A277622 Integers such that the first part "a", divided by the penultimate digit "b", leaves the last digit "c" as remainder (see the Example section for a detailed explanation).

110, 121, 131, 141, 151, 161, 171, 181, 191, 210, 220, 232, 242, 252, 262, 272, 282, 292, 310, 321, 330, 343, 353, 363, 373, 383, 393, 410, 420, 431, 440, 454, 464, 474, 484, 494, 510, 521, 532, 541, 550, 565, 575, 585, 595, 610, 620, 630, 642, 651, 660, 676, 686, 696, 710, 721, 731, 743, 752, 761, 770, 787, 797, 810

Offset

1,1

Comments

There are 8991 terms < 100000.

Links

Jean-Marc Falcoz, Table of n, a(n) for n = 1..8991

Example

Consider the first term, 110: the part "a" is "1", the penultimate digit "b" of "110" is "1" and the last digit "c" of "110" is "0"; we see indeed that 1/1 leaves a remainder of 0.
Consider the second term, 121: we see indeed that 1/2 leaves a remainder of 1 -- but no term between 110 and 121 has the same property (117, for instance, doesn't fit because 1/1 has 0 as remainder -- and not 7 as it should have to be part of the sequence).
Similarly, the integer 1983 is part of the sequence because 19/8 leaves a remainder of 3.

Mathematica

Select[Range[100, 1000], (u = Mod[#, 10]; d = Mod[(# - u)/10, 10]; d > 0 && u == Mod[Floor[#/100], d]) &] (* Giovanni Resta, Apr 11 2017 *)

Crossrefs

Sequence in context: A320121 A084042 A110735 * A101317 A274945 A274944

Adjacent sequences: A277619 A277620 A277621 * A277623 A277624 A277625

Keyword

nonn,base,easy

Author

Eric Angelini and Jean-Marc Falcoz, Apr 11 2017

Status

approved