Computing power

How difficult might it be to calculate term #1124577 of A179066? For me, very. I've been running my program now for a couple of days and have only managed some 25000 terms. I can probably see my way to 100000 terms but, short of a significantly more optimized version of my search routine (and even with that, the search beyond would prove taxing), there it will end. Remember also that I am running two other copies of Mathematica in the background (my ongoing Ruth-Aaron pairs tabulation) and these will eat up available computer-clock cycles, reducing the Mathematica kernel's CPU usage to some 60% of the available 2 MHz. Have I mentioned yet that I could really use a modern replacement for my seven-and-a-half-year-old machine?

Proof by contradiction

Yesterday I asked for the "final term" of a sequence. It may not be immediately obvious that there is for this sequence such a thing, but consider this: A179066 is defined in such a way that term #1124578 must have a digital root of 1 (the digital root of 1124578) and yet be composed of only the digits 0, 3, 6, and 9 (those digits that are not used in 1124578). But numbers made up of only 0s, 3s, 6s, and 9s will be evenly divisible by 3 or, alternatively, leave a remainder of 0, 3, or 6 when divided by 9 (making their digital roots 3, 6, or 9), thus contradicting our already established fact that the number that we are looking for has a digital root of 1. As defined, term #1124578 cannot exist and term #1124577, therefore, is likely the final term of this sequence.

28, 11, 12, 13, ...

Eric Angelini came up with this: Each successive term is the smallest positive integer not yet used, where both the term and its index have the same digital root (remainder when divided by 9) but do not have in common any of each other's base-ten digits. What might be this sequence's final term?

35000 Ruth-Aaron pairs

I've just added another 5000 Ruth-Aaron pairs to my ongoing tabulation thereof (bringing the total now to 35000 pairs). The project will likely consume the better part of this year. I realize that this would go much, much faster using a modern computer (my dual 2 GHz PowerPC G5 is now well over seven years old) but, in the absence of anyone else taking this on, at least I've got a decent start on it.

1+2+3/45/6/7*89

Using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in order, replace each comma with one of +, -, *, /, or a concatenation of adjacent numbers. Parentheses or other punctuation are not allowed and, as usual, multiplications and divisions are to be evaluated before additions and subtractions. My title today evaluates to a quantity fairly close to π. Can you come up with an expression that is even closer?

Countdown

Just how likely was my pre-midnight countdown? Consider the ordering of 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 and allow the commas to be replaced by one of +, -, *, /, or a concatenation of adjacent numbers (with the added proviso of no concatenation after the 10 to force the generally accepted start of the countdown). We exclude parentheses, so the mathematical statements must be evaluated in the standard order of operations (i.e., multiplications and divisions evaluated before additions and subtractions). Of the 1562500 (4*5^8) possible numerical outcomes, 486501 are unique. Of these, only 51249 are integers (31373 positive), the largest being 10*987654321. The smallest positive integers that do not appear are 1992, 2102, 2133, etc., so every year from 1993 to 2101 will have at least one solution: The largest number of solutions in this interval for a given year is 13 and the average, 616/109. Only two of the solutions start with 10/ (1997 = 10/9*87*6*5 - 43*21 and 2021 = 10/9*8*765/4 + 321). While 1992 is missing from our integer solutions, it may be approximated by the nearby 10 + 9 + 8 + 76*543/21 (as well, if we allow concatenation after the 10, it's 1098 + 765 + 4*32 + 1). You might like to look for one of the four solutions that evaluate to 377/120, the closest that we can get to the irrational number π.

An interesting bonus in last night's countdown exemplar is the symmetry of the operations: +, *, *, /, *, *, +. How common might that be? Out of the 616 solutions in the 1993-2101 interval, I count 12 that qualify:

1994 = 10*9 - 8*7 + 654*3 - 2*1
2001 = 10 - 9*87 + 65*43 - 21
2011 = 10 + 9*8*7/6*5*4 + 321
2020 = 10 - 9 + 8*7 + 654*3 + 2 - 1
2023 = 10*987/6 + 54/3*21
2024 = 10 + 9 + 8*7*6*5 + 4 + 321
2040 = 10 + 9 + 8*7 + 654*3 + 2 + 1
2072 = 10 + 9*8 + 7 + 654*3 + 21
2088 = 10*9*87/6/5/4*32*1
2093 = 10*98 + 7*6*5 + 43*21
2098 = 10 - 98*7 + 65*43 - 21 or 10 - 9*8 + 7 - 6 + 5*432 - 1

10+9*8*7/6*5*4+321

Happy New Year!

Goodbye

Larry

Unposed, Larry is sitting on a rock atop the hill southeast of St. Bernard's, Newfoundland, on 13 August 2003. The view is toward the northeast with, I think, Harbour Mille 7.5 km in the distance. St. Bernard's was Larry's adopted home since that is where his wife was from. He lived in Elmsdale, Nova Scotia. Larry is the one on the right in this lovely old fishing pose. He'll be sadly missed.

Outport

St. Bernard's

A very nice view of St. Bernard's taken on 14 August 2003 from atop the high hill northeast of the town. The house in which we stayed is circled in this detail. The settlement is the terminus of a twelve-kilometer road off the main highway going down the Burin Peninsula (unlike some Newfoundland outports that lack altogether direct vehicular access).

My abode #8

Our House

Built in 1904, John and Susan Adamson raised in this house four children: Ada, Della, Nelson, and Rhea. After Susan died in 1963, Ada continued to live here until her own death in 1973, after which Ray Jafelice and his family took over. It became our home on 3 October 1978.