I see that two versions of that sequence are already in OEIS: A092320 A126259. I avoided the standard spelling, because it's not clear how one should do it for arbitrarily large integers. Here's the list of terms <= 1000 (it's not in the OEIS): 4, 6, 12, 14, 32, 40, 42, 56, 66, 72, 91, 108, 110, 120, 126, 132, 154, 160, 162, 176, 180, 209, 210, 216, 220, 231, 260, 261, 273, 286, 308, 324, 360, 390, 396, 403, 429, 432, 444, 451, 468, 504, 506, 507, 528, 540, 576, 585, 588, 598, 605, 610, 612, 620, 621, 627, 637, 648, 649, 660, 666, 671, 682, 684, 720, 726, 728, 754, 756, 767, 770, 784, 792, 798, 832, 845, 852, 864, 871, 900, 902, 924, 936, 938, 946, 972, 975 Victor On Sun, Sep 25, 2011 at 3:18 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
FYI, my column "Four, Twenty Four,...", in Mathematical Intelligencer 24 #2 (Sprint 2002), is on a similar topic: instead I wrote number names in English and then *multiplied* the word lengths together, and checked whether they were equal to the original number. So after "four" comes "twenty four", with word lengths 6 and 4, and 6*4=24. I came up with a heuristic argument that there should only be finitely many numbers for which it worked, though. http://oeis.org/A058230
--Michael
On Sun, Sep 25, 2011 at 1:33 PM, Victor Miller <victorsmiller@gmail.com>wrote:
On the program cartalk last week, there was a puzzle of the following form:
A list of numbers was given, and one was asked what they had in common (I give the actual puzzler at the end). The answer was that each of these numbers was divisible by the number of letters (excluding spaces) in the standard spelling out of the number in words. This got me to thinking of the following modification:
Suppose that we map each positive integer into an integer by the following means: write the number in decimal (no leading zeros). The value of the function will be to total number of letters that one gets by mapping each digit to its English equivalent (e.g. 13 -> length("OneThree") = 8. Or more generally give a fixed map g : {0,1,...,9} -> positive integers, and define f(n) = sum_{d digits in n} g(d). So are there an infinite number of positive integers n such that n is divisible by g(n)? One can generalize this to other bases. If there are an infinite number of them how dense is the set?
Victor
Here's the original puzzler:
When my kids were in school, they, like all the other kids I guess, had to learn their numbers. So each day for homework, they would bring home a list of numbers on a piece of paper, and they were asked to write out the letters that spelled that number, right next to each of them. So the number seven would be there, there'd be a blank space, the kids would have to write S - E - V - E - N. And of course they were also asked which numbers were spelled out by the various combinations of letters, so they'd see S - I - X - T - Y and write Sixty, etc.
One day, son number two presented me with a list of numbers and he said, "These numbers are different. There's something special about them." Here are the numbers:
Four, Six, Twelve, Thirty, Thirty Three, Thirty Six, Forty, Forty Five, Fifty, Fifty Four, Fifty Six, Sixty, Seventy, Eighty One, Eighty Eight, Ninety, and a Hundred.
Now there are no other numbers between one and a hundred inclusive that share this same characteristic. There's something unusual about these numbers that son number two figured out. And I'll give you an additional hint that order does not matter. The best hint is that he determined that these numbers should be on the list perhaps from his homework assignment.
What is special about this list of numbers?
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