[seqfan] Re: a(n)*a(n+1) shows at least twice the same digit
Lars Blomberg
larsl.blomberg at comhem.se
Mon Dec 21 14:44:32 CET 2015
Hello,
Some results:
Let k be the number of factors in the product (and the number of identical digits required in the product)
k = 2
1, 11, 2, 22, 3, 33, 10, 20, 5, 23, 7, 16, 9, 13, 17, 15, 35, 19, 6, 24, 12, 21, 25, 4, 28, 8, 14, 18, 29, 31, 32, 34, 26, 38, 30, 36, 37, 27, 40, 45, 39, 41, 42, 44, 46, 47, 43, 52, 50, 48, 49, 51, 59, 56, 55, 57, 58, 54, 53, 61, 60, 65, 62, 67, 63, 68, 66
k = 3
1, 2, 111, 3, 10, 37, 6, 4, 93, 12, 20, 25, 8, 5, 50, 16, 13, 32, 34, 26, 88, 35, 61, 51, 7, 15, 53, 14, 30, 24, 31, 9, 40, 28, 45, 80, 29, 44, 19, 27, 39, 87, 11, 38, 54, 22, 63, 58, 52, 21, 55, 49, 42, 59, 18, 43, 56, 36, 57, 47, 72, 33, 65, 73, 75, 64, 23
k = 4
1, 2, 3, 1111, 5, 4, 10, 50, 15, 8, 17, 100, 25, 12, 6, 56, 125, 20, 7, 16, 45, 97, 35, 24, 30, 40, 66, 38, 142, 77, 14, 58, 81, 49, 11, 28, 39, 19, 55, 21, 51, 69, 120, 90, 27, 70, 59, 180, 53, 79, 41, 18, 89, 123, 119, 23, 68, 64, 22, 46, 34, 61, 36, 167, 92
k = 5
1, 2, 3, 4, 12037, 5, 20, 6, 40, 25, 9, 50, 8, 10, 28, 45, 100, 15, 11, 54, 55, 200, 30, 17, 18, 60, 195, 150, 12, 19, 16, 75, 110, 63, 13, 74, 7, 143, 21, 26, 37, 14, 165, 22, 65, 33, 35, 66, 182, 325, 44, 70, 39, 27, 125, 76, 23, 34, 105, 68, 180, 46, 120
k = 6
1, 2, 3, 4, 5, 25000, 6, 7, 8, 9, 10, 3375, 20, 14, 12, 15, 18, 375, 30, 11, 24, 45, 22, 125, 40, 16, 13, 25, 17, 85, 185, 160, 19, 48, 35, 53, 75, 80, 38, 49, 50, 21, 32, 144, 130, 159, 55, 96, 70, 27, 145, 150, 26, 66, 100, 28, 29, 115, 65, 33, 60, 42, 124
k = 7
1, 2, 3, 4, 5, 6, 125000, 7, 8, 9, 10, 11, 12, 3125, 17, 16, 14, 13, 22, 15, 625, 18, 40, 20, 19, 24, 25, 50, 23, 41, 56, 95, 32, 38, 500, 46, 33, 30, 75, 28, 21, 100, 39, 37, 63, 750, 60, 26, 80, 27, 44, 31, 1525, 69, 200, 104, 64, 45, 35, 65, 43, 70, 79, 48
k = 8
1, 2, 3, 4, 5, 6, 7, 200000, 10, 8, 9, 15, 12, 14, 16, 3125, 25, 18, 11, 13, 20, 17, 32, 500, 50, 27, 21, 26, 45, 19, 40, 144, 1500, 30, 35, 22, 24, 23, 37, 75, 285, 200, 57, 33, 28, 92, 29, 125, 43, 530, 44, 55, 56, 46, 34, 625, 42, 53, 60, 82, 38, 77, 65
k = 9
1, 2, 3, 4, 5, 6, 7, 8, 250000, 10, 12, 13, 11, 14, 9, 21, 15, 25000, 20, 18, 26, 28, 25, 16, 17, 22, 1250, 45, 23, 24, 40, 50, 27, 19, 30, 140, 75, 39, 48, 70, 137, 55, 35, 29, 67, 197, 195, 96, 175, 80, 66, 145, 91, 33, 60, 82, 125, 36, 32, 65, 150, 41, 37
k = 10
1, 2, 3, 4, 5, 6, 7, 8, 9, 2812500, 10, 20, 11, 12, 15, 13, 14, 16, 17, 15625, 23, 40, 25, 18, 19, 22, 24, 28, 27, 6500, 113, 83, 125, 36, 38, 45, 30, 34, 26, 209, 200, 50, 100, 43, 35, 41, 29, 21, 39, 55, 76, 500, 422, 166, 175, 56, 60, 31, 51, 59, 75, 211
Obviously, for k=2, when the product of 2 numbers has at least 11 digits there must be at least one digit that occurs twice.
So when the product reaches 11 digits we will have a(n)=n all the way to oo.
For k=3 we can have at most 2 occurrences of all digits except digit d and then 3 occurrences of digit d. So at least 2*9+3=21 digits are required in the product.
And in general (k-1)*9+k = 10k-9 digits are required in the product.
/Lars B
-----Ursprungligt meddelande-----
Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Eric Angelini
Skickat: den 21 december 2015 00:23
Till: Sequence Discussion list <seqfan at list.seqfan.eu>
Ämne: [seqfan] a(n)*a(n+1) shows at least twice the same digit
Hello SeqFans,
we want the lexico-first permutation of the integers > 0 such that a(n)*a(n+1) shows at least twice the same digit.
If I'm not wrong, here is the lexico-first permutation of the integers 1 to 38:
S = 1,11,2,22,3,33,10,20,5,23,7,16,9,13,17,15,35,19,6,24,12,21,
25,4,28,8,14,18,29,31,32,34,26,38,30,36,37,27,...
The first products are:
11,22,44,66,99,330,800,100,115,161,... and the last one above is 37*27=999
Best,
É.
(I guess there are similar permutations producing 3 identical digits in the result of a(n)*a(n+1)*a(n+2), or 4 identical digits in... etc.)
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