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%I #38 Feb 09 2020 02:57:52
%S 110,120,126,140,150,162,210,212,216,220,234,240,250,360,413,432,480,
%T 510,520,540,550,630,672,864,891,918,1010,1020,1040,1050,1062,1080,
%U 1100,1200,1250,1400,1500,2010,2012,2016,2020,2034,2040,2050,2072,2079,2080,2100,2112,2132,2176,2200,2250,2400,2500
%N Numbers k such that the product (m+n)*p, where m,n are the first and the last digits of k, and p is the number which is the part of k between m and n, is a divisor of k.
%C The sequence is infinite if one considers numbers like 10........010 (with N zeros between ones, N = 0, 1, 2, ...). The problem of looking for odd terms and zerofree terms remains. Still unclear is whether the sequence contains infinitely many zerofree terms. The proportions between numbers of N-digit numbers included in the sequence need to be explored.
%e 234 is divisible by 3*(2+4).
%e 1020 is divisible by 2*(1+0).
%e 1062 is divisible by 6*(1+2).
%e 1250 is divisible by 25*(1+0).
%t Select[Range[100, 2500], Mod[#1, If[#2 == 0, #1 - 1, #2] & @@ {#1, (First@ #2 + Last@ #2) FromDigits@ Most@ Rest@ #2}] == 0 & @@ {#, IntegerDigits@ #} &] (* _Michael De Vlieger_, Oct 11 2018 *)
%o (PARI) isok(n) = {d = digits(n); if (#d >= 3, x = d[1]; y = d[#d]; w = vector(#d-2, k, d[k+1]); z = fromdigits(w); if (z, return (!(n % (z*(x+y)))));); return (0);} \\ _Michel Marcus_, Oct 07 2018
%Y Cf. A320292 (zeroless terms).
%K nonn,base
%O 1,1
%A _Anton Deynega_, Oct 06 2018
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