not in OEIS, but that's not why I called you all here today.) Suppose the largest prime power of T(n) is p^e. Let W(n) be defined as np if p^e was found in the numerator of T(n), and n/p if it was found in the denominator. An example: Let n = 168 = 2^3.3.7; then S(n) = 480 = 2^5.3.5; and T(n) = 480/168 = 12/7 = 2^2.3.7^(-1). Of the three component prime powers, 3, 4, and 7, 7 is the largest and it is found in the denominator. So n' = n/7 = 24. The sequence I want to consider is the trajectory of 2 under iterated applications of W. It starts 2, 6, 12, 84, 168, 24 ... and of course is not in OEIS. Each term is obtained from the previous term by multiplying or dividing by a prime; to reiterate, the prime is the base of the largest prime power in T(n), and we multiply or divide according to which side of the fraction line we find it on. So let A(1) = 2, A(i) = W(A(i-1)). The sequence flops around randomly, growing and shrinking but mostly growing, and becoming more and more exuberantly composite. A(75) picks up the prime factor 2147483647 = 2^31 - 1. The sum of divisors of this is, of course, 2^31, and other multiply-even factors of A(75) drive the power of 2 up to 2^47. At this point the trajectory enters a long episode dominated by this large power of 2 in the numerator, doubling at each step. I was sure that before this reservoir of 2's was exhausted, we would pick up another multiply-even factor and the sequence would climb forever. I was mistaken. The glut of 2's ends at A(93); the "bull market" gives out and there is a crash in which we lose, first that big factor of 2147483647, and then lots of 2's and other detritus that we've been accumulating along the way. I'm up to A(116), which, in case you care, is 2^39.3^5.5^2.7^2.11.13^2.19^2.31^2.61.83.127.331.9719 . The exponent of 2 peaked at 47 and has been declining fitfully ever since. If it ever gets up to 60 it will acquire another Mersenne factor, sparking another bubble, but the exponent may bop around between 30 and 60 forever (with the sequence eventually entering a cycle). I don't know if I have the patience to find out. But the long spells of growth and decay are certainly intriguing. I was led to this sequence because one way to find multiply perfect numbers is to start with a candidate n, and repeatedly adjust the exponents of its prime factors in a way suggested by the non-integralness of T(n), the goal being to whittle T(n) down to an integer. One naturally tries to automate the process of selecting which prime to adjust, which leads to sequences like A(i).
A(i) enters a cycle at A(146)=A(136). I'm obscurely disappointed. The exponent of 2 oscillates between 37 and 38 in the cycle. Dynamics under W is still interesting. The smallest cycle starts with 4: 2^2 -> 2^2.7 -> 2^3.7 -> 2^3 -> 2^2. On Sat, Jul 16, 2016 at 1:30 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I apologize for the blurriness of the following prose. There are no difficult concepts here, but I haven't been able to express it in English very well. Read carefully; I assure you there is a reward here. This came from my amateur fossickings in the world of multiply-perfect numbers.
For n a positive integer, let S(n) be the sum of the divisors of n; I'm using S because I can't easily type a sigma.
The abundancy index of n, T(n), is the rational number S(n)/n.
Classical perfect numbers have T(n) = 2; the more general multiply-perfect numbers have integer T(n). All of this so far is standard.
Suppose T(n) = U(n)/V(n), with the fraction expressed in lowest terms. Factor U(n) and V(n) into prime powers. Of these prime powers, one of them is the largest. Sometimes it is from the numerator, perhaps from the denominator. (The sequence of n where the largest prime power in T(n) is found the denominator, in this way, is 5, 8, 11, 13, 14, 17, 19 ..., which is not in OEIS, but that's not why I called you all here today.)
Suppose the largest prime power of T(n) is p^e. Let W(n) be defined as np if p^e was found in the numerator of T(n), and n/p if it was found in the denominator.
An example: Let n = 168 = 2^3.3.7; then S(n) = 480 = 2^5.3.5; and T(n) = 480/168 = 12/7 = 2^2.3.7^(-1). Of the three component prime powers, 3, 4, and 7, 7 is the largest and it is found in the denominator. So n' = n/7 = 24.
The sequence I want to consider is the trajectory of 2 under iterated applications of W. It starts 2, 6, 12, 84, 168, 24 ... and of course is not in OEIS. Each term is obtained from the previous term by multiplying or dividing by a prime; to reiterate, the prime is the base of the largest prime power in T(n), and we multiply or divide according to which side of the fraction line we find it on.
So let A(1) = 2, A(i) = W(A(i-1)). The sequence flops around randomly, growing and shrinking but mostly growing, and becoming more and more exuberantly composite. A(75) picks up the prime factor 2147483647 = 2^31 - 1. The sum of divisors of this is, of course, 2^31, and other multiply-even factors of A(75) drive the power of 2 up to 2^47. At this point the trajectory enters a long episode dominated by this large power of 2 in the numerator, doubling at each step. I was sure that before this reservoir of 2's was exhausted, we would pick up another multiply-even factor and the sequence would climb forever. I was mistaken.
The glut of 2's ends at A(93); the "bull market" gives out and there is a crash in which we lose, first that big factor of 2147483647, and then lots of 2's and other detritus that we've been accumulating along the way. I'm up to A(116), which, in case you care, is
2^39.3^5.5^2.7^2.11.13^2.19^2.31^2.61.83.127.331.9719
. The exponent of 2 peaked at 47 and has been declining fitfully ever since. If it ever gets up to 60 it will acquire another Mersenne factor, sparking another bubble, but the exponent may bop around between 30 and 60 forever (with the sequence eventually entering a cycle). I don't know if I have the patience to find out. But the long spells of growth and decay are certainly intriguing.
I was led to this sequence because one way to find multiply perfect numbers is to start with a candidate n, and repeatedly adjust the exponents of its prime factors in a way suggested by the non-integralness of T(n), the goal being to whittle T(n) down to an integer. One naturally tries to automate the process of selecting which prime to adjust, which leads to sequences like A(i).
Allan, I've taken the liberty of creating 3 new entries (A274939-A274941) for the very interesting sequences that you mentioned. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Sat, Jul 16, 2016 at 8:30 AM, Allan Wechsler <acwacw@gmail.com> wrote:
A(i) enters a cycle at A(146)=A(136). I'm obscurely disappointed. The exponent of 2 oscillates between 37 and 38 in the cycle.
Dynamics under W is still interesting. The smallest cycle starts with 4: 2^2 -> 2^2.7 -> 2^3.7 -> 2^3 -> 2^2.
On Sat, Jul 16, 2016 at 1:30 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I apologize for the blurriness of the following prose. There are no difficult concepts here, but I haven't been able to express it in English very well. Read carefully; I assure you there is a reward here. This came from my amateur fossickings in the world of multiply-perfect numbers.
For n a positive integer, let S(n) be the sum of the divisors of n; I'm using S because I can't easily type a sigma.
The abundancy index of n, T(n), is the rational number S(n)/n.
Classical perfect numbers have T(n) = 2; the more general multiply-perfect numbers have integer T(n). All of this so far is standard.
Suppose T(n) = U(n)/V(n), with the fraction expressed in lowest terms. Factor U(n) and V(n) into prime powers. Of these prime powers, one of them is the largest. Sometimes it is from the numerator, perhaps from the denominator. (The sequence of n where the largest prime power in T(n) is found the denominator, in this way, is 5, 8, 11, 13, 14, 17, 19 ..., which is not in OEIS, but that's not why I called you all here today.)
Suppose the largest prime power of T(n) is p^e. Let W(n) be defined as np if p^e was found in the numerator of T(n), and n/p if it was found in the denominator.
An example: Let n = 168 = 2^3.3.7; then S(n) = 480 = 2^5.3.5; and T(n) = 480/168 = 12/7 = 2^2.3.7^(-1). Of the three component prime powers, 3, 4, and 7, 7 is the largest and it is found in the denominator. So n' = n/7 = 24.
The sequence I want to consider is the trajectory of 2 under iterated applications of W. It starts 2, 6, 12, 84, 168, 24 ... and of course is not in OEIS. Each term is obtained from the previous term by multiplying or dividing by a prime; to reiterate, the prime is the base of the largest prime power in T(n), and we multiply or divide according to which side of the fraction line we find it on.
So let A(1) = 2, A(i) = W(A(i-1)). The sequence flops around randomly, growing and shrinking but mostly growing, and becoming more and more exuberantly composite. A(75) picks up the prime factor 2147483647 = 2^31 - 1. The sum of divisors of this is, of course, 2^31, and other multiply-even factors of A(75) drive the power of 2 up to 2^47. At this point the trajectory enters a long episode dominated by this large power of 2 in the numerator, doubling at each step. I was sure that before this reservoir of 2's was exhausted, we would pick up another multiply-even factor and the sequence would climb forever. I was mistaken.
The glut of 2's ends at A(93); the "bull market" gives out and there is a crash in which we lose, first that big factor of 2147483647, and then lots of 2's and other detritus that we've been accumulating along the way. I'm up to A(116), which, in case you care, is
2^39.3^5.5^2.7^2.11.13^2.19^2.31^2.61.83.127.331.9719
. The exponent of 2 peaked at 47 and has been declining fitfully ever since. If it ever gets up to 60 it will acquire another Mersenne factor, sparking another bubble, but the exponent may bop around between 30 and 60 forever (with the sequence eventually entering a cycle). I don't know if I have the patience to find out. But the long spells of growth and decay are certainly intriguing.
I was led to this sequence because one way to find multiply perfect numbers is to start with a candidate n, and repeatedly adjust the exponents of its prime factors in a way suggested by the non-integralness of T(n), the goal being to whittle T(n) down to an integer. One naturally tries to automate the process of selecting which prime to adjust, which leads to sequences like A(i).
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Allan Wechsler -
Neil Sloane