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A072210
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a(1)=a(2)=1; a(n)=reverse(reverse(a(n-1))+reverse(a(n-2))) for n > 2.
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1
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1, 1, 2, 3, 5, 8, 31, 12, 43, 55, 98, 441, 332, 773, 16, 834, 994, 739, 6341, 3732, 9083, 2816, 1999, 37161, 46162, 73324, 10586, 838011, 933971, 771092, 615964, 396957, 9029221, 2098891, 1118123, 3107025, 4215248, 73123631, 16275022, 89398653, 95664775
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OFFSET
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1,3
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COMMENTS
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I call this sequence the Fibonacci mirror sequence for the following reason. For n>2, the expression "a(n)=a(n-1)+a(n-2)" is a valid equation if read backwards. For example, "a(9)=a(8)+a(7)" is "43=12+31", which read backwards is 13+21=34, a valid equation.
Reverse(a(n))=reverse(a(n-1))+reverse(a(n-2)). a(n) is the least natural number k such that reverse(k)=reverse(a(n-1))+reverse(a(n-2)).
(Added Jul 06 2002) Actually, the previous comments are true only if reverse(a(n-1))+reverse(a(n-2)) does not end in the digit 0. It ends in 0 for n = 15, but for no other n < 3 * 10^4. Mark Lewis claims that n = 15 is the only such value of n. He observes that the first fifteen terms of a(n) are the reverses of the first fifteen terms of the Fibonacci sequence. The later terms of a(n) are the reverses of the terms of the Fibonacci sequence starting with 377, 61 (excluding these initial two terms). Lewis' argument depends on his assertion that the (377,61)-sequence is, modulo 10, periodic with period 12 and with no zeros-one for which he, as yet, offers only empirical evidence.
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LINKS
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EXAMPLE
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a(9)=reverse(reverse(a(8))+reverse(a(7)))=reverse(21+13)=43.
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MATHEMATICA
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rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; r = {1, 1}; For[i = 1, i < 30, i++, l = Length[r]; r = Append[r, rev[rev[r[[l]]] + rev[r[[l - 1]]]]]]; r
rev[n_]:=FromDigits[Reverse[IntegerDigits[n]]]; nxt[{a_, b_}]:={b, rev[ rev[ a]+ rev[b]]}; Transpose[NestList[nxt, {1, 1}, 50]][[1]] (* Harvey P. Dale, Apr 25 2014 *)
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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