OBSERVATION: The #31, The Golden Scale, Prime Counting Function & Partition Numbers


by Raphie
Tags: counting, function, golden, numbers, observation, partition, prime, scale
Raphie
Raphie is offline
#1
Feb4-11, 10:19 PM
P: 153
The guiding premise of this thread is the following proposition: If fractals play a role in the behavior of partitions, then maybe, just maybe, they play a role also in the positioning of the primes; and if they do, then who is to say that the two, prime numbers and partition numbers, cannot at some point down the road be mathematically related in precise manner via, for instance, the prime counting function and/or various number progressions related to fractals and/or the division of n-dimensional spaces?

In other words, many seem to believe it impossible that we will ever be able to do more than simply estimate where the next prime may be found. I would contend that (although I certainly would not be the one to prove it..), in principle, it should be just as possible to locate the positions of the primes as it is to calculate the value of the partition numbers.

Thoughts, as well as any related numerical observations, more than welcome...

- RF

=================================================
In relation to recent discussion here...
Ken Ono and Hausdorff dimensions
http://www.physicsforums.com/showthread.php?t=468910

Quote Quote by chis View Post
Hi All, In Ken Onos lecture he mentions Hausdorff dimensions appertaining to prime numbers:
5,7,11 relate to 0 dimension and primes from 13 to 31 as 1 dimension.
and here...
relatively prime and independent confusion
http://www.physicsforums.com/showthread.php?t=467088

Quote Quote by al-mahed View Post
[tex]a^{10^{n+1}}-a^{10^n}\equiv\ 0\ mod\ 77[/tex]
[tex]a^{10^n}(a^{10^{n+1}-10^n}-1)\equiv\ 0\ mod\ 77[/tex]
[tex]a^{10^{n+1}-10^n}\equiv\ 1\ mod\ 77[/tex]
and [tex]10^{n+1}-10^n=10^n(10-1)=9\cdot\ 10^n[/tex]
since the above can be written as [tex]9\cdot\ 10^n=60k[/tex] for a natural k

[tex]a^{9\cdot\ 10^n}=a^{60k}\equiv\ 1\ mod\ 77[/tex]

by euler's theorem, if gcd(a,77)=1, since [tex]\varphi{(77)}=60[/tex] then

[tex]a^{60k}\equiv\ 1\ mod\ 77[/tex]

and it completes the proof
Quote Quote by Raphie View Post

Ken Ono cracks partition number mystery
http://www.physicsforums.com/showthread.php?t=465696

To avoid confusion, denote par_n as the n-th partition number. Then Ramanujan's congruences are...

par_(5k+4) == 0 (mod 5)
par_(7k+5) == 0 (mod 7)
par_(11k+6) == 0 (mod 11)
http://en.wikipedia.org/wiki/Ramanujan%27s_congruences
Note: 5, 7 & 11 are the 4th, 5th & 6th partition numbers and because they are all also prime (totient p) = (p-1), then their totient product: totient (5)*totient (7)*totient (11) = 240 = totient (5*7*11)

We can, in relation to your prior posting observe that...

par_(totient (77) + (1 + 77n)) = par_(60 + (1 +77n))
... I thought to pass along the following observation:

OBSERVATION
The #31, The Golden Scale, The Prime Counting Function & Partition Numbers

A SIMPLE ALGEBRAIC STATEMENT
20 + 2T_(n+1)
= 5^(n+2) + (n + 2)^2 - (5 + (n + 2))
n = (0 --> 7)

for...
T_n denotes the n-th Triangular Number

---------------------------------------------
5^2 + 2^2 - (5 + 2) = 029 - 07 = 22
---------------------------------------------
5^2 + 3^2 - (5 + 3) = 034 - 08 = 26
5^2 + 4^2 - (5 + 4) = 041 - 09 = 32
5^2 + 5^2 - (5 + 5) = 050 - 10 = 40
5^2 + 6^2 - (5 + 6) = 061 - 11 = 50
5^2 + 7^2 - (5 + 7) = 074 - 12 = 62
5^2 + 8^2 - (5 + 8) = 089 - 13 = 76
---------------------------------------------
5^2 + 9^2 - (5 + 9) = 106 - 14 = 92
---------------------------------------------
SUM = 400 = 20^2

106 - 29 = 77

NOW COMPARE, keeping in mind this progression....
The Golden Scale (Fibonacci 2,5)
2, 5, 7, 12, 19, 31, 50, 81....


for...
pi(n) - denotes the Prime Counting Function
par_n - denotes the n-th Partition Number


----------------------------------
pi_(par_(12)) = pi (077) = 21 = (5 - 2) * 7
----------------------------------
pi (par_(13)) = pi (101) = 26
pi (par_(14)) = pi (135) = 32
pi (par_(15)) = pi (176) = 40 = 90 - pi_(par_(16)))
pi (par_(16)) = pi (231) = 50 = 90 - pi_(par_(15)))
pi (par_(17)) = pi (297) = 62
pi (par_(18)) = pi (385) = 76
----------------------------------
pi_(par_(19)) = pi (490) = 93 = (5 - 2) * 31
----------------------------------
SUM = 400 = 20^2

pi (par_(16)) - pi (par_(15)) = 50 - 40 = 10 --> 10 = 2 * 5; 16 + 15 = 31
pi (par_(17)) - pi (par_(14)) = 62 - 32 = 30 --> 10 + 30 = 40 = 90 - 50; 17 + 14 = 31
pi (par_(18)) - pi (par_(13)) = 76 - 26 = 50 --> 10 + 30 + 50 = 90 = 40 + 50; 18 + 13 = 31
pi_(par_(19)) - pi_(par_(12)) = 93 - 21 = 72 --> 10 + 30 + 50 + 72 = 162 = 2 * 81; 19 + 12 = 31

also...
pi_(par_(20)) = pi (627) = 114
= pi_(par_(19)) + pi_(par_(12)) = 93 + 26 = 114

Are these relationships presented above (which, hopefully, I need not spell out...) "random" or "coincidental?" My reply: "Sure, they could be, but I find it rather unlikely." At the very least, it would be interesting to look at all combinations of partition numbers, the index numbers of which sum to 5, 7, 11, 13 or 31, associated with Hausdorff dimension 0 and 1, just to see what relationships one might find. Some interesting patterns could emerge in relation to discrete intervals, just as there are interesting patterns that emerge in regards to prime gaps, etc...
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Raphie
Raphie is offline
#2
Feb5-11, 05:02 PM
P: 153
A (POSSIBLY) RELATED OBSERVATION
(Fibonacci + 10) Indexed Palindromes & The Golden Scale for 1 and 2 dimensional numbers
Quote Quote by chis View Post
Hi All, In Ken Onos lecture he mentions Hausdorff dimensions appertaining to prime numbers:
5,7,11 relate to 0 dimension and primes from 13 to 31 as 1 dimension.
The Fibonacci Series (F_n): 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89...
The Golden Scale (G_n): 2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898...

Note: Golden Scale Numbers are the sum of 5 consecutive Fibonacci Numbers e.g (1+2 + 3 + 5 + 8 = 19 = G_4) and (13 + 21 + 34 + 55 + 89 = 212 = G_9)

Two Digits --> 1 Dimensional Number 10^1*z + 10^0*y
PALINDROME_(F_(n + 1) + 10) = G_(n - 3) + G_(n+2)
range: n = 0 --> 5
Corresponds with: Fibonacci Numbers from F_1 through F_6
Corresponds with: Golden Scale Numbers from G_-3 through G_7

n-th Palindrome (=10*F_(n+1) + F_(n+1))
(1+10) = 11 --> 11 --> 1 = 7 + 4 = G_2 + G_-3
(1+10) = 11 --> 11 --> 1 = 12 - 1 = G_3 + G_-2
(2+10) = 12 --> 22 --> 2 = 19 + 3 = G_4 + G_-1
(3+10) = 13 --> 33 --> 3 = 31+ 2 = G_5 + G_0
(5+10) = 15 --> 55 --> 5 = 50+ 5 = G_6 + G_1
(8+10) = 18 --> 88 --> 8 = 81+ 7 = G_7 + G_2

4, - 1, 3, 2, 5, 7, 12, 19, 31, 50, 81

Three Digits --> 2 Dimensional Number 10^2*z + 10^1*y + 10^0*x
PALINDROME_(F_(n + 7) + 10) = G_(n+8)
range: n = 0 --> 4
Corresponds with: Fibonacci Numbers from F_7 through F_11
Corresponds with: Golden Scale Numbers from G_8 through G_12

n-th Palindrome = G_n
(13+10) = 23 --> 131 = 131 = G_8
(21+10) = 31 --> 212 = G_9
(34+10) = 44 --> 343 = G_10
(55+10) = 65 --> 555 = G_11
(89+10) = 99 --> 898 = G_12

131, 212, 343, 555, 898

(Concatenate: 13+1, 21+2, 34+3, 55+5, 89+8, all Fibonacci numbers, or, alternatively, 1+31, 2+12, 3+43, 5+55, 8+98, all Fibonacci Numbers read backwards...)

It ought to be possible to extend this to 3 and 4 dimensions, but the patterns become less clear as numbers start to overlap, in similar manner to the decimal expansion of 1/89 (period = 44 --> 13 + 31), which embeds the Fibonacci Sequence...

1/89 = 0.01123595505617977528089887640449438202247191... (repeating)

- RF


P.S. Here's an excellent page that discusses the relationship between the Golden Scale and Music Theory...
On Rabbits, Mathematics and Musical Scales
John S. Allen

excerpt
Now let's see how these numbers build musical structures. The simplest musical interval is the octave, a 2/1 frequency ratio. The next simplest are the fifth, a 3/2 ratio (more or less, in various tunings) and its inversion, the fourth (4/3, more or less). The numbers in the musical Fibonacci series 2, 5, 7, 12, 19 ... all are generated by increasingly long series of musical fourths and fifths, as Sir James Jeans described and shown in the table below.
http://www.bikexprt.com/tunings/fibonaci.htm
Raphie
Raphie is offline
#3
Feb9-11, 05:57 PM
P: 153
A few rather odd equivalencies in relation to the topic of this thread...

Note first that par_0 is the first Partition Number greater than a Triangular Number of same index # (1 vs. 0) and that par_13 is the second one (101 vs. 91). Subsequent to par_13, all partition numbers are greater than triangular numbers of same index #, but for the range 1 through 12 the Triangular Numbers are greater than or equal to the partition numbers.

e.g.
00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 013 = n
-------------------------------------------------------------
01, 01, 02, 03, 05, 07, 11, 15, 22, 30, 42, 56, 77, 101 = nth Partition number
00, 01, 03, 06, 10, 15, 21, 28, 36, 45, 55, 66, 78, 091 = nth Triangular number

Now take the following partition numbers:

par_0 = 1
par_6 = 11
par_12 = 77
par_13 = 101

0 + 6 + 12 + 13 = 31

Then for pi(n) the Prime Counting Function, phi(n) the Euler Totient Function, p_n the nth prime, par_n the nth partition number, and T_n the nth Triangular number, bearing in mind that...

T_00 + p'_00 - par_00 = 00 + 01 - 001 = 00
T_13 + p'_13 - par_13 = 91 + 41 - 101 = 31


p'_x denotes an Integer| 0 < d(n) < 3

... then, the following equivalencies hold...

101 + 11 = 91 + 21 = T_13 + T_06 = par_13 + par_06 = 112

0 + pi (31 - 11) + phi (31 - 11) + (31 - 11) = 0 + 8 + 08 + 20 = 26
0 + pi (31 - 12) + phi (31 - 12) + (31 - 12) = 0 + 8 + 18 + 19 = 45
0 + pi (31 - 13) + phi (31 - 13) + (31 - 13) = 0 + 7 + 06 + 18 = 31

26 + 45 + 31 = 112

1 + 77 = 0 + 78 = par_0 + par_12 = T_0 + T_12 = 78

0 + pi (00 + 11) + phi (00 + 11) + (00 + 11) = 0 + 5 + 10 + 11 = 26
0 + pi (00 + 12) + phi (00 + 12) + (00 + 12) = 0 + 5 + 04 + 12 = 21
0 + pi (00 + 13) + phi (00 + 13) + (00 + 13) = 0 + 6 + 12 + 13 = 31

26 + 21 + 31 = 78

112 + 78 = T_(13 + 6) = T_19 = 190
000 + 78 = T_(0 + 12) = T_12 = 078

Additionally:
p_(11 + 15) = p_26 = par_13 = 101
p_(par_6 + par_7) = par (6 + 7) = 101

This prime partition pair is the last one | p_2x = par_x

- RF


A RELATED OBSERVATION
A Prime / Mersenne / (Ramanujan) Triangular Number Convolution
http://www.physicsforums.com/showthread.php?t=452148

For p'_x denotes an Integer| 0 < d(n) < 3, M_x a Mersenne Number and T_n a Triangular Number, then...

((p'_x * p'_2x) * (M_x - |T_x - 1|)) / ((T_(M_x) - T_|T_x - 1|) is in N

for |T_x - 1| = 1, 0, 2, 5, 90 and x = 0, 1, 2, 3, 13

EXPANSION
((01 * 001) * (0000 - 01)) / ((0000^2 + 0000)/2 - (01^2 + 01)/2) = 01
((02 * 003) * (0001 - 00)) / ((0001^2 + 0001)/2 - (00^2 + 00)/2) = 06
((03 * 007) * (0003 - 02)) / ((0003^2 + 0003)/2 - (02^2 + 02)/2) = 07
((05 * 013) * (0007 - 05)) / ((0007^2 + 0007)/2 - (05^2 + 05)/2) = 10
((41 * 101) * (8191 - 90)) / ((8191^2 + 8191)/2 - (90^2 + 90)/2) = 01
For quick verification, anyone interested can just copy and paste these equations into the compute box of Wolfram Alpha at http://www3.wolframalpha.com/

0, 1, 2, 5, 90 are the index numbers associated with the Ramanujan-Nagell Triangular Numbers 0, 1, 3, 15, 4095, which, as Î have previously noted, map, whether coincidentally or no, in 1:1 manner with Maximal Sphere Packings for dimension sigma (M_n(mod 5)) = 0, 1, 4, 8, 24 by the formula 2*T_z*sigma (M_n(mod 5)), for T_z denotes a Ramanujan-Nagell Triangular Number and sigma(n) the sum of divisor function.

2* 0 *0 = 0 = K_0
2* 1 *1 = 2 = K_1
2* 3 *4 = 24 = K_4
2* 15 *8 = 240 = K_8
2* 4095 * 24 = 196560 = K_24

RELATED PAPER
Kissing Numbers, Sphere Packings, and Some Unexpected Proofs
Florian Pfender & Günter M. Ziegler (2004)
http://mathdl.maa.org/mathDL/22/?pa=...nt&nodeId=3065

Raphie
Raphie is offline
#4
Mar1-11, 02:35 AM
P: 153

OBSERVATION: The #31, The Golden Scale, Prime Counting Function & Partition Numbers


As Fibonacci Numbers (F_n) are to Lucas Numbers (L_n), The Golden Scale (G_n) is to the "Ionian Scale" (I_n) (coinage) .

F_(n-1) + F_(n+1) = L_n

F_n = 0, 1, 01, 02, 03, 05, 008, 013...
L_n = 2, 1, 03, 04, 07, 11, 018, 029...

G_(n-1) + G_(n+1) = I_n

G_n = 2, 5, 07, 12, 19, 31, 050, 081...
I_n = 8, 9, 17, 26, 43, 69, 112, 181...

There are all kinds of "nifty" identities one can come up with simply via a bit of hands on observation of the "behavior" of these number sets, with the safety and surety of knowing one could prove such identities, if need be, via induction. For instance, I_n - I_-n*(-1)^n always = 10*F_n and I_n + I_-n*(-1)^n = 8*L_n

e.g.
9 + 1 = 10 * F_1 = 10 * 1
9 - 1 = 8 * L_1 = 8 * 1

17 - 7 = 10 * F_2 = 10 * 1
17 + 7 = 8 * L_2 = 8 * 3

26 + -6 = 10 * F_3 = 10 * 2
26 - -6 = 8 * L_3 = 8 * 4

Far more dicey would be to suggest that, for instance, it is not "accident" that F_7 = p_(6) = 13 and I_7 = p_(6^2 + 6) = 181, and 2*T_13 - 1 = 181. 181^2, of course, being the greatest n | n^2 = 2^(y+2) - 7 (In other words, (I_7)^2 = 2^(13+2) - 7). In order to even begin to make such a statement in a manner that more conservative minds might find responsible, it would first have to be generally accepted that a) the precise positioning of the primes is not random (the hypothetical assumption upon which this thread is based...), b) that such lack of randomness might have something to do with fractals, and c) that one such class of relevant fractal progression might be number progressions based upon the Golden ratio.

- RF

Note: Golden Scale Numbers are equal to the sum of 5 consecutive Fibonacci Numbers.
Goongyae
Goongyae is offline
#5
Mar1-11, 12:23 PM
P: 70
Very interesting data Ralphie!

>If fractals play a role in the behavior of partitions, then maybe, just maybe, they play a role also in the positioning of the primes

There is definitely some structure to the primes. Primes occur based on frequencies evident in the explicit formula: http://en.wikipedia.org/wiki/Explicit_formula

There is also the sum-of-divisors function which obeys Euler's recursive pattern I linked to earlier: d(x) = d(x-1) + d(x-2) - d(x-5) - d(x-7) + ... where the terms alternate as ++--++--... and the numbers are generalized pentagonal numbers, and where d(0) if it appears must be replaced by x itself.
Raphie
Raphie is offline
#6
Mar2-11, 02:53 PM
P: 153
Quote Quote by Goongyae View Post
and where d(0) if it appears must be replaced by x itself.
Do you happen to have any data on where d(0) appears? It would be interesting to see what kind of patterns might arise. And it would be super interesting if it somehow turned out that one could establish linkage with the zeta function. In general, the "feel" I get from Euler's recursive formula is akin to a clock that recallibrates (i.e. resets) itself each time the seconds, minutes or hours indicator makes one full revolution.

As an FYI, the sum of divisors of n, sigma d(n), up to...

42 --> 168 = 4*(42); 42 = 6^2 + 6 = totient (7)^2 + totient (7)
930 --> 6510 = 7*(930); 930 = 30^2 + 30 = totient (31)^2 + totient (31)

2^7 - 1 = (p_31) = p_(p_11) = p_(2^pi(11) - 1)

A long standing question I have had is if there is any greater Pronic number (n^2 + n) than 930, such that k*(n^2+n) = the sum of d(n) up to n^2 + n. 42 and 930, of course, are also expressible, respectively, as: 7^2 - 7 and 31^2 - 31. Interesting, since the Dirichlet Divisor function can be related to points on a lattice and...

7*4 = 28 = K_4 + 4 = Dimensions of D4 --> 1/1 * 2nd Perfect Number (28; Totient 28 = 12 = 1/2 * K_4)
31*8 = 248 = K_8 + 8 = Dimensions of E8 --> 1/2 * 3rd Perfect Number (496; Totient 496 = 240 = 1/1 * K_8)

Note: 6 & 30 are themselves also both pronic numbers. 2^2 + 2 & 5^2 + 5, respectively.

- RF


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