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Affichage des articles du 2023

Triples for the new year

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(Dall-e creation ) The hereunder sequence  S  is both a succession of triples of monotonically increasing numbers and a succession of triples of monotonically increasing digits. S  is the lexicographically earliest such sequence with distinct number-triples: S = 0,1,2,0,1,3,0,1,4,0,1,5,0,1,6,0,1,7,0,1,8,0,1,9,0,2,3,0,2,4,0,2,5,0,2,6,0,2,7,0,2,8,0,2,9,0,3,4,0,3,5,0,3,6,0,3,7,0,3,8,0,3,9,0,4,5,0,4,6,0,4,7,0,4,8,0,4,9,0,5,6,0,5,7,0,5,8,0,5,9,0,6,7,0,6,8,0,6,9,0,7,8,0,7,9,0,8,9,1,2,3,1,2,4,1,2,5,1,2,6,1,2,7,1,2,8,1,2,9,1,3,4,1,3,5,1,3,6,1,3,7,1,3,8,1,3,9,1,4,5,1,4,6,1,4,7,1,4,8,1,4,9,1,5,6,1,5,7,1,5,8,1,5,9,1,6,7,1,6,8,1,6,9,1,7,8,1,7,9,1,8,9,2,3,4,2,3,5,2,3,6,2,3,7,2,3,8,2,3,9,2,4,5,2,4,6,2,4,7,2,4,8,2,4,9,2,5,6,2,5,7,2,5,8,2,5,9,2,6,7,2,6,8,2,6,9,2,7,8,2,7,9,2,8,9,3,4,5,3,4,6,3,4,7,3,4,8,3,4,9,3,5,6,3,5,7,3,5,8,3,5,9,3,6,7,3,6,8,3,6,9,3,7,8,3,7,9,3,8,9,4,5,6,4,5,7,4,5,8,4,5,9,4,6,7,4,6,8,4,6,9,4,7,8,4,7,9,4,8,9,5,6,7,5,6,8,5,6,9,5,7,8,5,7,9,5,8,9,6,7,8,6,7,9,6,8,9,7,...

First digits after the decimal point

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(Dall-e creation ) If a , b and c are three successive terms of a sequence S, we want that a is what we see immediately after the decimal point when b is divided by c (we do not take into account the leading zeros between the decimal point and a ).  I guess S could start like this (S being the lexico-earliest sequence of distinct positive terms): S = 1,2,11,5,42,4,94,38,401,1029,2560,... 1   is what we see after the decimal point when 2 is divided by 11 (indeed 2/11 = 0. 1 8181818181 ...) 2   is what we see after the decimal point when 11 is divided by 5 (indeed 11/5 = 2. 2 ) 11   is what we see after the decimal point when 5 is divided by 42 (indeed 5/42 = 0. 11 904761904 ...) 5   is what we see after the decimal point when 42 is divided by 4 (indeed 42/4 = 10. 5 ) 42   is what we see after the decimal point when 4 is divided by 94 (indeed 4/94 =  0.0 42 55319148 ...) 4   is what we see after the decimal point when 94 is divided...

7 times and 9 times (and more)

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(Dall-e creation ) Keeping at every step the last digit of 7*a(n) reproduces the sequence S digit by digit. We want of course S to be the lexicographically earliest sequence of distinct terms >0. S = 5, 13, 9, 7, 1, 3, 19, 23, 17, 6, 29, 33, 11, 8, 16, 27, 39, 49, 43, 53, 4, 63, 18, 26, 21, 59, 37, 2, 47, 12, 69, 15, 79, 22, 28, 89, 73, 14, 36, 38, 46, 83, 25, 57, 99, 31, 56, 32, 41, 93, 66, 48, 67, 103, 35, 51, 77, 76, 86, 96, 24, 34, 87, 61, 109, 113, 42, 119, 58, 129, 44, 52, 68, 54, 139, 106, 45, 55, 71, 97, 107, 149, 123, 65, 78, 159, 116, 62, 133, 117, 169, 88, 98, 72, 64, 108, 81, 143, 10, 179, ... ___________________________________________________________________________________ Keeping at every step the last digit of 9*a(n) reproduces the sequence T digit by digit. We want of course T... etc. T = 5, 19, 1, 9, 11, 29, 39, 8, 21, 7, 31, 2, 18, 49, 3, 17, 59, 28, 69, 12, 6, 41, 27, 79, 13, 15, 51, 38, 22, 4, 61, 89, 48, 14, 16, 99, 58, 23, 33, 71, 109, 37, 119, 25, 35, 129, ...

More (odd/even) chunks

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(Dall-e creation )   The sequence S can be cut into chunks of integers that have the same parity (odd/even). The size of the successive chunks is given by the successive digits of S. No digit 0 is present in S. S = 1,2,12,3,4,14,5,21,23,6,16,18,32,7,8,34,36,38,9,25,27,29,41,212,52,121,214,54,123,43,45,216,56,58,72,74,76,125,… (Dall-e creation )

A sum with visible digits

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(Dall-e creation ) We want the digits of a(n) and a(n+1) to be visible in the sum a(n) + a(n+1). We forbid any term > 9 made of a single repeated digit (like 11, 222, 44, 3333, etc.) What could be the lexico-first sequence starting with a(1) = 1? We have a lot of doubts about the hereunder results – but who knows? S = 1, 100, 2, 200, 3, 300, 4, 400, 5, 500, 6, 600, 7, 700, 8, 800, 9,   900, 10, 99, 192, 1299 , ... ____________________ Same date update , half an hour before midnight (Brussels, Belgium) Giorgos Kalogeropoulos was quick to correct and extend the sequence: S =  1, 100, 2, 200, 3, 300, 4, 400, 5, 500, 6, 600, 7, 700, 8, 800, 9, 89, 899, 19, 90, 890, 199, 92, 1000, 10, 900, 20, 1001, 21, 1311, 12, 1010, 210, 818, 13, 1818, 24, 2000, 14, 1828, 280, 1801, 6800, 30, 1002, 40, 1003, 50, 454, 95, 499, 455, 94, 399, 910, 109, 920, 1099, 91, 98, 799, 1188, 1628, 11188, 820, 1288, 9001, 60, 1004, 70, 1005, 80, 1006, 9090, 15, 1838, 1180, 28, 1814, 1334, 11740, 1307, 1710...

A Xmas tree with numbers

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(Dall-e creation ) The hereunder Xmas tree has only 6 rows – shame on us! Can it be improved? Pick any digit d in the tree: d is absent from rows immediately above or below d . A term is always the absolute difference of the two nearest numbers under it. Example 12  is the absolute difference between 65 and 53, and no  1  or  2  is visible in the rows sandwiching  12 .                2              1   3            4   5   2          7   3   8   6        2   9  12    4  10     58  56  65  53  57  67 _______________________ December 20st, 2023 update Giorgos Kalogeropoulos was quick to beat my attempt with 7 rows (and 28 integers whose sum is 1464)           ...

English numbered chunks

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(Dall-e creation ) This self-describing sequence is the lexicographically earliest of its kind: distinct positive integers and chunks of letters always ending with a precise letter as separator (when the said integers are written in American English): 1,4,6,100,3,5,7,8,9,10,11,12,13,15,16,17,18,19,51,1000,14,40,50,101,20,2,23,106,103,105,200,26,110,60,109,61,25,27,28,29,21,30,33,35,36,37,38,39,53,55,56,57,58,59,63,65,66,67,68,69,70,73,75,76,77,78,79,80,83,85,86,87,88,89,90,93,95,96,97,98,99,300,303,305,306,307,308,309,310,311,312,313,315,316,317,318,319,320,... The sequence of words is: One, four, six, one hundred, three, five, seven, eight, nine, ten, eleven, twelve, thirteen, fifteen, sixteen, seventeen, eighteen, nineteen, fifty-one, one thousand, fourteen, forty, fifty, one hundred one, twenty, two, twenty-three, one hundred six, one hundred three, onehundred five, two hundred, twenty-six, one hundred ten, sixty, one hundred nine, sixty-one, twenty-five, twenty-seven, twenty-eight...

7 octobre 2023, autres horreurs

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n' = #div + sumdiv + 1

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– Start the sequence S with an integer  n (like 2023) – compute the number A of divisors of n (2023 has A =  6 divisors, which are 1, 7, 17, 119, 289 and 2023), – make the sum B of the said divisors ( B = 2456 here) – replace n by n' = A + B + 1  ( n' = 2463 =  6  +  2456  + 1  here)  – iterate until a prime is reached (2463 —> 3293 —> 3425 —> 4285 —> 5153 is prime) – restart the iteration at that point with the smallest unused integer so far (which would be 1 here). Question #1 How does the graph of S look if we start S with a(1) = 1? S = 1,3,2,4,11,5,6,17,7,8,20,49,61,9,17,10,23,11,12,35,43,13,14,29,15,29,16,37,18,46,77,101,19,21,37,22,... Explanation 1 —> 3 (prime, extend now S with 2) 2 (prime, extend now S with 4) 4 —> 11 (prime, extend now S with 5) 5  (prime, extend now S with 6) 6 —> 17  (prime, extend now S with 7) 7  (prime, extend now S with 8) 8 —> 20 —> 49 —> 61 (prime, et...

Erase the progression

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Today we are growing numbers. Start with n . Check if there are 3 or more adjacent digits building a “+1 progression”, or a “-1 progression”. Like here (in red),  445 123 6 or there,  445 321 6 . If yes, erase the progression and concatenate what is left: 445 123 6 –> 445 123 6 –> 4456 445 321 6 –> 445 321 6 –> 4456 Check again and erase again if necessary: 4 456 –> 4 456 –> 4  (transforming 4451236 into 4 is considered as a single step). Now that there are no more such progressions in the result, we multiply it by 2: 4 * 2 –> 8 And we iterate. We start with 1 for a first try (this will help us improving the rules): S = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 33554, 67108, 134216, 268432, 268, 536, 1072, 2144, 4288, 8576, 17152, 34304, 68608, 137216, 274432, 274, 548, 1096, 2192, 4384, 8768, 8, ... We see that 1, 2, 4 a...