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2, 6, 1, 2, 0, 0, 0, 7, 4, 0, 4, 3, 4, 5, 2, 6, 0, 6, 4, 4, 3, 7, 3, 7, 1, 1, 3, 0, 9, 5, 4, 4, 5, 6, 7, 2, 4, 3, 3, 4, 0, 4, 5, 8, 7, 3, 7, 0, 9, 3, 8, 2, 6, 6, 0, 9, 3, 5, 1, 0, 8, 0, 6, 0, 5, 1, 5, 6, 0, 4, 1, 0, 8, 8, 7, 4, 9, 3, 0, 1, 3, 6, 2, 5, 1, 3, 6
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OFFSET
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1,1
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COMMENTS
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With this constant f(1) and using the formula f(n+1) = (floor(f(n))*(f(n))) - ((floor(f(n)))^2 - floor(f(n))) it is possible to obtain the prime numbers repeated exactly a number of times corresponding to the position of the prime number. That is, 2 once, 3 twice, 5 thrice, etc.
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LINKS
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FORMULA
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Equals 2 + (3-2)/(2) + (5-3)/(2*3^2) + (7-5)/(2*3^2*5^3) + (11-7)/(2*3^2*5^3*7^4) + ...
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EXAMPLE
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2.61200074043...
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MATHEMATICA
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imax:=87; First[RealDigits[N[2+Sum[(Prime[i]-Prime[i-1])/Product[Prime[j-1]^(j-1), {j, 2, i}], {i, 2, imax}], imax]]] (* Stefano Spezia, Dec 16 2020 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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