User:Fjackson/sandbox

Properties

The largest number that always divides the product abcd is 12. The quadruple with the minimal product is (1, 2, 2, 3).

Similar to a Pythagorean triple which generates a distinct right triangle, a Pythagorean quadruple will generate a distinct Heronian triangle . If a, b, c, d is a Pythagorean quadruple with ${\textstyle a^{2}+b^{2}+c^{2}=d^{2}}$ it will generate a Heronian triangle with sides x, y, z as follows:-

$x=d^{2}-a^{2}$

$y=d^{2}-b^{2}$

$z=d^{2}-c^{2}$.

It will have a semiperimeter ${\textstyle s=d^{2}}$, an area ${\textstyle A=abcd}$ and an inradius ${\textstyle r=abc/d}$.

The exradii will be:-

${\textstyle r_{x}=bcd/a}$

${\textstyle r_{y}=acd/b}$

${\textstyle r_{z}=abd/c}$.

The circumradius will be ${\textstyle R=(d^{2}-a^{2})(d^{2}-b^{2})(d^{2}-c^{2})/(4abcd)=abcd(1/a^{2}+1/b^{2}+1/c^{2}-1/d^{2})/4}$.

The ordered sequence of areas of this class of Heronian triangles can be found at (sequence A367737 in the OEIS).

Properties

• Any odd number of the form 2m+1, where m is an integer and m>1, can be the odd leg of a primitive Pythagorean triple [PPT]. See almost-isosceles PPT section below. However, only even numbers divisible by 4 can be the even leg of a PPT. This is because Euclid's formula for the even leg given above is 2mn and one of m or n must be even.
• The hypotenuse c is the sum of two squares. This requires all of its prime factors to be primes of the form 4n + 1. Therefore c is of the form 4n + 1. A sequence of possible hypotenuse numbers for a PPT can be found at (sequence A008846 in the OEIS).

Examples

As a consequence of the definition, 3 is an Ulam number (1+2); and 4 is an Ulam number (1+3). (Here 2+2 is not a second representation of 4, because the previous terms must be distinct.) The integer 5 is not an Ulam number, because 5 = 1 + 4 = 2 + 3. The first few terms are

1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, 38, 47, 48, 53, 57, 62, 69, 72, 77, 82, 87, 97, 99, 102, 106, 114, 126, 131, 138, 145, 148, 155, 175, 177, 180, 182, 189, 197, 206, 209, 219, 221, 236, 238, 241, 243, 253, 258, 260, 273, 282, ... (sequence A002858 in the OEIS).

There are infinitely many Ulam numbers. For, after the first n numbers in the sequence have already been determined, it is always possible to extend the sequence by one more element: U_{n − 1} + U_n is uniquely represented as a sum of two of the first n numbers, and there may be other smaller numbers that are also uniquely represented in this way, so the next element can be chosen as the smallest of these uniquely representable numbers.

Ulam is said to have conjectured that the numbers have zero density, but they seem to have a density of approximately 0.07398.

Properties

Apart from 1 + 2 = 3 any subsequent Ulam number cannot be the sum of its two prior consecutive Ulam numbers.

Proof: Assume that for n > 2, U_n−1 + U_n = U_{n + 1} is the required sum in only one way then so does U_n−2 + U_n produce a sum in only one way and it falls between U_n and U_{n + 1}. This contradicts the condition that U_{n + 1} is the next smallest Ulam number.

For n > 2, any three consecutive Ulam numbers (U_n−1, U_n, U_{n + 1}) as integer sides will form a triangle.

Proof: The previous property states that for n > 2, U_n−2 + U_n ≥ U_{n + 1}. Consequently U_n−1 + U_n > U_{n + 1} and because U_n−1 < U_n < U_{n + 1} the triangle inequality is satisfied.

The sequence of Ulam numbers forms a complete sequence.

Proof: By definition U_n = U_j + U_k where j < k < n and is the smallest integer that is the sum of two distinct smaller Ulam numbers in exactly one way. This means that for all U_n with n > 3, the greatest value that U_j can have is U_n−3 and the greatest value that U_k can have is U_n−1 .

Hence U_n ≤ U_n−1 + U_n−3 < 2U_n−1 and U₁ = 1, U₂ = 2, U₃ = 3. This is a sufficient condition for Ulam numbers to be a complete sequence.

For every integer n > 1 there is always at least one Ulam number U_j such that n ≤ U_j < 2n.

Proof: It has been proved that there are infinitely many Ulam numbers and they start at 1. Therefore for every integer n > 1 it is possible to find j such that U_j−1 ≤ n ≤ U_j. From the proof above for n > 3, U_j ≤ U_j−1 + U_j−3 < 2U_j−1. Therefore n ≤ U_j < 2U_j−1 ≤ 2n. Also for n = 2 and 3 the property is true by calculation.

In any sequence of 5 consecutive positive integers there can be a maximum of 2 Ulam numbers.

Proof: Assume that the sequence {i, i+1,..., i+4} has its first value i = U_j an Ulam number then it is possible that i+1 is the next Ulam number U_j+1. Now consider i+2, this cannot be the next Ulam number U_j+2 because it is not a unique sum of two previous terms. i+2 = U_j+1+U₁ = U_j+U₂ . A similar argument exists for i+3 and i+4.

Inequalities

Ulam numbers are pseudo-random and too irregular to have tight bounds. Nevertheless from the properties above, namely, at worst the next Ulam number U_n+1 ≤ U_n + U_n-2 and in any five consecutive positive integers at most two can be Ulam numbers, it can be stated that

5/2n-7 ≤ U_n ≤ N_n+1 for n > 0.

where N_n are the numbers in Narayana’s cows sequence: 1,1,1,2,3,4,6,9,13,19,... with the recurrence relation N_n = N_n-1 +N_n-3 that starts at N₀.

References

• Cassaigne, Julien; Finch, Steven R. (1995), "A class of 1-additive sequences and quadratic recurrences", Experimental Mathematics, 4 (1): 49–60, doi:10.1080/10586458.1995.10504307, MR 1359417
• Finch, Steven R. (1992), "On the regularity of certain 1-additive sequences", Journal of Combinatorial Theory, Series A, 60 (1): 123–130, doi:10.1016/0097-3165(92)90042-S, MR 1156652
• Guy, Richard (2004), Unsolved Problems in Number Theory (3rd ed.), Springer-Verlag, pp. 166–167, ISBN 0-387-20860-7
• Queneau, Raymond (1972), "Sur les suites s-additives", Journal of Combinatorial Theory, Series A (in French), 12 (1): 31–71, doi:10.1016/0097-3165(72)90083-0, MR 0302597
• Recaman, Bernardo (1973), "Questions on a sequence of Ulam", American Mathematical Monthly, 80 (8): 919–920, doi:10.2307/2319404, JSTOR 2319404, MR 1537172
• Schmerl, James; Spiegel, Eugene (1994), "The regularity of some 1-additive sequences", Journal of Combinatorial Theory, Series A, 66 (1): 172–175, doi:10.1016/0097-3165(94)90058-2, MR 1273299
• Ulam, Stanislaw (1964a), "Combinatorial analysis in infinite sets and some physical theories", SIAM Review, 6: 343–355, doi:10.1137/1006090, JSTOR 2027963, MR 0170832
• Ulam, Stanislaw (1964b), Problems in Modern Mathematics, New York: John Wiley & Sons, Inc, p. xi, MR 0280310
• Steinerberger, Stefan (2015), A Hidden Signal in the Ulam sequence, Experimental Mathematics, arXiv:1507.00267, Bibcode:2015arXiv150700267S

Almost-isosceles Pythagorean triples

No Pythagorean triples are isosceles, because the ratio of the hypotenuse to either other side is √2, but √2 cannot be expressed as the ratio of 2 integers.

There are, however, right-angled triangles with integral sides for which the lengths of the non-hypotenuse sides differ by one, such as,

$3^{2}+4^{2}=5^{2}$

$20^{2}+21^{2}=29^{2}$

and an infinite number of others. They can be completely parameterized as,

$({\tfrac {x-1}{2}})^{2}+({\tfrac {x+1}{2}})^{2}=y^{2}$

where {x, y} are the solutions to the Pell equation $x^{2}-2y^{2}=-1$.

If a, b, c are the sides of this type of primitive Pythagorean triple then the solution to the Pell equation above is given by the recursive formula

$a_{n}=6a_{n-1}-a_{n-2}+2$ with $a_{1}=3$ and $a_{2}=20$

$b_{n}=6b_{n-1}-b_{n-2}-3$ with $b_{1}=4$ and $b_{2}=21$

$c_{n}=6c_{n-1}-c_{n-2}$ with $c_{1}=5$ and $c_{2}=29$.

When it is the longer non-hypotenuse side and hypotenuse that differ by one, such as in

$5^{2}+12^{2}=13^{2}$

$7^{2}+24^{2}=25^{2}$

then the complete solution for the primitive Pythagorean triple a, b, c is

$a=2m+1,\quad b=2m^{2}+2m,\quad c=2m^{2}+2m+1$

and

$(2m+1)^{2}+(2m^{2}+2m)^{2}=(2m^{2}+2m+1)^{2}$

where integer $m>0$ is the generating parameter.

It also shows that all odd numbers (greater than 1) appear in a primitive Pythagorean triple.

Another property of this type of almost-isosceles primitive Pythagorean triple is that the sides are related such that

$a^{b}+b^{a}=Kc$

for some integer $K$. Or in other words $a^{b}+b^{a}$ is divisible by $c$ such as in

$(5^{12}+12^{5})/13=18799189$.

Mersenne numbers in nature and elsewhere

In computer science, unsigned n-bit integers can be used to express numbers up to M_n. Signed (n + 1)-bit integers can express values between −(M_n + 1) and M_n, using the two's complement representation.

In the mathematical problem Tower of Hanoi, solving a puzzle with an n-disc tower requires M_n steps, assuming no mistakes are made. The number of rice grains on the whole chessboard in the wheat and chessboard problem is M₆₄.

The asteroid with minor planet number 8191 is named 8191 Mersenne after Marin Mersenne, because 8191 is a Mersenne prime (3 Juno, 7 Iris, 31 Euphrosyne and 127 Johanna having been discovered and named during the 19th century).

In geometry, an integer right triangle that is primitive and has its even leg a power of 2 ( ≥ 4 ) generates a unique right triangle such that its inradius is always a Mersenne number. For example if the even leg is 2^n + 1 then because it is primitive it constrains the odd leg to be 4ⁿ − 1, the hypotenuse to be 4ⁿ + 1 and its inradius to be 2ⁿ − 1.

Furthermore, if a Mersenne number is used as the side of a primitive right triangle then M₀ and M₁ give the Mersenne numbers 0 and 1, neither of which can be the side of a primitive right triangle. For n > 1, all M_n are congruent to 3 mod 4. Consequently, no Mersenne number can be the hypotenuse of a primitive right triangle. This leaves the odd leg as the only candidate. However, unless the Mersenne number is prime, the right triangle is not unique. The number of primitive right triangles whose odd leg is M_n starting at n = 2 is 1, 1, 2, 1, 2, 1, 4, 2, 4, 2, 8, ... (sequence A297294 in the OEIS).

Occurrences and counts of integers within primitive Pythagorean triples

Not all integers can be the leg or hypotenuse of a primitive Pythagorean triple. For example only integers congruent to 1 modulo 4 and divisible only by primes congruent to 1 modulo 4 can be a hypotenuse. However 1 cannot be a hypotenuse because the legs of its right triangle would not be integral as they would have to be <1. The first integer that can be a hypotenuse is therefore 5.

Similarly, 1 or 2 cannot be a leg. 1 is odd and it has to be represented by the Euclidean parameter m^2-n^2=1 where m>n>0 and m, n have differing parities. This is not possible. Also 2 is even and has to be represented by the Euclidean parameter 2mn=2. Again this is not possible. The first integer that can be a leg is 3.

Finally neither a leg nor hypotenuse can be congruent to 2 modulo 4. This is because the hypotenuse is always 1 modulo 4 and the leg has to be represented by 2mn or m^2-n^2 neither of which can be 2 modulo 4.

Consequently, it is possible to define two functions L(s) and H(s) where s is an integer side length and L(s), H(s) are counts of primitive Pythagorean triples that contains s as either a leg - L(s) or a hypotenuse - H(s). Given the integer s and its prime factorisation L(s) and H(s) are derived as follows:-

Variant to the congruent number problem

General rational triangles

A generalisation to the congruent number problem is to determine whether triangles with rational sides (rational or Heron triangle) not restricted to right triangles have integer areas. It has been proved that every positive number is the area of some rational triangle.

This gives rise to the term t-congruent number. A t-congruent number is the integer area of a rational triangle that has one of its internal angles θ such that t = tan θ/2. Consequently the generalisation above can be re-stated as "Any square-free natural number n can be realised as a t-congruent number for some integer t.

Other special rational triangles

Congruent numbers stem from rational right triangles. So is there any other special rational triangle that can generate a variation on congruent numbers? Two cases have been studied.

Prime counting function - Inequalities

In his well-known notebooks, Ramanujan proves that the inequality

$\pi (x)^{2}<{\frac {ex}{\log x}}\pi {\bigg (}{\frac {x}{e}}{\bigg )}$

holds for all sufficiently large values of $x$.

Dudek and Platt have recently shown that this is true if $x\geq e^{9400}$. Moreover, they prove that on the assumption of the Riemann hypothesis, the largest integer counterexample to the above inequality is at $x=38,358,837,682$.

Adding Test Latex

This answer has been updated with a generalization of the identity $2\sum _{0}^{n}m+(n+1)=(n+1)^{2}$

By the binomial theorem

$(n+1)^{p}-n^{p}=1+\sum _{r=1}^{r=p-1}{\binom {p}{r}}n^{r}$

summing both sides over n starting at 1 gives

$(n+1)^{p}-1=n+\sum _{r=1}^{r=p-1}{\binom {p}{r}}\sum _{m=1}^{m=n}m^{r}$

Hence for $p=2,3,4,5,...$

$2\sum _{0}^{n}m+(n+1)=(n+1)^{2}$

$3\sum _{0}^{n}m^{2}+3\sum _{0}^{n}m+(n+1)=(n+1)^{3}$

$4\sum _{0}^{n}m^{3}+6\sum _{0}^{n}m^{2}+4\sum _{0}^{n}m+(n+1)=(n+1)^{4}$

$5\sum _{0}^{n}m^{4}+10\sum _{0}^{n}m^{3}+10\sum _{0}^{n}m^{2}+5\sum _{0}^{n}m+(n+1)=(n+1)^{5}$ etc.

This also provides an iterative process for determining the formulas of the power sums.

Collatz steps for Mersenne Primes above a-45

Above a-45, the ranking of Mersenne primes is provisional. So, the formal sequence of Mersenne primes is only verifiable up to a-45. Consequently, the formal sequence of Collatz steps for Mersenne primes is only verifiable up to s(a-45).

The comment section of OEIS A181777 has all 6 unranked steps for a-46 to a-51 and your value for a-48 is confirmed.

Apologies for a tardy reply but have been away visiting friends.

Practical number

$p_{i}\leq 1+\sigma (p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}}\dots p_{i-1}^{\alpha _{i-1}})=1+\sigma (p_{1}^{\alpha _{1}})\sigma (p_{2}^{\alpha _{2}})\dots \sigma (p_{i-1}^{\alpha _{i-1}})=1+\prod _{j=1}^{i-1}{\frac {p_{j}^{\alpha _{j}+1}-1}{p_{j}-1}},$