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A072210
a(1)=a(2)=1; a(n)=reverse(reverse(a(n-1))+reverse(a(n-2))) for n > 2.
1
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
OFFSET
1,3
COMMENTS
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.
EXAMPLE
a(9)=reverse(reverse(a(8))+reverse(a(7)))=reverse(21+13)=43.
MATHEMATICA
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 *)
CROSSREFS
Sequence in context: A041331 A290033 A004091 * A273003 A014258 A237575
KEYWORD
base,nonn
AUTHOR
Joseph L. Pe, Jul 03 2002
EXTENSIONS
More terms from Harvey P. Dale, Apr 25 2014
STATUS
approved