20220106, 12:06  #12  
Aug 2020
Brasil
2^{5} Posts 
Quote:
a. Let´s consider only the prime numbers. Then, we have: 1. A100995(prime) = 1 2. A032741(prime) = 1. 3. A014963(prime) = prime Then, A100994(prime) = A014963(prime) ^ A100995(prime) = A014963(prime) ^ A032741(prime) = prime, or A100994(prime) = prime^1 = prime^1. b. Now, let´s consider only the composites that are prime power (prime^m) numbers. Then, we have: 1. A100995(prime^m) = m 2. A032741(prime^m) = m = number of proper divisors of prime^m. 3. A014963(prime^m) = prime Then, A100994(prime^m) = A014963(prime^m) ^ A100995(prime^m) = A014963(prime^m) ^ A032741(prime^m) = prime^m, or A100994(prime^m) = prime^m = prime^m. c. Finally, let´s consider only the composites that are not prime power (not prime^m) numbers. Then, we have: 1. A100995(not prime^m) = 0 2. A032741(not prime^m) = number of proper divisors of (not prime^m) = d(not prime^m) – 1. 3. A014963(not prime^m) = 1 Then, A100994(not prime^m) = A014963(not prime^m) ^ A100995(not prime^m) = A014963(not prime^m) ^ A032741(not prime^m), or A100994(not prime^m) = 1^0 = 1^(d(not prime^m) – 1) = 1. So, we can say for any integer: A100994 = A014963^A100995 = A014963^A032741. Now, when we consider the OEIS offset, I think we should write in OEIS: A100994(n) = A014963(n)^A100995(n) = A014963(n)^A032741(n+1).  P.S.: In my opinion, it is controversial to define a(0) = 0 for A032741 or to have definition a(0)=1 for A325765. This is because the number zero has uncountable negative and positive divisors. 

20220106, 12:15  #13  
Aug 2020
Brasil
40_{8} Posts 
Quote:
The complete and more general equation in https://oeis.org/A056737 would be n=k(k±m). When the equation allows ±m, then it is correct “KEYWORD nonn” for a(n)=m. The correct example would be: a(8)=2 because 8=2(2+2), and because 8=4(42). We could complete the example with a square and a prime: a(9)=0 because 9=3(3+0)=3(30). a(10)=3 because 10=2(2+3)=5(53). a(11)=10 because 11=1(1+10)=11(1110). a(12)=1 because 12=3(3+1)=4(41). And so on… For each a(n)=m>0 always exist two different values for k resulting in the same n. For each a(n)=m=0 always exist one value for k resulting in the same n. This reflect that every quadratic equation always has two roots. The roots are equal only on square numbers. So, we should also take out the words "nonnegative" and "positive" from the name. We should also take out the word "integer” because we are only dealing with sums of integers. The following title would be more suit: "Minimum m such that n=k(k±m)". This reflects exactly the difference between the 2 divisors of the central pair of complementary divisors, as per a(n)=A056737(n)=A033677(n)A033676(n).  Omar E. Pol, Jun 21, 2009. The general formulae are a(n)=A056737(n)=A033677(n)A033676(n). Because, n=A033676(n)*A033677(n). Then, n=A033676(n)*(A033676(n)+A056737(n)) n=A033677(n)*(A033677(n)A056737(n)). Or n=A033676(n)*(A033676(n)+a(n))
n=A033677(n)*(A033677(n)a(n)). Last fiddled with by Charles Kusniec on 20220106 at 12:46 

20220109, 11:40  #14  
Aug 2020
Brasil
2^{5} Posts 
Understanding A056737 and the quadratic sequences: not so "quibbles".
Quote:
Apart from egos blinding us, thinking strictly logically, and hoping someone will return back the original word "questions", we can already discover some unpublished mathematical reasonings from what we have in this thread. Let us start with the first reasoning. When we say that A056737(n)=A033677(n)A033676(n), this means:
As a direct consequence we may expand with this recurring reasoning/algorithm:
And so on… Generalizing: 1. We can express integers in the form of \(n=x(x+even)\) substituting \(x=y(even/2)\) resulting in \(n=(yeven/2)(y+even/2)\) which equals \(n=y^2(even/2)^2=square sequencesquare number\). Once \(x\) can be either A033676(n) or A033677(n), then let us call the quadratic sequence of the numbers in the form of \(x(x±b_m)\) for \(b_m=A056737(n)=even\) as the “square sequence A000290 minus a square number \((even/2)^2\)”. 2. We can express integers in the form of \(n=x(x+odd)\) substituting \(x=y(odd1)/2\) resulting in \(n=(y(odd1)/2)(y+(odd+1)/2)\) which equals \(n=y^2+y((odd1)/2)((odd+1)/2)=oblong sequenceoblong number\). Once \(x\) can be either A033676(n) or A033677(n), then let us call the quadratic sequence of the numbers in the form of \(x(x±b_m)\) for \(b_m=A056737(n)=(odd1)/2\) as the “oblong sequence A002378 minus an oblong number \(((odd1)/2)((odd+1)/2)\)”. The square and oblong numbers that are subtracted from their respective square and oblong sequences form a new sequence given by https://oeis.org/A002620. See below the table of equivalence of the first 32 sequences in OEIS. We have a column with the hyperbolic equations and two columns with two equivalent quadratic equations, where one has offset not zero and the other offset zero. The equations in each row always generate the same sequence. Changes only the offset: A fast explanation about this table: When we express the quadratic sequences in the form of \(n=x(x±b_m)\), then we have an offset \(f\) given by the vertex (inflection point) \(ip=∓b_m/2\). When we express the same quadratic sequence in the form of (square sequence minus square number) or (oblong sequence minus oblong number) the offset is zero. Offset zero for \(b_m=even\) we have \(ip=0\) and for \(b_m=odd\) \(ip=±0.5\). In any way, they generate the very same sequence of integer numbers. Now one (from many others) Because we can cover all integers with the quadratic sequences in the form of \(x(x±b_m )\), and since we have an infinitude of infinite quadratic sequences versus just one infinite number line with all integers, then how do we get the bijection between them? 

20220109, 16:51  #15 
Mar 2019
11011010_{2} Posts 
I am a Knower of 4 corner
simultaneous 24 hour Days that occur within a single 4 corner rotation of Earth. 
20220109, 17:54  #16  
Aug 2020
Brasil
2^{5} Posts 
Quote:
I am not a native English speaker and unfortunately I have no idea how what you wrote relates to the bijection question between the number line and the quadratic sequences in the form of \(x(x+b_m)\). 

20220110, 02:55  #17 
"Curtis"
Feb 2005
Riverside, CA
5,153 Posts 
Since we pointed out that nobody cares enough to give replies about your quibbles, you should expect any replies to be off topic. I, for one, appreciate the poetry.

20220110, 10:03  #18  
Aug 2020
Brasil
100000_{2} Posts 
Quote:
Moderator note: Readers should carefully consider the fact that, in sending an Email, you are giving the recipient your Email address. Last fiddled with by Dr Sardonicus on 20220110 at 20:13 Reason: As indicated 

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