- #1
Raphie
- 151
- 0
[Edit] Apologies for the typo in the title. "Trinagular" should read "Triangular." [/Edit]
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No x > 8, for x an element of N, satisfies the following equivalency: 3/2 * Tetra_(x-1) = 2^y -2 for any y an element of N
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for Tetra_x is the x-th Tetrahedral Number --> (n*(n+1)*(n+2))/3!
This Conjecture has been computer checked to 100,000,000 and I have been told that not only is a proof possible, but it is so "simple" a high schooler could do it. I'm not sure that's the case for your standard fare high schooler, but I do presume that whatever means were used to prove the finiteness of Ramanujan-Nagell Squares and Triangular Numbers would come into play. Not being familiar with the ins and outs of unique factorization domains, which I know were involved in that proof, I look forward to any responses and have included complex expressions for terms that I see as being possibly related to the issue at hand...
Regarding notation and format:
A) I have included 0 and 1 terms (--> n (modulo 2) and/or additive and multiplicative identities) where appropriate to demonstrate similar form
B) I have also included leading zeroes, so that one may compare apples to apples reading "down" as well as left to right.
================================================================
THE RAMANUJAN-NAGELL TRIANGULAR (T_x) NUMBERS
{0,1,3,15,4095}
Z_1
------------------
4/2 * T_x = 2^y -2
--> (x^(2+0) - x^(0+0))/4
--> ((((x - 1)/2) - sqrt (-((x - 1)/2))) * (((x - 1)/2) + sqrt (-((x - 1)/2))) - 0*x)^(2/2) - 0*x)/1^2
A Simple Expression
(001^(2+0) - 001^(0+0))/4 = 0000 = 2/1*T_00 = 2^01 - 1
(003^(2+0) - 003^(0+0))/4 = 0002 = 2/1*T_01 = 2^02 - 1
(005^(2+0) - 005^(0+0))/4 = 0006 = 2/1*T_02 = 2^03 - 1
(011^(2+0) - 011^(0+0))/4 = 0030 = 2/1*T_05 = 2^05 - 1
(181^(2+0) - 181^(0+0))/4 = 8190 = 2/1*T_90 = 2^13 - 1
An Equivalent Complex Expression
((00 - sqrt (-00)) * (00 + sqrt (-00)) - 0)^(2/2) - 0)/1^2 = 0000 = 2/1*T_00 = 2^01 - 1
((01 - sqrt (-01)) * (01 + sqrt (-01)) - 0)^(2/2) - 0)/1^2 = 0002 = 2/1*T_01 = 2^02 - 1
((02 - sqrt (-02)) * (02 + sqrt (-02)) - 0)^(2/2) - 0)/1^2 = 0006 = 2/1*T_02 = 2^03 - 1
((05 - sqrt (-05)) * (05 + sqrt (-05)) - 0)^(2/2) - 0)/1^2 = 0030 = 2/1*T_05 = 2^05 - 1
((90 - sqrt (-90)) * (90 + sqrt (-90)) - 0)^(2/2) - 0)/1^2 = 8190 = 2/1*T_90 = 2^13 - 1
e.g.
((90 - sqrt (-90)) = 90 - 9.48683298 i
((90 + sqrt(-90)) = 90 + 9.48683298 iA ONE TO ONE RAMANUJAN-NAGELL TETRAHEDRAL (Tetra_x) COUNTERPART (?)
{0,0,4,20,84}
Z_2
------------------
3/2 * Tetra_(x-1) = 2^y -2
--> (x^(2+1) - x^(0+1))/4
--> ((((x - 0)/1) - sqrt (-((x - 0)/1))) * (((x - 0)/1) + sqrt (-((x - 0)/1))) - 1*x)^(3/2) - 1*x)/2^2
A Simple Expression
(000^(2+1) - 000^(0+1))/4 = 000 = 3/2 * Tetra_-1 = 2^1 - 1
(001^(2+1) - 000^(0+1))/4 = 000 = 3/2 * Tetra_0 = 2^1 - 1
(003^(2+1) - 003^(0+1))/4 = 006 = 3/2 * Tetra_2 = 2^3 - 1
(005^(2+1) - 005^(0+1))/4 = 030 = 3/2 * Tetra_4 = 2^5 - 1
(008^(2+1) - 008^(0+1))/4 = 126 = 3/2 * Tetra_7 = 2^7 - 1
An Equivalent Complex Expression
((00 - sqrt (-00)) * (00 + sqrt (-00)) - 0)^(3/2) - 0)/2^2 = 000 = 3/2 * Tetra_-1 = 2^1 - 1
((01 - sqrt (-01)) * (01 + sqrt (-01)) - 1)^(3/2) - 1)/2^2 = 000 = 3/2 * Tetra_0 = 2^1 - 1
((03 - sqrt (-03)) * (03 + sqrt (-03)) - 3)^(3/2) - 3)/2^2 = 006 = 3/2 * Tetra_2 = 2^3 - 1
((05 - sqrt (-05)) * (05 + sqrt (-05)) - 5)^(3/2) - 5)/2^2 = 030 = 3/2 * Tetra_4 = 2^5 - 1
((08 - sqrt (-08)) * (08 + sqrt (-08)) - 8)^(3/2) - 8)/2^2 = 126 = 3/2 * Tetra_7 = 2^7 - 1
e.g.
8 - sqrt(-8) = 8 - 2.82842712 i
8 + sqrt(-8) = 8 + 2.82842712 i
Z_0 = Z_1 UNION Z_2 (includes both 0's from Z_2, but no repeats between sets)
Z_0 --> {0,0,2,6,30,126,8190}
y_0 = y_1 UNION y_2
y_0 --> {1,1,2,3,5,7,13} --> n an element of N such that d'(p' - 1) = n
This set includes 1 repeated twice, plus the first 5 Mersenne Prime Exponents, the first 12 of which are...
{2,3,5,7,13,17,19,31,61,89,107,127...}
Background Note #1:
2,3,5,7, & 13 are the set of Mersenne Prime exponents highlighted by Frampton and Kephart in their 1999 paper:
Mersenne Primes, Polygonal Anomalies and String Theory Classification
http://arxiv.org/abs/hep-th/9904212
REASON FOR INTEREST:
If there is anything to the above conjecture, then this would bring me one step closer to building a reasonable case in support of an heuristically constructed hypothesis that predicts a 1:1 correspondence between "Frampton-Kephart" numbers and the Maximal Kissing Numbers of lattices in Dimension 1,4,8,14 and 24, with 1764 targeted as the Maximal Sphere Packing for Dimension 14 (126 * 14 = 1764).
e.g.
X_00 = 0000 * 00 = 000000 = K_00
X_00 = 0000 * 00 = 000000 = K_00
X_01 = 0002 * 01 = 000002 = K_01
X_04 = 0006 * 04 = 000024 = K_04
X_08 = 0030 * 08 = 000240 = K_08
X_14 = 0126 * 14 = 001764 = ?
X_24 = 8190 * 24 = 196560 = K_24
... for 0, 2, 6, 30 & 8190 constitute the elements of Z_1, (aka "Twice the Ramanujan-Nagell Triangular Numbers")
Background Note #2:
Z_1 INTERSECTION Z_2 = Z_0(-)
--> {0,6,30}
Let k = 3n + 1 --> {1, 4, 7}
Let x = (Z_0(-))^2 + (Z_0(-))^1 --> {0, 42, 930}
Then...
k*x = SUM [d(x)] for range (0,x)
Note: The Number of divisors of zero are considered to be 0
000 = 0^2 + 0
000 = ((0 - sqrt (-0)) * (0 + sqrt (-0)) - sqrt (-(0 - sqrt (-0)) * (0 + sqrt (-0)))) * ((0 - sqrt (-0)) * (0 + sqrt (-0)) + sqrt (-(0 - sqrt (-0)) * (0 + sqrt (-0))))
1*000 = 0000 = SUM [d(x)] for range n = 0 --> 000
042 = 6^2 + 6
042 = ((2 - sqrt (-2)) * (2 + sqrt (-2)) - sqrt (-(2 - sqrt (-2)) * (2 + sqrt (-2)))) * ((2 - sqrt (-2)) * (2 + sqrt (-2)) + sqrt (-(2 - sqrt (-2)) * (2 + sqrt (-2))))
4*042 = 0168 = SUM [d(x)] for range n = 0 --> 042
930 = 30^2 + 30
930 = ((5 - sqrt (-5)) * (5 + sqrt (-5)) - sqrt (-(5 - sqrt (-5)) * (5 + sqrt (-5)))) * ((5 - sqrt (-5)) * (5 + sqrt (-5)) + sqrt (-(5 - sqrt (-5)) * (5 + sqrt (-5))))
7*930 = 6510 = SUM [d(x)] for range n = 0 --> 930Related Sequence (extended to include 0):
Sum of divisor function d(n) (A000005) up to n is divisible by n
[0] 1, 4, 5, 15, 42, 44, 47, 121, 336, 340, 347, 930...
Sloane's A050226: http://www.research.att.com/~njas/sequences/A050226
Proper title for progression extension that includes 0:
Sum of divisor function d(n) (A000005) up to n is divisible by n!/|n-1|!
Background Note #3:
Conjecture 1.1 is a follow-up conjecture to...
CONJECTURE # 1.0
n an element of N such that d'(p' - 1) = n
n --> {0,1,2,3,4,6}
p' --> {1,2,3,5,7,13}
More: https://www.physicsforums.com/showthread.php?t=442527
====================================================================
No x > 8, for x an element of N, satisfies the following equivalency: 3/2 * Tetra_(x-1) = 2^y -2 for any y an element of N
====================================================================
for Tetra_x is the x-th Tetrahedral Number --> (n*(n+1)*(n+2))/3!
This Conjecture has been computer checked to 100,000,000 and I have been told that not only is a proof possible, but it is so "simple" a high schooler could do it. I'm not sure that's the case for your standard fare high schooler, but I do presume that whatever means were used to prove the finiteness of Ramanujan-Nagell Squares and Triangular Numbers would come into play. Not being familiar with the ins and outs of unique factorization domains, which I know were involved in that proof, I look forward to any responses and have included complex expressions for terms that I see as being possibly related to the issue at hand...
Regarding notation and format:
A) I have included 0 and 1 terms (--> n (modulo 2) and/or additive and multiplicative identities) where appropriate to demonstrate similar form
B) I have also included leading zeroes, so that one may compare apples to apples reading "down" as well as left to right.
================================================================
THE RAMANUJAN-NAGELL TRIANGULAR (T_x) NUMBERS
{0,1,3,15,4095}
Z_1
------------------
4/2 * T_x = 2^y -2
--> (x^(2+0) - x^(0+0))/4
--> ((((x - 1)/2) - sqrt (-((x - 1)/2))) * (((x - 1)/2) + sqrt (-((x - 1)/2))) - 0*x)^(2/2) - 0*x)/1^2
A Simple Expression
(001^(2+0) - 001^(0+0))/4 = 0000 = 2/1*T_00 = 2^01 - 1
(003^(2+0) - 003^(0+0))/4 = 0002 = 2/1*T_01 = 2^02 - 1
(005^(2+0) - 005^(0+0))/4 = 0006 = 2/1*T_02 = 2^03 - 1
(011^(2+0) - 011^(0+0))/4 = 0030 = 2/1*T_05 = 2^05 - 1
(181^(2+0) - 181^(0+0))/4 = 8190 = 2/1*T_90 = 2^13 - 1
An Equivalent Complex Expression
((00 - sqrt (-00)) * (00 + sqrt (-00)) - 0)^(2/2) - 0)/1^2 = 0000 = 2/1*T_00 = 2^01 - 1
((01 - sqrt (-01)) * (01 + sqrt (-01)) - 0)^(2/2) - 0)/1^2 = 0002 = 2/1*T_01 = 2^02 - 1
((02 - sqrt (-02)) * (02 + sqrt (-02)) - 0)^(2/2) - 0)/1^2 = 0006 = 2/1*T_02 = 2^03 - 1
((05 - sqrt (-05)) * (05 + sqrt (-05)) - 0)^(2/2) - 0)/1^2 = 0030 = 2/1*T_05 = 2^05 - 1
((90 - sqrt (-90)) * (90 + sqrt (-90)) - 0)^(2/2) - 0)/1^2 = 8190 = 2/1*T_90 = 2^13 - 1
e.g.
((90 - sqrt (-90)) = 90 - 9.48683298 i
((90 + sqrt(-90)) = 90 + 9.48683298 iA ONE TO ONE RAMANUJAN-NAGELL TETRAHEDRAL (Tetra_x) COUNTERPART (?)
{0,0,4,20,84}
Z_2
------------------
3/2 * Tetra_(x-1) = 2^y -2
--> (x^(2+1) - x^(0+1))/4
--> ((((x - 0)/1) - sqrt (-((x - 0)/1))) * (((x - 0)/1) + sqrt (-((x - 0)/1))) - 1*x)^(3/2) - 1*x)/2^2
A Simple Expression
(000^(2+1) - 000^(0+1))/4 = 000 = 3/2 * Tetra_-1 = 2^1 - 1
(001^(2+1) - 000^(0+1))/4 = 000 = 3/2 * Tetra_0 = 2^1 - 1
(003^(2+1) - 003^(0+1))/4 = 006 = 3/2 * Tetra_2 = 2^3 - 1
(005^(2+1) - 005^(0+1))/4 = 030 = 3/2 * Tetra_4 = 2^5 - 1
(008^(2+1) - 008^(0+1))/4 = 126 = 3/2 * Tetra_7 = 2^7 - 1
An Equivalent Complex Expression
((00 - sqrt (-00)) * (00 + sqrt (-00)) - 0)^(3/2) - 0)/2^2 = 000 = 3/2 * Tetra_-1 = 2^1 - 1
((01 - sqrt (-01)) * (01 + sqrt (-01)) - 1)^(3/2) - 1)/2^2 = 000 = 3/2 * Tetra_0 = 2^1 - 1
((03 - sqrt (-03)) * (03 + sqrt (-03)) - 3)^(3/2) - 3)/2^2 = 006 = 3/2 * Tetra_2 = 2^3 - 1
((05 - sqrt (-05)) * (05 + sqrt (-05)) - 5)^(3/2) - 5)/2^2 = 030 = 3/2 * Tetra_4 = 2^5 - 1
((08 - sqrt (-08)) * (08 + sqrt (-08)) - 8)^(3/2) - 8)/2^2 = 126 = 3/2 * Tetra_7 = 2^7 - 1
e.g.
8 - sqrt(-8) = 8 - 2.82842712 i
8 + sqrt(-8) = 8 + 2.82842712 i
Z_0 = Z_1 UNION Z_2 (includes both 0's from Z_2, but no repeats between sets)
Z_0 --> {0,0,2,6,30,126,8190}
y_0 = y_1 UNION y_2
y_0 --> {1,1,2,3,5,7,13} --> n an element of N such that d'(p' - 1) = n
This set includes 1 repeated twice, plus the first 5 Mersenne Prime Exponents, the first 12 of which are...
{2,3,5,7,13,17,19,31,61,89,107,127...}
Background Note #1:
2,3,5,7, & 13 are the set of Mersenne Prime exponents highlighted by Frampton and Kephart in their 1999 paper:
Mersenne Primes, Polygonal Anomalies and String Theory Classification
http://arxiv.org/abs/hep-th/9904212
REASON FOR INTEREST:
If there is anything to the above conjecture, then this would bring me one step closer to building a reasonable case in support of an heuristically constructed hypothesis that predicts a 1:1 correspondence between "Frampton-Kephart" numbers and the Maximal Kissing Numbers of lattices in Dimension 1,4,8,14 and 24, with 1764 targeted as the Maximal Sphere Packing for Dimension 14 (126 * 14 = 1764).
e.g.
X_00 = 0000 * 00 = 000000 = K_00
X_00 = 0000 * 00 = 000000 = K_00
X_01 = 0002 * 01 = 000002 = K_01
X_04 = 0006 * 04 = 000024 = K_04
X_08 = 0030 * 08 = 000240 = K_08
X_14 = 0126 * 14 = 001764 = ?
X_24 = 8190 * 24 = 196560 = K_24
... for 0, 2, 6, 30 & 8190 constitute the elements of Z_1, (aka "Twice the Ramanujan-Nagell Triangular Numbers")
Background Note #2:
Z_1 INTERSECTION Z_2 = Z_0(-)
--> {0,6,30}
Let k = 3n + 1 --> {1, 4, 7}
Let x = (Z_0(-))^2 + (Z_0(-))^1 --> {0, 42, 930}
Then...
k*x = SUM [d(x)] for range (0,x)
Note: The Number of divisors of zero are considered to be 0
000 = 0^2 + 0
000 = ((0 - sqrt (-0)) * (0 + sqrt (-0)) - sqrt (-(0 - sqrt (-0)) * (0 + sqrt (-0)))) * ((0 - sqrt (-0)) * (0 + sqrt (-0)) + sqrt (-(0 - sqrt (-0)) * (0 + sqrt (-0))))
1*000 = 0000 = SUM [d(x)] for range n = 0 --> 000
042 = 6^2 + 6
042 = ((2 - sqrt (-2)) * (2 + sqrt (-2)) - sqrt (-(2 - sqrt (-2)) * (2 + sqrt (-2)))) * ((2 - sqrt (-2)) * (2 + sqrt (-2)) + sqrt (-(2 - sqrt (-2)) * (2 + sqrt (-2))))
4*042 = 0168 = SUM [d(x)] for range n = 0 --> 042
930 = 30^2 + 30
930 = ((5 - sqrt (-5)) * (5 + sqrt (-5)) - sqrt (-(5 - sqrt (-5)) * (5 + sqrt (-5)))) * ((5 - sqrt (-5)) * (5 + sqrt (-5)) + sqrt (-(5 - sqrt (-5)) * (5 + sqrt (-5))))
7*930 = 6510 = SUM [d(x)] for range n = 0 --> 930Related Sequence (extended to include 0):
Sum of divisor function d(n) (A000005) up to n is divisible by n
[0] 1, 4, 5, 15, 42, 44, 47, 121, 336, 340, 347, 930...
Sloane's A050226: http://www.research.att.com/~njas/sequences/A050226
Proper title for progression extension that includes 0:
Sum of divisor function d(n) (A000005) up to n is divisible by n!/|n-1|!
Background Note #3:
Conjecture 1.1 is a follow-up conjecture to...
CONJECTURE # 1.0
n an element of N such that d'(p' - 1) = n
n --> {0,1,2,3,4,6}
p' --> {1,2,3,5,7,13}
More: https://www.physicsforums.com/showthread.php?t=442527
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