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A276435
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Smallest integer z > 1 that satisfies all the integer congruences z mod i = b(i) mod i, where b(i) is the i-th composite between prime(k) and prime(k+1), 0 < i < prime(k+1) - prime(k) - 1, and k > 1 is the smallest integer not used by previous terms of the sequence.
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2
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23, 31, 47, 53, 13, 23, 89, 113, 11, 139, 31, 37, 47, 53, 181, 199, 211, 53, 241, 11, 17, 23, 31, 283, 293, 317, 31, 337, 53, 359, 7, 13, 23, 389, 401, 409, 421, 13, 23, 29, 467, 59, 71, 23, 509, 523, 547, 17, 23, 31, 577, 47, 53, 7, 619, 631, 47, 53, 661, 17, 263, 691, 281, 709, 299, 7, 13, 323
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OFFSET
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1,1
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COMMENTS
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1) All lists of consecutive composites are checked in order. The first list is empty since for k=1 there are no numbers between prime(k)=2 and prime(k+1)=3.
2) The second list of composites corresponds to k=2 and is made by numbers between prime(k)=3 and prime(k+1)=5 which consists of the number 4 only. But for this number the smallest number z that satisfies "z mod 1 = 4 mod 1" is z=0 and since this is not greater than 1, z=0 is not in the sequence.
3) To find the first element of the sequence we must keep on increasing k until the list of i composites {C1,C2,C3,...,Ci} between prime(k) and prime(k+1) is such that the smallest number z that satisfies the integer congruences "z mod 1 = C1 mod 1", "z mod 2 = C1 mod 2", "z mod 3 = C3 mod 3", ... "z mod i = Ci mod i", is greater than 1.
The first k for which the previous condition is satisfied is k=9, as explained in the Example section. So the first term of the sequence is a(1)=23.
4) Following this algorithm we find that a(n=2)=31 with k=11, and a(n=11) =31 with k=36. The value 31 is also assigned to a(n=23) obtained with k=58. In fact for k=58 we have the i=5 composites 272, 273,274, 275 and 276 (between prime(58)=271 and prime(59)=277), and z=31 is the smallest number greater than 1 that satisfies the corresponding five congruences.
The four scatter plots of the first elements obtained exploring respectively the first 10^2, 10^3, 10^4 and 10^5 primes shows similar peculiar patterns which suggests self-similarity and quite probably a fractal structure (see plots in link).
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LINKS
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EXAMPLE
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The first term is obtained with k = 9 for which prime(9)=23, prime(10)=29, and its corresponding i=5 intermediate composites are 24, 25, 26, 27 and 28, then a(1) = z = 23 because this is the smallest integer that satisfies all five integer congruences:
z mod 1 = 24 mod 1,
z mod 2 = 25 mod 2,
z mod 3 = 26 mod 3,
z mod 4 = 27 mod 4,
z mod 5 = 28 mod 5.
With k = 10 the corresponding primes are prime(10)=29 and prime(11)=31, and the only (i=1) corresponding intermediate composite is 30, then the smallest z that satisfies z mod 1 = 30 mod 1 is z=0 which is not greater than 1 and then, k=10 cannot be used to generate a term of the sequence.
The second term is obtained with k = 11 for which prime(11)=31, prime(12)=37, and the corresponding i=5 intermediate composites are 32, 33, 34, 35 and 36, then a(2) = z = 31 because this is the smallest integer that satisfies:
z mod 1 = 32 mod 1,
z mod 2 = 33 mod 2,
z mod 3 = 34 mod 3,
z mod 4 = 35 mod 4,
z mod 5 = 36 mod 5.
a(7) = 89 is obtained with k = 24 for which prime(24)=89, prime(25)=97, and its corresponding i=7 intermediate composites are 90,91,...,95,96, then a(7) = z = 89 because this is the smallest integer that satisfies:
z mod 1 = 90 mod 1,
z mod 2 = 91 mod 2,
...
z mod 6 = 95 mod 6,
z mod 7 = 96 mod 7.
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MATHEMATICA
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nmax = 132 (* last explorative prime index *);
Select[Table[ChineseRemainder[Range[Prime[k]+1, Prime[k+1]-1], Table[j, {j, 1, Prime[k+1]-Prime[k]-1}]], {k, 2, nmax}], #>1&]
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PROG
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(PARI) lista(nn) = {for (n=1, nn, p = prime(n); nbc = nextprime(p+1) - p - 1; if (nbc, v = vector(nbc, k, p+k); vm = vector(nbc, k, Mod(v[k], k)); ij = lift(chinese(vm)); if (ij>1, print1(ij, ", ")); ); ); } \\ Michel Marcus, Sep 11 2016
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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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