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Infographic created using NotebookLM |
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Table 1: permalink |
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Infographic created using NotebookLM |
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Table 1: permalink |
number factorisation divisors1 1 12 2 24 2^2 36 2 * 3 444 2^2 * 11 666 2 * 3 * 11 8252 2^2 * 3^2 * 7 182112 2^6 * 3 * 11 282772 2^2 * 3^2 * 7 * 11 366336 2^6 * 3^2 * 11 4227972 2^2 * 3^3 * 7 * 37 4848384 2^8 * 3^3 * 7 72
count n multiple1 1 12 2 13 3 14 4 15 5 16 6 17 7 18 8 19 9 110 0 011 11 112 252 2113 494 3814 252 1815 525 3516 272 1717 272 1618 252 1419 171 920 0 021 252 1222 22 123 161 724 696 2925 525 2126 494 1927 999 3728 252 929 232 830 0 031 434 1432 2112 6633 33 134 272 835 525 1536 252 737 111 338 494 1339 585 1540 0 041 656 1642 252 643 989 2344 44 145 585 1346 414 947 141 348 2112 4449 343 750 0 051 969 1952 676 1353 212 454 27972 51855 55 156 616 1157 171 358 232 459 767 1360 0 061 26962 442
27972 requires 13 steps to reach the palindrome 4964444694 under the Reverse and Add algorithm.
27972 is a decagonal or ten-sided number and is the 84th decagonal number and the second non-trivial palindromic decagonal number after 232. See Figure 1 where 232 is shown but not 27972.
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Figure 1: source |
27972 is a member of OEIS A356854: palindromes that can be written in more than one way as the sum of two distinct palindromic primes. In the case of 27972 we have:
10501 + 17471 = 2797211311 + 16661 = 2797211411 + 16561 = 2797212421 + 15551 = 27972
It's been 560 days since my last \( \textbf{weird} \) day. Today I turned \( \textbf{27790} \) days old and once again it's weird. I've written about this type of number before in blog posts titled Weird Numbers on August 4th 2019 and also in an earlier post titled Zumkellar Numbers, Half Zumkellar Numbers and Pseudoperfect Numbers on November 22nd 2018.
Most abundant numbers are pseudoperfect meaning that a subset of their proper divisors can be chosen so that their sum is equal to the number. If the sum of all the proper divisors of a number equals the number itself, then the number is said to be perfect. The first few perfect numbers are 6, 28, 496, 8128 and 33550336. An example of an abundant number that is pseudoperfect is 24 with proper divisors of 1, 2, 3, 4, 6, 8 and 12. Adding these gives 36 so the number is clearly abundant. However, if we add the proper divisors without including 12, we reach 24 and so the number is pseudoperfect.
There are two types of weird numbers: primitive and non-primitive. A non-primitive weird number is a number that is a multiple of a weird number. The primitive weird numbers up to 40000 are 70, 836, 4030, 5830, 7192, 7912, 9272, 10792 and 17272. Here is a list of all weird numbers (primitive and non-primitive) up to 40000:
70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110, 12530, 12670, 13370, 13510, 13790, 13930, 14770, 15610, 15890, 16030, 16310, 16730, 16870, 17272, 17570, 17990, 18410, 18830, 18970, 19390, 19670, 19810, 20510, 21490, 21770, 21910, 22190, 23170, 23590, 24290, 24430, 24710, 25130, 25690, 26110, 26530, 26810, 27230, 27790, 28070, 28630, 29330, 29470, 30170, 30310, 30730, 31010, 31430, 31990, 32270, 32410, 32690, 33530, 34090, 34370, 34930, 35210, 35630, 36470, 36610, 37870, 38290, 38990, 39410, 39830, 39970
All the numbers above with the exception of the primitive weird numbers in red are multiples of 70. Any weird number multiplied by a prime number that is greater than the sum of the divisors of the numbers is itself weird. Thus 70 has a sum of divisors of 144 and the first prime above this is 149. 70 x 149 produces the weird number 10430.
As can be seen, my next weird number (28070) is 280 days away and after that there is a gap of 560 days to my next weird number (28630). The gaps of course are determined by the distance between successive prime numbers. Thus:$$ \begin{align} 27790 &=70 \times 397\\28070 &=70 \times 401 \\28630 &=70 \times 409 \end{align} $$Here is a fuller list of primitive weird numbers (link):
70, 836, 4030, 5830, 7192, 7912, 9272, 10792, 17272, 45356, 73616, 83312, 91388, 113072, 243892, 254012, 338572, 343876, 388076, 519712, 539744, 555616, 682592, 786208, 1188256, 1229152, 1713592, 1901728, 2081824, 2189024, 3963968, 4128448
While this post is a little repetitive, I feel it's important to acknowledge the occurrence of weird numbers when they occur because of their rarity.
Being born on the 3rd April 1949, my date of birth is often represented as 3 - 4 - 49. These numbers when concatenated form the prime number 3449. I was reminded of this number because of the factorisation of the number associated with my diurnal age today, 27592.$$27592 = 2^3 \times 3449 = 8 \times 3449$$So today my life can be divided into exactly eight equal parts, each of them 3449 days long which is about 9.44 years. The previous multiple$$7 \times 3449 = 24143$$occurred on May 10th 2015 when I was still working at the Shanghai Singapore International School. The next multiple$$9 \times 3449 = 31041$$ will fall on March 29th 2034, shortly before my 84th birthday (if I make it that far).
3449 forms the initial prime of a Cunningham chain of the first type with length exactly 3 and so: $$ 3449 \rightarrow 2 \times 3449 + 1 = 6889 \text{ (prime)} \\ 6889 \rightarrow 2 \times 6889 +1 = 13799 \text{ (prime)}$$Primes with this property form OEIS A059762. Another prime-related property of 3449 qualifies it for membership in OEIS A088483:
For 3449, the twin primes are \(11899049\) and \(11899051\).
3449 is also a home prime with a homeliness of 3 because:$$ \begin{align} 611 &= 13 \times 47 \rightarrow 1347\\ 1347 &= 3 \times 449 \rightarrow 3449 \end{align}$$Not all primes are home primes of course. Take 613 as an example of a prime that is not a home prime because it cannot be formed by the concatenation of the prime factors of any number (the prime factors need to be concatenated in ascending order).
3449 is also a member of OEIS A153116:
Here the two primes are \(11895589\) and \(11895613\). Additionally:$$ \text{period of}\frac{1}{3449}=\frac{3449-1}{8} = 431$$This property qualifies 3449 for membership of OEIS A056213:
3449 is a Sophie Germain prime because:$$2 \times 3449+1=6899 \text{ is prime}$$3449 also features in so-called "Golden Semiprimes" and this qualifies it for membership in OEIS A108544:
Consider the Diophantine equation:$$a^4+b^4+c^4+d^4=(a+b+c+d)^4$$It's clear that no set of positive numbers will satisfy this equation, at least one of the numbers will need to be negative. It's also clear that if we find a solution \(n=a+b+c+d\) then this will generate an infinity of solutions between all multiples of this number will satisfy the equation:$$ \begin{align} (na)^4+(nb)^4+(nc)^4+(nd)^4 &= (n *(a+b+c+d))^4\\n^4 a^4 +n^4 b^4 + n^4 c^4 + n^4 d^4 &= n^4 (a+b+c+d)^4 \\a^4+b^4+c^4+d^4 &= (a+b+c+d)^4 \end{align} $$where \(n\) can be any integer, positive or negative, or even zero. So what is the first number that satisfies the equation? In 1964, somebody named Brudno found the number 5491 that satisfies where 5491 is written as 955 + 1700 - 2364 + 5400 and we have:$$955^4+1700^4+(-2364)^4+5400^4\\=(955 + 1700 - 2364 + 5400)^4$$In terms of positive numbers, this means that the progressive multiples of 5491 are also solutions viz. 10982, 16473, 21964, 27455, 32946, 38437, 43928, 49419, ... etc..
Are there other numbers? Well somebody named Wroblewski found 51361 to be a solution when it written as 48150 - 31764 + 27385 + 7590. Thus:$$48150^4 +(-31764)^4+27385^$+7590^4\\=(48150 - 31764 + 27385 + 7590)^4$$At this point, it's relevant to include this quote from the comments to OEIS A138760: numbers \(n\) such that \(n^4\) is a sum of 4th powers of four nonzero integers whose sum is \(n\):
Any multiple of a member is also a member. A member that is not a multiple of another member is called primitive. Using elliptic curves, Jacobi and Madden prove that there are infinitely many primitive members. According to them, the only primitive members less than 222,000 are 5491 (due to Brudno) and 51361 (due to Wroblewski).
The comments then list a number greater than 220,000 and that number is 1347505009. The reason I came across these numbers is that today my diurnal age is 27455 and that is the fifth multiple of 5491.
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Figure 1 |
2023 = (2+0+2+3)(22+02+22+32)2
It also turns out that the prime factors, when concatenated, form a palindrome.
A063416 | Multiples of 7 whose sum of digits is equal to 7. |
| A070258 | Smallest of 3 consecutive numbers each divisible by a square. |
I'm impressed in a negative way. Except for points 2 and 6, every mathematical statement that you made was wrong. 2023 is not a prime number and so on. How can you get things so wrong?
I apologize for any errors in my previous response. As a language model, I can provide information and answer questions to the best of my ability based on the knowledge and information that I have been trained on. However, I am not infallible and it is possible for me to make mistakes or provide incorrect information.In fact, 2023 is not a prime number. It can be divided evenly by 3, 11, and 61, so it is a composite number. I apologize for any confusion this may have caused.I hope this clears up any misunderstandings and I apologize for any errors in my previous response. Please feel free to ask me any other questions you may have.