Theoretics on Hailstone Numbers

Residue Theory and the Collatz Conjecture (Hailstone Numbers Problem)
By: Steven Zheng
15/02/2011

Introduction
In 1937, mathematician Lothar Collatz asked whether any positive integer will always reach the number one after a succession of operations are applied, one being to divide the number when it is even, two being to triple the number then add one when it is odd. Until this day, the conjecture is still unsolved but most mathematicians expect the answer to be affirmative. Current computer assisted research has verified the first 5.48×10^18 positive integers reach one. (2009).

Purpose
To prove the Collatz conjecture by verifying the following:
1)Any integer will not diverge into the infinite under the operations of the Collatz conjecture.
2)Digit loops do not exist before reaching the integer one.

Common Notions
1) An even number divided by two will result an even number if it is not composed of an even number followed by two (ex. 62, 122, and 4082).
2) An odd multiplied by an odd number then summed by an odd number is even.
3) Powers of two are hailstone numbers under the conditions of the Collatz Conjecture.

Assumptions
1) Any integer will not diverge into the infinite under the operations of the Collatz conjecture.
2) Digit loops do not exist before reaching the integer one.

Axioms
1) One is the smallest positive integer.
2) Every positive odd number can be described by the equation 2n+1.
Every positive even number can be described by the equation 2n+2.

Parameters
1) The Collatz conjecture is restricted to the positive integers.
2) The two operations are n/2 for every even number and 3n+1 for any odd number.
3) When the number sequence reaches one, stop.
4) The order type of the positive integers is ω (a definite minimum but no definite maximum).
5) The cardinality of the positive integers is〖 ℵ〗_0.
6) The number one is not accounted in any set of positive integers because it is the theorized “last number” in all sequences.

Hypothesis
If every odd and even positive integer greater than one will become a power of two under the conditions of the Collatz Conjecture, then every positive integer will be a hailstone number.

Part 1: Turbulent Operations
Definition: Turbulent operations are a set of operations that prevent a counting set to diverge to the infinite.

Let there be a counting set of integers {Z} and an operation *.
{Z}* will never approach ±∞.

Postulates
1) Turbulent operations can be made not turbulent by a change of the set and vice versa. (Obvious/ Strong)
Proof- Let a set of real numbers {R}. When {R is placed under the operation * = n/2, where n is an element of {R}, {R} will approach the infinitely small. When the set {R} changes to a set of positive integers {Z+}, the minimum element is restricted to the number one, therefore the operation terminates when it reaches one.
2) None turbulent operations can be made turbulent by introducing a complementary non-turbulent operation. (Weak)
Proof-In the Collatz conjecture, there exists two operations: n/2 for every even positive integer and 3n+1 for every odd positive integer.
3n+1 is a non-turbulent operation in the set {Z+}
n/2 is a turbulent operation in the set {Z+}
For every sufficiently large odd number in the set {Z+}, the operation 3n+1 will force n to become an even number. For every even number, it is guaranteed that the number can be divided by two at least once.

Complementary Argument
Let n be a positive odd integer.
(3n+1)/2>n
(3n+1)/4(3n+1)/2^(m+1) There are an infinite number of odd numbers that when the Collatz operation 3n+1 is applied, the new even number can be divided by two more than once. Thus, the number n will be reduced to a value less than itself, preventing it from reaching infinity.

Part 2
2^n numbers - positive odd integer continuum
Statement 1: If all positive odd integers eventually reach a 2^n number, then the sequence will not diverge into the infinite because a 2^n number will reach the number one in (n – 1) steps.
Statement 2: If all positive odd integers eventually reach a 2^n number, then a digit loop will not occur until it reaches one because a 2^n will become a smaller 2^n number each step.
The new focus is now the universal link between all positive odd numbers and powers of two.

Part 3: Theory of Residues and the continuum
Definition
1) Residue numbers are numbers that any value within the realms of a set can be evaluated by all turbulent operations.
2) The residue operation is a reduction (division) operation that differs from the modular operation where a separate operation is first applied to a value.
Division: N/ m
Modular: N≡a(mod m)
Residue: (N)* ≡R.N.(mod m)
Let the set of R.N. be defined by the function R(x).
Derivation of R(x) = 6x+4 for the Collatz conjecture
It is known that any odd number multiplied by three then added to one will result an even product.
Fact: The number four is the smallest residue number because 8/ 2 = 4 and 3(1) + 1 = 4.
The operation * is 3n+1 for all odd numbers.
All odd integers can be defined by the equation 2(x) +1
3(odd number)+1≡4 (mod m),where m is 3
3(2x+1)+1≡4 (mod 3), x∈{Z+}
R(x)=6x+4
1 3 5 7 9 11 13 15 17 [R(x)-1]/3
4 10 16 22 28 34 40 46 52 R(x)
Notice every positive odd number is directly linked to a residue number under the operation 3n+1.
Questions
Question 1: Do all residue numbers lead to a different residue number?
Answer: Yes
Proof- 1§ [6(even number) + 4]/2 = [6(2n+2) + 4]/2 = 6n + 8 ,6n+8 ∈Even numbers
(6n + 8)/2 = 3n + 4: n = even number (2k + 2) or odd number (2k + 1)
Even number: 3(2k + 2) + 4 = 6k + 10, 6k+10≡4 (mod 6)
Odd number: 3(2k + 1) + 4 = 6k + 7,6n+7 ∈Odd numbers
3(6k + 7) + 1 = 18k + 22, 18k+22≡4 (mod 6)
2§ [6(odd number) + 4]/2 = [6(2n+1) + 4]/2 = 6n + 5 ,6n+5 ∈Odd numbers
3(6n+5) + 1 = 18n + 16, 18n+16≡4 (mod 6)
QED
Question 2: Do all residue numbers lead to a power of two?
Answer: Indeterminate
Some powers of two are residue numbers.
Let 2^n be set S and R(x)=6x+4 be set T
S∩T=4,16,64,256,1024,4096,16384 …, (2^(2k+2),k∈{W})
S-T=2,8,32,128,512,2048,8192,…, (2^(2k+1),k∈{W})
Congruence of R(x) = 6x + 4 and 2^n (The Final Continuum)
Proposition 1
(6x+4≡2^x)=(2^(2n+2),n∈{Z+})
2^(2n+2)=6(∑_0^(k=n-1)▒〖2^(2k+1))+4,〗 where n∈{Z+}
Proof- Use base two arithmetic
2^(2n+2)=(0*2^0 )+(0*2^1 )+(0*2^2 )+⋯+(1*2^n ),where n is an even integer
Or 1000…000 (base 2)
(∑_0^(k=n-1)▒〖2^(2k+1))= 〗 (0*2^0 )+(1*2^1 )+(0*2^2 )+⋯+(0*2^(n-1) )+(1*2^n )where n is an odd integer
Or 1010…1010 (base 2)

Let x[y] →n denote x followed by string y n times to the right.
Let n←[x]y denote y followed by string x n times to the left.
Let [x] →n denote the string x repeated n times to the right.
Let n←[x] denote the string x repeated n times to the left.

Then 2^(2n+2)=(1[0]→2n+2)base 2 and (∑_0^(k=n-1)▒〖2^(2k+1))= 〗 (n←[10])base 2
(n←[10])base 2*(110)base 2=(2n←[1]00)base 2
(2n←[1]00)base 2+(100)base 2=(1[0]→2n+2)base 2
Therefore
2^(2n+2)=6(∑_0^(k=n-1)▒〖2^(2k+1))+4,〗 where n∈{Z+}
QED
This proof unites every power of two to the specific type of R(x) described above because every even power of two is even, thus the division by two is dictated by the second operation of the Collatz conjecture, resulting in an odd power of two.
Proposition 2
Every odd power of two is not congruent to 1 modulo 3 but 2 modulo 3.
Proof- Use base two arithmetic
2^(2n+1)=(1[0]→2n+1)base 2
(1[0]→2n+1)base 2-(1)base 2=([1]→2n+1)base 2
([1]→2n+1)base2 / (11)base 2 =([n←10])base 2 Remainder 1
Therefore 2^(2n+1)≡2(mod 3)
QED
The significance of this proof states that no positive odd integer corresponds to an odd power of two in one step of the Collatz conjecture.
Proposition 3
All residue numbers R(x) divided by two are congruent to 2 modulo 3.
Proof
R(x)=6x+4
R(x)/2=3x+2
2^(2n+1)=3(∑_0^(k=n-1)▒〖2^(2k+1))+2,〗 where n∈{Z+}
QED
Specific values of residue numbers divided by two are odd powers of two.
As proven above, a residue number R(x) will always lead to a different residue number if it is not equal to an even power of two. Since there is infinitude of residue numbers and powers of two, all residue numbers that are not an even power of two will eventually become one under the operations of the Collatz conjecture.

Summary
1) Every positive even number will become an odd number when it follows the operation so of the Collatz conjecture. Consequently, all positive even integers are unified with the set of positive odd integers.
2)6n+4,n∈{Z+} The function of all positive odd numbers of the Collatz conjecture, called the residue numbers. Hence the function binds every odd integers to a residue number.
3) Every residue number will lead to a different residue number following the operations of the Collatz number.
2^(2n+2)=6(∑_0^(k=n-1)▒〖2^(2k+1))+4 〗and 2^(2n+1)=3(∑_0^(k=n-1)▒〖2^(2k+1))+2,〗 where n∈{Z+}
4) There exists an infinite number of residue numbers and powers of two. According to point number three, every residue number will lead to a different residue number under the Collatz conjecture. Therefore, every residue number will eventually become an even power of two.
5) Every power of two where the power is a positive integer will terminate at the number one and never diverge into the infinite when it follows the operations of the Collatz conjecture.

Conclusion
Intuitively, according to the summary, every positive integer greater than one will reach the number one under the turbulent operations of the Collatz conjecture before it enters an eternal loop of digits and without diverging into the infinite.
Sources
Stewart, Ian. Professor Stewart’s Hoard of Mathematical Treasures. London: PROFILE BOOKS LTD. 2009
Alex Lopez-Ortiz. Collatz Problem. University of Waterloo. 28 Jan 2011 http://www.cs.uwaterloo.ca/~alopez-o/math-faq/node55.html
Collatz Problem. Wolfram Mathworld. 28 Jan. 2011 http://mathworld.wolfram.com/CollatzProblem.html

Sequences and Series of Factorials

Factorial Sequences and Series
By: Steven Tao Zheng
23/03/2011
Print Notes: Sorry for the bad layout of equations, the original equations editor tool on Microsoft Word did not show when I pasted my document. If you want a copy with better quality, email me at eulerians@hotmail.com.
Legend
_a^b▒ = from limits a to b (of an integral or sum)
〖〗= type of brackets
lim┬(n→∞)⁡= the limit as n approaches infinity
t_n = t subscript n

Abstract
Arithmetic sequences and series deal with the growth and decay of patterns that have a common difference. Geometric sequences and series deal with the growth and decay of patterns that have a common rate. This introduction to factorial sequences and series is a definitive collection of theories and theorems that deal with a structured but unstable growth through the implementation of analytical reasoning. The purpose of this study is to derive a simple formula using analysis to build a standing theory on factorial sequences and series.

Introduction
Consider the following problem: 1! +2! +3! +⋯+N!
To this date, this simple yet elusive combinatorial problem has yet provided a simple formula for evaluation (a formula without the usage of special non-elementary functions).

Theory
Axioms
1)The value of n and its factorial are natural numbers.
2)For factorials n≤6, the order of magnitude might be shared by more than one value of n.
3)The order of magnitude for factorials where n is greater than six are distinct, meaning they are different exponentiations of ten.
4)For every value of n, there is only one value for n!.

1.0 Sequence of Factorials
Mathematical sequences are sets of numbers whose behaviour follows a set pattern. When dealing with factorial sequences, only the natural numbers play a role, making the entire sequence discrete. [Axiom 1]
Let t_n describe a factorial sequence.
t_n t_n=k f (t_n) =c (t_n)
t_1 1 f (t_1) =c
t_2 2 f (t_2) =2c
t_3 6 f (t_3) =6c
t_n n! f (t_n )=n!c

1.1 Finding the nth term of a factorial sequence
Suppose a problem involving finding the nth term of a factorial sequence looked like this:
(f (t_n))/ (f (t_3)) =20
f (t_n )=20*3!c
f (t_n )=5!c
n!c=5!c
n!=5!
n=5
Small factorials can be easily determined.

However for a problem like this one:
(f (t_n))/ (f (t_7)) =2.413593262×〖10〗^13
(n=19)
The problem presented above is difficult to solve mentally.

1.2 Finding the quotient of two terms of sequence
(f(t_18))/(f(t_5))= 18!c/5!c=5.335311421×〖10〗^13
Conceptually, the nth terms’ are identical to the value of n and the constants cancel.

1.3 Stirling’s Approximation and Augmentations
Stirling’s Approximation states lim┬(n→∞)⁡〖((√2π)*n^(n+1) *e^(-n))/n!〗 = 1

Therefore n!≈ (√2π)*n^(n+1)* e^(-n)

Despite the Stirling’s Approximations asymptotic approach, the accuracy is not precise. However, the accuracy increases as the value of n increases.

1.4 Analytical Refinement of the Sterling’s Approximation
From the famous Stirling’s approximation, we can deduce that:
(√2π)*n^(n+0.5)* e^(-n)*√2π)*n^(n+1) e^(-n)
However, we want an approximation that is precise to at least three decimals for all values greater than n=14. Since the Stirling’s Approximation increases in precision as n increases, so does any proper augmentation.

(√2π)* n^(n+0.5)* e^(-n)*k≈n! would lead to a closer approximation where 1
The value of k should asymptotically approach 1 as n increases as well as maintaining at least three decimals of precision for n factorial beyond n=14.

Let k= e^c
If lim┬(c→∞)⁡〖e^c 〗=1
Then c=1/(m(n)+q) , where m and q are constants.

Then e^(1/12n) is close to the precision range

Verification
5!=120
5!≈(√2π)*5^(5+0.5) *e^(-5) *e^(1/5m)
5!≈118.019168e^(1/5m)
m=11 5! = 120.2584396 δ=+0.2584396
m=12 5! = 120.0026371 δ=+0.0026371
m=13 5! = 119.8488862 δ= -0.1511138

12 is a good approximate for the coefficient m.

e^(1/12n)
5!≈(√2π)*5^(5+0.5) *e^(-5)* e^(1/(60+b))
0
δ=n!-[(√2π)* n^(n+0.5) *e^(-5) *e^(1/(12n+b))]
b=0.1 5! = 119.9993093 δ= -6.90685×〖10〗^(-4)
b=0.5 5! = 119.9861089 δ= -0.0138911
b=0.9 5! = 119.9730835 δ= -0.0269165

20!≈(√2π)〖20〗^(20+0.5) e^(-20) e^(1/(240+b))
0
20!=2.432902008×〖10〗^18
b=0.1 20! = 2.432898630×〖10〗^18 δ=+3.378×〖10〗^12
b=0.5 20! = 2.432881777×〖10〗^18 δ=+2.0231×〖10〗^13

Therefore e^c≈e^(1⁄((12n+0.1) ))

Refined Stirling’s Approximation
n!≈(√2π)*n^(n+0.5) *e^(-n) *e^(1/(12n+0.1))

1.5 Solving sequences with increased precision
n!≈(√2π)n^(n+0.5) e^(-n) e^(1/(12n+0.1))
(f(t_n))/(f(t_p))=c Solve for f(t_n ) when f(t_p ) is known.
f(t_n )=c*f(t_p)
Let k=c*f(t_n)/(√2π)
(f(t_n))/(f(t_p))=c Solve for f(t_p ) when f(t_n ) is known.
f(t_p )=1/c*f(t_n)

Let k=f(t_n)/(c*√2π)
Let f(t_n )≈〖(n/e)〗^n since the value of e^(1⁄(12n+0.1 ))decreases as n increases, making it ultimately negligible.
ln⁡〖〖(n/e)〗^n 〗=ln⁡k
n[ln⁡(n)-ln⁡(e)]=ln⁡k
n[ln(n)-1]=ln⁡ k

n[ln⁡(n)-1]< ln⁡ k<(n+1)(ln⁡(n+1)-1)
Therefore (n+0.5)((ln (n+0.5))-1)≈ln⁡ k
Using Newton’s Method, the value of n can be solved.
f(n)=(n+0.5)((ln⁡(n+0.5))-1)-ln⁡ k
f^' (n)=ln⁡(n+0.5)
This class of function may implement the zero as an initial estimate.
x_1=0-f(0)/(f^' (0) )
x_2=x_1-f(x_1 )/(f^' (x_1 ) )
x_3=x_2-f(x_2 )/(f^' (x_2 ) )
x_n=x_(n-1)-f(x_(n-1) )/(f^' (x_(n-1) ) )

1.6 The Concept of One to One Correspondence for above methods
The methods above for approximating the values of factorials and solving sequences have one common property: one to one correspondence. For every value n, there is only one distinct value of n factorial. The refined approximations are designed to be precise to the point the one to one correspondence is clear. [Axiom 4]

Factorial Series
Factorial series are sums of consecutive factorials as dictated by the factorial sequence. This formula is derived using Gamma functions.
Equation 1
n!+(n+1)!+(n+2)!+(n+3)!+⋯+(n+p)!=∑_(k=n)^(n+p)▒k!
Equation 2
∑_(k=n)^(n+p)▒k!=Г(n+1)+Г(n+2)+Г(n+3)+Г(n+4)+⋯+Г(n+p+1)
Equation 3
Г(n+1)+Г(n+2)+Г(n+3)+Г(n+4)+⋯+Г(n+p+1)=∫_0^∞▒〖e^(-t) t^n 〗 dt+∫_0^∞▒〖e^(-t) t^(n+1) 〗 dt+∫_0^∞▒〖e^(-t) t^(n+2) 〗 dt+∫_0^∞▒〖e^(-t) t^(n+3) 〗 dt+⋯+∫_0^∞▒〖e^(-t) t^(n+p) 〗 dt
Equation 4
∫_0^∞▒〖〖(e〗^(-t) t^n+e^(-t) t^(n+1) 〗+e^(-t) t^(n+2)+e^(-t) t^(n+3)+⋯+e^(-t) t^(n+p))dt=∑_(k=n)^(n+p)▒k!
Equation 5
∫_0^∞▒〖〖e^(-t) (t〗^n+t^(n+1) 〗+t^(n+2)+t^(n+3)+⋯+t^(n+p))dt=∑_(k=n)^(n+p)▒k!
Equation 6
∫_0^∞▒〖e^(-t) t^n ((t^(p+1)-1)/(t-1))〗 dt=∑_(k=n)^(n+p)▒k!
Equation 7 (Final)
∫_0^∞▒(t^(n+p+1)-t^n)/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒k!
Consequently,
∫_0^∞▒(t^n-t^(n+p+1))/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒〖-(k!)〗

Factorial Series Theorem or Factorial Sum Function
c∫_0^∞▒(t^(n+p+1)-t^n)/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒〖c(k!)〗
The variables of change in the summation of consecutive factorials are c, n and p, where c is a constant, n is the initial term and p is the difference of terms from n.

2.1 Solving Factorial Series
Finding the value of a factorial series where the values c, n and p are known.
∑_(k=n)^(n+p)▒〖c(k!)〗=Q
If the numbers are small for n and p, then manual calculations can be easily executed. For larger numbers, the values of c, n and p should be directly substituted to the Factorial Series Theorem and possibly using numerical methods of integration to compute its true value.
Solving for c when Q and ∑_(k=n)^(n+p)▒k! Are known.
∑_(k=n)^(n+p)▒〖c(k!)〗=Q
c∑_(k=n)^(n+p)▒k!=Q
c=Q/(∑_(k=n)^(n+p)▒k!)
When consecutive factorials are summed, the largest term in the series is the easiest to solve for and the smallest term is the hardest to solve for because the order of magnitude for the largest term is responsible for most of the series’ make-up, whereas the smaller terms may be negligible.
Solving for n+p when Q and c are known
∑_(k=n)^(n+p)▒c(k!) =cQ
∑_(k=n)^(n+p)▒k!=Q
The order of magnitude of Q should be closest to if not identical with (n+p)!. Therefore, we can solve for n+p, with Newton’s method introduced in section 1.5.

2.2 Factorial Difference Series
Factorial difference series are differences of consecutive factorials as dictated by the factorial sequence. This type of problem is considered a combinatorial decay problem.
Equation 1
n!-(n+1)!-(n+2)!-(n+3)!-…-(n+p)!=n!-∑_(k=n)^(n+p-1)▒〖(k+1)!〗
Equation 2
n!-∑_(k=n)^(n+p)▒(k+1)!=Г(n+1)-Г(n+2)-Г(n+3)-Г(n+4)-…-Г(n+p+1)
Equation 3
∫_0^∞▒〖〖e^(-t) (t〗^n-t^(n+1) 〗-t^(n+2)-t^(n+3)-…-t^(n+p))dt=n!-∑_(k=n)^(n+p-1)▒(k+1)!
a(1-r^2-r^3-…-r^n )=a(1-(r+r^2+r^3+⋯+r^n-1))
a(1-r^2-r^3-…-r^n )=a(1-((r^n-1)/(r-1)-1)
Therefore,a(1-r^2-r^3-…-r^n )=a(2-(r^n-1)/(r-1))
Equation 4 (Final)
∫_0^∞▒〖e^(-t) t^n (2-(t^(p+1)-1)/(t-1))〗 dt=n!-∑_(k=n)^(n+p-1)▒(k+1)!

Factorial Difference Series Theorem
c∫_0^∞▒〖e^(-t) t^n (2-(t^(p+1)-1)/(t-1))〗 dt=c(n!-∑_(k=n)^(n+p)▒(k+1)!)
Where c is a constant.

2.3 Additional Theorems
1.
n!-(n-1)!-(n-2)!-(n-3)!-…-(n-p)!=n!-∑_(k=n)^(n-p+1)▒〖(k-1)!〗

c∫_0^∞▒〖e^(-t) t^n (2-(〖(1/t)〗^(p+1)-1)/((1/t)-1))〗 dt=c(n!-∑_(k=n)^(n-p+1)▒(k-1)!)
This theorem is based on the fact:
a(r^n-r^(n-1)-r^(n-2)-…-r^1 )=a(1-((1/r)+(1/r)^2+(1/r)^3+⋯+(1/r)^n-1))
Therefore,a(r^n-r^(n-1)-r^(n-2)-…-r^1 )=a(2-((1/r)^n-1)/((1/r)-1))

3.1Arithmetic Method for Evaluating Large Factorial Sums Part 1
Integration of the Factorial Sum Function
Although the formulas above are simple compared to the formulas of their predecessors, they are still considered implicit, simply because the formulas are expressed as improper integrals. The validity and precision of these formulas are total, but the formulas fail to explicitly give numerical results.
∫_0^∞▒(t^(n+p+1)-t^n)/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒k!
This expression can be written as
∫_0^∞▒〖e^(-t) t^n (1+t+t^2+t^3+⋯+t^p)〗 dt=∑_(k=n)^(n+p)▒k!
Using integration by parts:
Let u=t^n+t^(n+1)+t^(n+2)+⋯+t^(n+p)
du=[nt^(n-1)+(n+1) t^n+(n+2) t^(n+1)+⋯+(n+p) t^(n+p) ]dt
dv=e^(-t) dt
v=-e^(-t)
Suppose the objective is to find the indefinite integral of this expression, the employment of successive integration by parts will be required.
∫▒〖udv=uv-∫▒vdu〗
∫▒〖e^(-t) t^n (1+t+t^2+t^3+⋯+t^p)dt〗= -e^(-t) (t^n+t^(n+1)+t^(n+2)+⋯+t^(n+p) )+ ∫▒[nt^(n-1)+(n+1) t^n+(n+2) t^(n+1)+⋯+(n+p) t^(n+p) ]dt
This method terminates by the p+1 derivative. Hence in the end, the integral becomes:
∫▒〖e^(-t) t^n (1+t+t^2+t^3+⋯+t^p )dt〗= -e^(-t) (u+u^'+u^''+⋯+u^(n+p+1)')
Arithmetically speaking, the value of,
∑_(k=n)^(n+p)▒k!=-e^(-t) (∑_(x=0)^(x=n+p+1)▒〖u^((x)') 〗)
Example
∫_0^∞▒(t^5-t^2)/(e^t (t-1)) dt=2!+3!+4!=32
∫▒〖(t^5-t^2)/(e^t (t-1)) dt= -e^(-t) (〗 t^4+5t^3+16t^2+32t+32)
This implies that the value of ∫_0^∞▒(t^5-t^2)/(e^t (t-1)) dt is equal to the absolute value of the y-intercept of ∫▒〖(t^5-t^2)/(e^t (t-1)) dt〗.

3.2 Arithmetic Method for Evaluating Large Factorial Sums Part 2
The Ascent Base Notation Part 2(An arithmetic convention)
Base Two
In binary (base two), the number 1101 is equal to 13 in base ten.
1101=1(2^3 )+1(2^2 )+0(2^1 )+1(2^0 )=13
There exists a shortcut to the evaluation of base two.
(((((1*2)+1)*2)+0)*2)+1=13
Ascent Base
Notation
1_(n+p)…1_(n+2) 1_(n+1) 0_n 0_(n-1)…0_etc
Examples
5!+4!+3!+2!+1!=1_5 1_4 1_3 1_2 1_1=153
(((((((((1*5)+1)*4)+1)*3)+1)*2)+1)*1)+1=153
6!+5!+4!=1_6 1_5 0_4 0_3 0_2 0_1=864
((((1*6)+1)*5)+1)*4!=864

Significance
By transforming the sum of consecutive factorials into the consecutive sums of Gamma Functions, it is evident that there exists a link between factorials and geometric progression.
Sources
Wrede, Robert C. and Spiegel Murray. Schaum’s Outlines Advanced Calculus. McGraw-Hill Companies, Inc. Second Edition, 2002.
Credible Online Sources
Factorial Sums. Wolfram Mathworld. 24 Feb. 2011 http://mathworld.wolfram.com/FactorialSums.html