mersenneforum.org General quibbles about some OEIS sequences
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2022-01-06, 12:06   #12
Charles Kusniec

Aug 2020
Brasil

25 Posts

Quote:
 Originally Posted by Charles Kusniec Just to ask if you agree it is more intuitive also to define A100994(n) = A014963(n)^A325765(n) or A100994(n) = A014963(n)^A032741(n)?
To clarify this question:

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.

2022-01-06, 12:15   #13
Charles Kusniec

Aug 2020
Brasil

3210 Posts

Quote:
 Originally Posted by Charles Kusniec I will start with the sequence A056737. In my opinion, the complete and more general equation would be n = k*(k+-m). I would also take out the words "nonnegative" and "positive". To me, the following title would be more comprehensive: "Minimum integer m such that n = k*(k+-m) for integer k." or even just "Minimum m such that n = k*(k+-m)." I also think the example is half done. The complete one would be: a(8) = 2 because 8 = 2*(2+2) = 4*(4-2). This shows that every quadratic equation always has two roots. Do you agree?
To clarify this question:

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(4-2).
We could complete the example with a square and a prime:
a(9)=0 because 9=3(3+0)=3(3-0).
a(10)=3 because 10=2(2+3)=5(5-3).
a(11)=10 because 11=1(1+10)=11(11-10).
a(12)=1 because 12=3(3+1)=4(4-1).
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 2022-01-06 at 12:46

2022-01-09, 11:40   #14
Charles Kusniec

Aug 2020
Brasil

3210 Posts
Understanding A056737 and the quadratic sequences: not so "quibbles".

Quote:
 Originally Posted by Charles Kusniec To clarify this question: 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)).
As a victimmember of this forum, I feel bad because someone changed the name of this thread from general "questions" to general "quibbles".

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.

When we say that A056737(n)=A033677(n)-A033676(n), this means:
• All different integers n that have the same value of A056737(n), belong to the same quadratic sequence in the form of
n=A033676(n)*(A033676(n)+A056737(n))
or, in the form of
n=A033677(n)*(A033677(n)-A056737(n)).
• We can cover all integers with the above quadratic sequences.

As a direct consequence we may expand with this recurring reasoning/algorithm:
1. We can express integers in the form of n=x(x+0) substituting x=y-0 resulting in n=(y-0)(y+0) which equals n=y^2. These integers belong to the sequence https://oeis.org/A000290.
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±A056737) for A056737(n)=0 as the "square minus 0" sequence.

2. We can express integers in the form of n=x(x+1) substituting x=y-0 resulting in n=(y-0)(y-0+1) which equals n=y^2+y. These integers belong to the sequence https://oeis.org/A002378.
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±A056737) for A056737(n)=1 as the "oblong minus 0" sequence.

3. We can express integers in the form of n=x(x+2) substituting x=y-1 a resulting in s n=(y-1)(y+1) which equals n=y^2-1. These integers belong to the sequence https://oeis.org/A005563.
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±A056737) for A056737(n)=2 as the "square minus 1" sequence.

4. We can express integers in the form of n=x(x+3) substituting x=y-1 resulting in n=(y-1)(y+2) which equals n=y^2+y-2. These integers belong to the sequence https://oeis.org/A028552.
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±A056737) for A056737(n)=3 as the "oblong minus 2" sequence.

5. We can express integers in the form of n=x(x+4) substituting x=y-2 resulting in n=(y-2)(y+2) which equals n=y^2-4. These integers belong to the sequence https://oeis.org/A028347.
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±A056737) for A056737(n)=4 as the "square minus 4" sequence.

6. We can express integers in the form of n=x(x+5) substituting x=y-2 resulting in n=(y-2)(y+3) which equals n=y^2+y-6. These integers belong to the sequence https://oeis.org/A028557.
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±A056737) for A056737(n)=5 as the "oblong minus 6" sequence.

7. We can express integers in the form of n=x(x+6) substituting x=y-3 resulting in n=(y-3)(y+3) which equals n=y^2-9. These integers belong to the sequence https://oeis.org/A028560.
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±A056737) for A056737(n)=6 as the "square minus 9" sequence.

8. We can express integers in the form of n=x(x+7) substituting x=y-3 resulting in n=(y-3)(y+4) which equals n=y^2+y-12. These integers belong to the sequence https://oeis.org/A028563.
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±A056737) for A056737(n)=7 as the "oblong minus 12" sequence.

And so on…

Generalizing:

1. We can express integers in the form of $$n=x(x+even)$$ substituting $$x=y-(even/2)$$ resulting in $$n=(y-even/2)(y+even/2)$$ which equals $$n=y^2-(even/2)^2=square sequence-square number$$.
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 $$x=y-(odd-1)/2$$ resulting in $$n=(y-(odd-1)/2)(y+(odd+1)/2)$$ which equals $$n=y^2+y-((odd-1)/2)((odd+1)/2)=oblong sequence-oblong number$$.
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 $$b_m=A056737(n)=(odd-1)/2$$ as the “oblong sequence A002378 minus an oblong number $$((odd-1)/2)((odd+1)/2)$$”.

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:

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) “quibble”question for the next reasoning:
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?

 2022-01-09, 16:51 #15 mathwiz   Mar 2019 21810 Posts I am a Knower of 4 corner simultaneous 24 hour Days that occur within a single 4 corner rotation of Earth.
2022-01-09, 17:54   #16
Charles Kusniec

Aug 2020
Brasil

25 Posts

Quote:
 Originally Posted by mathwiz I am a Knower of 4 corner simultaneous 24 hour Days that occur within a single 4 corner rotation of Earth.
Dear mathwiz,

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 $$x(x+-b_m)$$.

 2022-01-10, 02:55 #17 VBCurtis     "Curtis" Feb 2005 Riverside, CA 515610 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.
2022-01-10, 10:03   #18
Charles Kusniec

Aug 2020
Brasil

25 Posts

Quote:
 Originally Posted by VBCurtis 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.
For those who appreciate discovering new things in mathematics, please contact me charleskusniec at iclould dot com. Regards,

Last fiddled with by Dr Sardonicus on 2022-01-10 at 20:13 Reason: As indicated

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