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A272856
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Greatest length of a chain of consecutive primes p==1 (mod 3) for which A261029 (2*prime(n)*p) is 4-i if prime(n) == i (mod 3), where i=1,2.
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1
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5, 7, 16, 19, 32, 35, 50, 76, 81, 108, 140, 139, 171, 206, 254, 259, 305, 346, 349, 404, 449, 504, 582, 634, 645, 699, 707, 772, 930, 1006, 1078, 1097, 1258, 1271, 1362, 1448, 1529, 1633, 1737, 1752, 1951, 1970, 2064, 2082, 2310, 2550, 2659, 2672, 2783, 2917
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OFFSET
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3,1
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LINKS
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EXAMPLE
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Let n=3; then prime(n)=5. Since 5 == 2 (mod 3), i=2. So a(3) is the greatest length of a chain of consecutive primes p == 1 (mod 3) for which A261029(10*p) = 4 - 2 = 2. So these primes are in A272381. The first term is 7, and we have the chain of consecutive primes == 1 (mod 3): {7, 13, 19, 31, 37}. Since the following prime 43 == 1 (mod 3) is not in A272381, the chain ends and its length is 5. The second chain is the singleton {71}. So a(3)=5.
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MATHEMATICA
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a261029[n_]:=a261029[n]={x, y, z}/.{ToRules[Reduce[x^3+y^3+z^3-3 x y z==n&&0<=x<=y<=z&&z>=x+1, Integers]]}/.{x, y, z}->{};
data={};
Do[p=Prime[n];
primes=Select[Prime[Range[1+PrimePi[(2p)^2]]], Mod[#, 3]==1&];
tmp=Map[{#, Length[a261029[2 # p]]}&, primes];
AppendTo[data, {{n, 2p, 1+Mod[2p, 3]}, {{Length[#], Max[Map[Length, Select[Split[Differences[Flatten[Map[Position[primes, #, 1, 1]&, #]]]], #[[1]]==1&]]+1]}, #}&[Map[#[[1]]&, Select[tmp, #[[2]]==(1+Mod[2p, 3])&]]]}]; Print[Last[data]], {n, 3, 10}]
Map[Length[a261029[#]]&, Range[0, 20]] (* A261029 *)
Last[Last[data[[1]]]] (* A272381 *)
Last[Last[data[[2]]]] (* A272382 *)
Last[Last[data[[3]]]] (* A272384 *)
Last[Last[data[[4]]]] (* A272404 *)
Last[Last[data[[5]]]] (* A272406 *)
Last[Last[data[[6]]]] (* A272407 *)
Last[Last[data[[7]]]] (* A272409 *)
Map[#[[2]][[1]][[1]]&, data] (* A268665 *)
Map[#[[2]][[1]][[2]]&, data] (* A272856 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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