A143014 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.

1, 3, 9, 27, 189, 2457, 12285, 159705, 9103185, 2030010255, 11000625571845, 187010634721365, 45069562967848965, 188943190838905598464005, 3169167002067055110614170009605, 53875839035139936880440890163285

Offset

1,2

Comments

All terms are odd.

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).

Links

Table of n, a(n) for n=1..16.

Maple

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

Mathematica

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 *)

Prog

(PARI) isok(ka) = my(b=binary(ka)); b==Vecrev(b);
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

Crossrefs

Cf. A143015, A143016.

Cf. A006995. - R. J. Mathar, Aug 08 2008

Sequence in context: A254334 A028855 A299597 * A202907 A032261 A300981

Adjacent sequences: A143011 A143012 A143013 * A143015 A143016 A143017

Keyword

base,nonn

Author

Leroy Quet, Jul 15 2008

Extensions

a(6)-a(13) added by R. J. Mathar, Aug 08 2008

a(14)-a(16) from Ray Chandler, Jun 21 2009

Status

approved