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%I #14 Oct 26 2017 14:39:29
%S 1,3,9,27,189,2457,12285,159705,9103185,2030010255,11000625571845,
%T 187010634721365,45069562967848965,188943190838905598464005,
%U 3169167002067055110614170009605,53875839035139936880440890163285
%N a(1) = 1. a(n) = the smallest multiple of a(n-1), a(n) > a(n-1), such that a(n) in binary is a palindrome.
%C All terms are odd.
%C There are an infinite number of terms. Proof: (2^m + 1)*a(n) is a palindrome, where m is >= the number of binary digits in a(n). So a(n+1) <= (2^m + 1)*a(n).
%p isA006995 := proc(n) local dgs,i ; dgs := convert(n,base,2) ; for i from 1 to nops(dgs)/2 do if op(i,dgs) <> op(-i,dgs) then RETURN(false) ; fi; od: RETURN(true) ; end: A143014 := proc(n) option remember ; local m,a ; if n = 1 then 1; else for m from 2 do a := m*A143014(n-1) ; if isA006995(a) then RETURN(a) ; fi; od: fi ; end: for n from 1 to 100 do printf("%d,",A143014(n)) ; od: # _R. J. Mathar_, Aug 08 2008
%t Fold[Function[{a, n}, Append[a, SelectFirst[Range[2^(n + 2)] Last[a], And[# > Last[a], PalindromeQ[IntegerDigits[#, 2]]] &]]] @@ {#1, #2} &, {1}, Range[2, 13]] (* _Michael De Vlieger_, Oct 25 2017 *)
%o (PARI) isok(ka) = my(b=binary(ka)); b==Vecrev(b);
%o lista(nn) = {print1(a=1, ", "); for (n=2, nn, k=2; while (! isok(k*a), k++); a *= k; print1(k, ", "););} \\ _Michel Marcus_, Oct 26 2017
%Y Cf. A143015, A143016.
%Y Cf. A006995. - _R. J. Mathar_, Aug 08 2008
%K base,nonn
%O 1,2
%A _Leroy Quet_, Jul 15 2008
%E a(6)-a(13) added by _R. J. Mathar_, Aug 08 2008
%E a(14)-a(16) from _Ray Chandler_, Jun 21 2009
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