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A246792
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Smallest number m such that for 0 < k < n+1, np(m+k-1) = np(m)-k+1, where np(t) is number of primes p with prime(t) < p < prime(t)^(1 + 1/t).
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3
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1, 7, 25, 25, 181, 208, 208, 1867, 14345, 19609, 40918, 40918, 620326, 2552265, 2552265, 7225612, 7225612, 16679492, 33772734, 33772734, 33772734, 620326386, 1516416904, 1516416904, 4764006481, 5272314878, 21423652192
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OFFSET
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1,2
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COMMENTS
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According to the definition, numbers np(a(n)), np(a(n)+1), ..., np(a(n)+n-2) and np(a(n)+n-1) are n consecutive numbers in descending order.
See A251736 for the corresponding values of np.
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LINKS
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EXAMPLE
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a(15) = 2552265, since np(2552265) = 24, np(2552265+1) = 23 , ..., np(2552265+13) = 11, np(2552265+14) = 10 are 15 consecutive numbers in descending order.
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MATHEMATICA
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np[t_] := np[t] = Length[Select[Range[Prime[t]+1, Prime[t]^(1+1/t)], PrimeQ]]; a[1]=1; a[n_] := a[n] = (For[m = a[n-1], c = Table[np[m+k-1], {k, n}]; c != Reverse[Range[Min[c], Max[c]]], m++]; m); Do[Print[a[n]], {n, 15}]
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PROG
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(PARI) np(n) = primepi(prime(n)^(1+1/n))-n;
isok(m, n) = {for (k=1, n, if (np(m+k-1) != np(m)-k+1, return (0)); ); return (1); }
a(n) = {m = 1; while (! isok(m, n), m++); m; } \\ Michel Marcus, Dec 07 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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