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Primes modulo five partitioned sequentially in groups of four are then given the number of the order in which such groupings come up sequentially in an n X n X n X n vector.
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%I #12 Jan 13 2022 01:24:43

%S 74,237,74,178,47,211,122,32,26,177,157,159,194,178,42,37,95,220,181,

%T 241,188,74,212,179,23,198,178,157,39,202,132,220,133,230,195,127,122,

%U 38,62,26,25,124,12,195,211,110,179,130,138,178,248,178,40,18,144,159

%N Primes modulo five partitioned sequentially in groups of four are then given the number of the order in which such groupings come up sequentially in an n X n X n X n vector.

%C There are 256 ways you can get a one-dimensional tile of colors {red, green, blue, yellow}: these are used to index the appearance of modulo five appearance of such orders in the primes. The most popular occurrence is {4,5,2,3} at 5 for the first 256 prime partitions after the prime 5.

%F a0[m] = {1 + Mod[Prime[m], 5],1 + Mod[Prime[m+1], 5],1 + Mod[Prime[m+2], 5]1 + Mod[Prime[m+3], 5]} b0[n]={x[n],y[n],z[n],w[n]} a(m) = a0[m] indexed by b0[n].

%e a[[4]] -> b[[178]] = {4, 5, 2, 3}.

%t a = Partition[Table[1 + Mod[Prime[n], 5], {n, 4, 260}], 4]; b = Union[Flatten[Table[{x, y, z, t}, {x, 2, 5}, {y, 2, 5}, {z, 2, 5}, {t, 2, 5}], 3]]; f[n_, m_] := If[a[[m]] - b[[n]] == {0, 0, 0, 0}, n, {}] c=Flatten[Table[f[m, n], {n, Length[a]}, {m, Length[b]}]]

%K nonn,uned

%O 1,1

%A _Roger L. Bagula_, Aug 29 2006