My son, Alexey Radul, wrote a program that finds the largest numbers to start a sequence in the Online Encyclopedia of Integer Sequences (OEIS). To my surprise, the top ten are all numbers consisting of ones only. The largest number is 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111: a number with 128 ones. The sequence is A095646: a(n) is 128 written in base n. It starts with base 1, or more precisely, the unary expansion, which indeed requires 128 ones to express the number 128.
I decided to expand my top list to 50. Again, most of the numbers are bunches of ones: 48 out of the top 50 numbers are unary expansions of numbers 81 through 128. There are two more numbers in the top 50 that are different and belong to awesome sequences.
The first awesome sequence is sequence A033290: Ten consecutive primes in arithmetic progression. It starts with the number 100996972469714247637786655587969840329509324689190041803603417758904341703348882159067229719, which has 93 digits. This number takes 37th place on my list.
The second awesome sequence is sequence A291042: One powerful arithmetic progression with a nontrivial difference and maximal length. The sequence corresponds to a cool puzzle that appeared in the American Mathematical Monthly in 2000. The question was, “What is the length of the longest non-constant arithmetic progression of integers with the property that the kth term is a perfect kth power?” The answer is 5. John P. Robertson proved that such progression can’t have 6 terms and provided an example of a sequence with 5 terms, which is the sequence in the OEIS.
Here is how to construct this sequence. Start with an arithmetic progression 1, 9, 17, 25, 33, and multiply each term by 32453011241720: the result is also an arithmetic progression. The first term is trivially a first power. The second term is 32653011241720 = (31351511121710)2 and a square. The third term is 32453011241721 = (38510118177)3 and a cube. The fourth term is 32453211241720 = (3658116175)3 and a fourth power. The fifth term is 32553011251720 = (3556115174)5 and a fifth power.
The sequence starts with the number 10529630094750052867957659797284314695762718513641400204044879414141178131103515625. It has 83 digits, and it takes 48th place on my list.Share: