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A061797
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Smallest k such that k*n has even digits and is a palindrome or becomes a palindrome when 0's are added on the left.
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2
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1, 2, 1, 2, 1, 4, 1, 98, 1, 74, 2, 2, 5, 154, 49, 4, 5, 38, 37, 34, 1, 286, 1, 36, 25, 8, 77, 329144, 31, 16, 2, 28, 25, 2, 19, 196, 23, 6, 17, 154, 1, 542, 143, 1602, 1, 148, 18, 6, 88, 14, 4, 824, 77, 8, 164572, 4, 143, 1198, 8, 1154, 1, 1126, 14, 962, 66, 308, 1, 998
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OFFSET
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0,2
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COMMENTS
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Every integer n has a multiple of the form 99...9900...00. To see that n has a multiple that's a palindrome (allowing 0's on the left) with even digits, let 9n divide 99...9900...00; then n divides 22...2200...00. - Dean Hickerson, Jun 29 2001
a(81), if it exists, is greater than 5 million. - Harvey P. Dale, Dec 19 2021
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LINKS
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EXAMPLE
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a(12) = 5 since 5*12 = 60 (i.e., "060") is a palindrome.
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MATHEMATICA
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a[n_] := For[k = 1, True, k++, id = IntegerDigits[k*n]; If[AllTrue[id, EvenQ], rid = Reverse[id]; If[id == rid || (id //. {d__, 0} :> {d}) == (rid //. {0, d__} :> {d}), Return[k]]]]; a[0] = 1; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Apr 01 2016 *)
skpal[n_]:=Module[{k=1}, While[Count[IntegerDigits[k n], _?OddQ]>0 || (!PalindromeQ[(k n)/10^IntegerExponent[n k]]), k++]; k]; Array[skpal, 70, 0] (* Harvey P. Dale, Dec 19 2021 *)
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PROG
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(ARIBAS): stop := 500000; for n := 0 to 75 do k := 1; test := true; while test and k < stop do m := omit_trailzeros(n*k); if test := not all_even(m) or m <> int_reverse(m) then inc(k); end; end; if k < stop then write(k, " "); else write(-1, " "); end; end;
(Haskell)
a061797 0 = 1
a061797 n = head [k | k <- [1..], let x = k * n,
all (`elem` "02468") $ show x, a136522 (a004151 x) == 1]
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CROSSREFS
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KEYWORD
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nonn,base,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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