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A102084
a(1) = 0; for n>0, write 2n=p+q (p, q prime), p*q maximal; then a(n)=p*q (see A073046).
8
0, 4, 9, 15, 25, 35, 49, 55, 77, 91, 121, 143, 169, 187, 221, 247, 289, 323, 361, 391, 437, 403, 529, 551, 589, 667, 713, 703, 841, 899, 961, 943, 1073, 1147, 1189, 1271, 1369, 1363, 1517, 1591, 1681, 1763, 1849, 1927, 2021, 1891, 2209, 2279, 2257, 2491
OFFSET
1,2
COMMENTS
For n>1, largest semiprime whose sum of prime factors = 2n. Assumes the Goldbach conjecture is true. Also the largest semiprime <= n^2.
Also the greatest integer x such that x' = 2*n, or 0 if there is no such x, where x' is the arithmetic derivative (A003415). Bisection of A099303. The only even number without an anti-derivative is 2. All terms are <= n^2, with equality only when n is prime. In fact a(n) = n^2 - k^2, where k is the least number such that both n-k and n+k are prime; k = A047160(n). It appears that the anti-derivatives of even numbers are overwhelmingly semiprimes of the form n^2 - k^2. For example, 1000 has 28 anti-derivatives, all of this form. Sequence A189763 lists the even numbers that have anti-derivatives not of this form. - T. D. Noe, Apr 27 2011
FORMULA
a(n) = n^2 - A047160(n)^2. - Jason Kimberley, Jun 26 2012
EXAMPLE
n=13: 2n = 26; 26 = 23 + 3 = 19 + 7 = 13 + 13; 13*13 = maximal => p*q = 13*13 = 169.
MATHEMATICA
f[n_] := Block[{pf = FactorInteger[n]}, If[Plus @@ Last /@ pf == 2, If[ Length[pf] == 2, Plus @@ First /@ pf, 2pf[[1, 1]]], 0]]; t = Table[0, {51}]; Do[a = f[n]; If[ EvenQ[a] && 0 < a < 104, t[[a/2]] = n], {n, 2540}]; t (* Robert G. Wilson v, Jun 14 2005 *)
Table[k = 0; While[k < n && (! PrimeQ[n - k] || ! PrimeQ[n + k]), k++]; If[k == n, 0, (n - k)*(n + k)], {n, 100}] (* T. D. Noe, Apr 27 2011 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael Taktikos, Feb 16 2005
EXTENSIONS
Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar
STATUS
approved