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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
 1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 61, 75, 76, 77, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 LINKS Jean-Marc Falcoz, Table of n, a(n) for n = 1..40000 EXAMPLE 1 is in the sequence because 10/(1+0) is the integer 10; 2 is in the sequence because 21/(2+1) is the integer 7; 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); 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; 5 is in the sequence because 54/(5+4) is the integer 6; 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. 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. . . . 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. 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. Etc. CROSSREFS Cf. A316534 where the concatenation k and k+1 is considered (instead of k and k-1 here). Sequence in context: A265567 A265551 A359999 * A283206 A084545 A069908 Adjacent sequences: A316535 A316536 A316537 * A316539 A316540 A316541 KEYWORD base,nonn AUTHOR Eric Angelini and Jean-Marc Falcoz, Jul 06 2018 STATUS approved

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