Anagrammatic sums

Éric Angelini asked about anagrammatic sums on MathFun on January 12: "Let a + b = c and a < b < c and a, b, c = anagrams of each other." Halfway down his sausage article, he lists the Gilles Esposito-Farèse calculation for 3- to 5-digit results: 1 @ 3-digit, 25 @ 4-digit, and 648 @ 5-digit. In that spirit, here are 17338 @ 6-digit results.

The idea for these has been around a few years. Claudio Meller's A160851 appears to be a (currently) somewhat misguided attempt at enumeration, while Rajesh Bhowmick's A203024, fleshed out by Charles Greathouse, provides a seemingly complete listing of sums, including 9449 6-digit terms. My 17338 6-digit results yield only 9443 distinct sums. Why six fewer?

Apparently 6-digit sums are the first that allow the sums to be twice one of the addends (i.e., a = b). In A023086 we see that there are twelve such. It turns out that six of these are the six that are not in my 17338 sums (because I did not allow a = b):

251748 = 2 * 125874
257148 = 2 * 128574
285174 = 2 * 142587
285714 = 2 * 142857
517482 = 2 * 258741
825174 = 2 * 412587

The other six are included because they had an alternate solution:

517428 = 2 * 258714 = 241587 + 275841

571428 = 2 * 285714 = 142857 + 428571

571482 = 2 * 285741 = 158724 + 412758

825714 = 2 * 412857 = 241587 + 584127

851742 = 2 * 425871 = 127584 + 724158

857142 = 2 * 428571 = 142857 + 714285 = 275418 + 581724 = 285714 + 571428

Mysterious lights in St. John's Cemetery on the Humber

Photo taken from the middle of Denison Park, looking south, the evening darkness fast approaching, two bright lights appear in St. John's Cemetery on the Humber and, after a short while, go out. I suppose one hundred years ago when the park and surrounding area were still all fields, one might have thought such an occurrence — especially at night — to be mysterious, especially if the lights reappeared in the same location at irregular intervals — but never on for very long.

Of course I've seen this phenomenon innumerable times over the last forty years, as have many of the local residents. I've never given it much thought because the explanation was always a tad obvious, all the more so if one was in the cemetery when the lights came on. I was reminded of it only because I came across an article by Clark Kim in the 30 October 2013 edition of the York Guardian:

"There was a story that Denison Cemetery was haunted," said [Cherri] Hurst [of the Weston Historical Society]. It was also known as St. John’s Cemetery on the Humber where members of the Denison family are buried. 

Lights would mysteriously be seen in the cemetery at night. It turned out those lights were turned on and off by whiskey runners trying to hide their stashes of alcohol.

"Weston was a dry town for a long time. It was a small town. You couldn’t get away with stuff," she said.

Note the emphasis on the lights being turned on and off. Behind the south end of the cemetery is an almost 200-metre stretch of much lower ground before it rises again to where West Park Healthcare Centre has its presence. I don't know what the road layout was one hundred years ago, but today there is a curved West Park Receiving road off Emmett Ave. that presents a gated "employee entrance" a little along. Beyond the gate, one arrives at a 50-metre stretch of straight road that then abruptly turns right. That straight stretch is what allows the lights to "turn on" and the sharp right, to "turn off". In the following Google map, a yellow arrow extends the straight stretch of road over the low ground, through the cemetery, through Denison Park, stopping at the park's north end (at Lippincott St. W.):

Inundation

A lot of rain here on Saturday had the "Raymore" island on the east side of the Humber (just above the weir; photo taken on Friday) ...

... somewhat inundated with river water on Sunday:

By this morning, the torrent had subsided:

The Fu Yao fruit bag mystery

Yesterday morning, walking Bodie up the north side of Clouston Ave., I saw a bag by a tree, across the street at the southwest corner of Clouston Ave. and Centre Rd. Thinking that it was trash, I crossed the road, determined to pick it up so as to deposit it in the trash bin at Denison Park on my way back. It wasn't trash but, rather, a plastic bag full of fruit. Nestled against the tree were three Canadian quarters. I left the bag of fruit but decided to take the money home. This morning, I had another look:

The bag was from Fu Yao Supermarket which has two locations in Toronto, neither of them anywhere near the northwest part of the city where I reside. Inside the bag were nine apples, a few grapes, and some sort of flattish bread-like item:

I'm at a loss to imagine a scenario that could account for all this.

10*987/6+54+321

Happy New Year!

True probabilities in self-referential statements

On December 8, Éric Angelini asked in an online MathFun forum for a solution to a self-referential sentence in which randomly picking four letters from the statement yielding the four letters F, O, U, and R, was a probability 𝜶/𝜷, where 𝜶 and 𝜷 were the English number names for the numeric quantities that allowed the sentence to be true. After determining that picking one letter depleted the letter availability of the next pick by one, I settled on my template being: "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are 𝜶 out of 𝜷."

Mathematica has a function for converting integers into English words so all I had to do was run through a bunch of them and test the resulting sentences for truthfulness. Here are nine solutions:

1. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are two out of two hundred nineteen thousand six hundred eighty-seven."

2. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are three out of two hundred ninety-two thousand nine hundred sixteen."

3. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are three out of one hundred ninety thousand six hundred fifty."

4. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are six out of seven hundred twenty-one thousand seven hundred ninety-one."

5. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are twelve out of one million seven hundred sixty-six thousand six hundred twenty-two."

6. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are eighteen out of one million six hundred thousand two hundred."

7. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are twenty-one out of two million ninety-seven thousand twenty-four."

8. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are thirty-five out of two million six hundred sixty-seven thousand four hundred twenty."

9. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are seventy-one out of nine million nine hundred twenty thousand two hundred sixty-two."

I believe that this exhausts the possibilities for 𝜶 up to 100 and 𝜷 up to 10000000. But only for this particular template. One may easily alter the template by adding/subtracting/changing words without altering the essential thrust of what the sentence is saying. Each of these new statements would have its own set of solutions.

Of the above nine solutions, the first two are special: The ratio 𝜶/𝜷 is in its lowest terms. In the other seven the two numbers share a common factor, so they can be reduced. Of course we mustn't do that because that would rob the statements of their truthfulness.

If you are interested in verifying the nine statements, this will help:

 1.  {2,219687}  {5,11,6,8}  <132>   5*11*6*8/(132*131*130*129) = 2/219687
 2.  {3,292916}  {5,11,6,9}  <132>   5*11*6*9/(132*131*130*129) = 3/292916
 3.  {3,190650}  {7,10,6,9}  <126>   7*10*6*9/(126*125*124*123) = 1/63550
 4.  {6,721791}  {5,11,6,8}  <135>   5*11*6*8/(135*134*133*132) = 2/240597
 5. {12,1766622} {5,12,6,8}  <145>   5*12*6*8/(145*144*143*142) = 2/294437
 6. {18,1600200} {5,12,6,8}  <128>   5*12*6*8/(128*127*126*125) = 1/88900
 7. {21,2097024} {6,13,5,7}  <130>   6*13*5*7/(130*129*128*127) = 7/699008
 8. {35,2667420} {7,12,7,10} <147>  7*12*7*10/(147*146*145*144) = 1/76212
 9. {71,9920262} {5,13,6,8}  <146>   5*13*6*8/(146*145*144*143) = 1/139722

After {𝜶, 𝜷} are the letter counts of {F,O,U,R}, followed by the statement's <total-letter-count>, followed by the probability calculation and 𝜶/𝜷 in its lowest terms.

Sapphire

Today is the sixth Friday, December 13th anniversary of our marriage. The photograph of the somber newlyweds was taken by my friend, Alfy Marcuzzi.

A330365

The idea for OEIS sequence A330365 came from Éric Angelini who asked me if I was willing to check and extend it. I was happy to. On the OEIS page for the sequence one can click "graph" and get (at the bottom) a logarithmic scatterplot. I have a much prettier version:

Click on the picture for a larger display of it. The blue points are the values at odd indices; the red points, values at even indices. Green lines connect adjacent values.

My 400th Leyland prime find

This latest addition to my table came in just after midnight. It was (at the time) asterisked because I had been unable to submit it (as well as seven previous others) to PRPtop, where submissions were timing out on the "captcha" (I'm not a robot) requirement. This was annoying to me for a number of reasons. I had to worry about Norbert Schneider rediscovering one of these already-found primes and I also use the PRPtop numbers to double-check my total counts.

Update: Finally, on November 25, the submissions started to work again.

Revenant gems

Éric Angelini's revenant numbers became OEIS sequence A328095. In Éric's article he mentions this "gem" found by Jean-Marc Falcoz, 62227496:

6*2*2*2*7*4*9*6 = 72576 * 62227496 = 4516222749696

Subsequently, Chai Wah Wu found 6886826188:

6*8*8*6*8*2*6*1*8*8 = 14155776 * 6886826188 = 97488368868261888

Allowing zeros in the product in any but the units position, I suggested 3691262781:

3*6*9*1*2*6*2*7*8*1 = 217728 * 3691262781 = 803691262781568

Whereupon Giovanni Resta came up with the significantly larger 774773248793:

7*7*4*7*7*3*2*4*8*7*9*3 = 348509952 * 774773248793 = 270016187747732487936