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 A316538 Numbers k such that k concatenated with k-1 and then divided by 2k-1 produces an integer after some divisions explained in the Example section. 3

%I #18 Jul 15 2018 13:58:43

%S 1,2,3,4,5,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,

%T 30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,61,75,

%U 76,77,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121

%N Numbers k such that k concatenated with k-1 and then divided by 2k-1 produces an integer after some divisions explained in the Example section.

%H Jean-Marc Falcoz, <a href="/A316538/b316538.txt">Table of n, a(n) for n = 1..40000</a>

%e 1 is in the sequence because 10/(1+0) is the integer 10;

%e 2 is in the sequence because 21/(2+1) is the integer 7;

%e 3 is in the sequence though 32/(3+2) is not an integer, because if we do floor(32/(3+2)) we get 6, and if we use this 6 in floor(26/(2+6)) we get 3 and 63/(6+3) is an integer (here 7);

%e 4 is in the sequence though 43/(4+3) is not an integer, because if we apply the "floor" trick again, we will end in an integer: floor(43/(4+3)) = 6, and 36/(3+6)) is the integer 4;

%e 5 is in the sequence because 54/(5+4) is the integer 6;

%e 6 is not in the sequence because 65/(6+5) is not an integer and even if we repeatedly apply the "floor" trick, we will be stuck in the loop 65/(6+5): floor(65/(6+5)) = 5, then floor(55/(5+5)) = 5, then again floor(55/(5+5)) = 5, etc. So 6 will never produce an integer at the end.

%e 7 is not in the sequence because 76/(7+6) is not an integer and even if we repeatedly apply the "floor" trick, we will be stuck in the same loop as 6: floor(76/(7+6) = 5, then floor(65/(6+5)) = 5, then floor(55/(5+5)) = 5, then again floor(55/(5+5)) = 5, etc. So 7 will never produce an integer at the end.

%e . . .

%e 10 is not in the sequence because 109/(10+9) is not an integer and even if we repeatedly apply the "floor" trick, we will be stuck in the same loop as 6 and 7: floor(109/(10+9)) = 5, then floor(95/(9+5)) = 6, then floor(56/(5+6)) = 5, then floor(65/(6+5)) = 5, then floor(55/(5+5)) = 5, then again floor(55/(5+5)) = 5, etc. So 10 will never produce an integer at the end.

%e 11 is in the sequence though 1110/(11+10) is not an integer, but if we repeatedly apply the "floor" trick, we will produce an integer at the end (here 19): floor(1110/(11+10) = 52, then floor(1052/(10+52)) = 16, then floor(5216/(52+16)) = 76, then floor(1676/(16+76)) = 18, then floor(7618/(76+18)) = 81, and 1881/(18+81) is the integer 19.

%e Etc.

%Y Cf. A316534 where the concatenation k and k+1 is considered (instead of k and k-1 here).

%K base,nonn

%O 1,2

%A _Eric Angelini_ and _Jean-Marc Falcoz_, Jul 06 2018

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