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A062406
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Cardinality of the (ordered) list L_n defined inductively by: L_1 = {2}; L_(n+1) = L_n - {p}, where p is the first member of L_n, from left to right, such that prime(n+1) can be appended to the end or beginning of p so that the neighboring digits are equal, if p exists; append prime(n+1) to the end of L_n, otherwise.
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2
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1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 6, 5, 4, 5, 6, 5, 4, 3, 4, 5, 6, 7, 6, 7, 8, 9, 8, 9, 10, 11, 12, 13, 12, 11, 12, 13, 14, 13, 14, 15, 16, 17, 18, 17, 16, 17, 18, 19, 18, 19, 18, 19, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 11, 10
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OFFSET
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1,2
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COMMENTS
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It appears that a(n) is growing slowly on the average. (A moving average filter applied to the sequence will show an upward trend.) Probably a(n) > 0 for all n, but lacking a proof, one is never sure. For example, L_9 = {5} comes perilously close to extinction. It would be interesting to have a closed-form expression giving, at least asymptotically, the value of a(n).
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LINKS
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EXAMPLE
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L_n for n = 1..5 are: {2}, {2,3}, {2,3,5}, {2,3,5,7}, {2,3,5,7,11}. The first five values of the sequence are then 1,2,3,4,5, respectively. For n = 6, Prime(6) = 13 can be appended to the beginning of 3 in L_5 so that the neighboring digits (i.e., 3's) are equal, so eliminate 3 from L_5 to get L_6 = {2,5,7,11}. Hence a(6) = 4.
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PROG
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(Sage)
def a_list(n) :
res = []
curlist = []
for m in range(1, n+1) :
d = nth_prime(m).digits()
ds = (d[-1], d[0])
remitem = next((i for i in range(len(curlist)) if curlist[i][0] == ds[1] or curlist[i][1] == ds[0]), None)
if remitem != None : del curlist[remitem]
else : curlist.append(ds)
res.append(len(curlist))
return res
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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Extended and corrected and definition edited by Eric M. Schmidt, Oct 19 2016
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STATUS
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approved
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