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%I #17 Apr 26 2024 05:18:26
%S 2,4,6,10,15,22,33,44,55,68,85,102,119,145,174,203,232,261,296,333,
%T 370,410,451,492,533,590,649,708,767,826,885,944,1005,1072,1139,1207,
%U 1278,1358,1455,1552,1649,1746,1843,1940,2037,2134,2231,2328,2425,2540,2667
%N a(n) = max(prime(i)*(n+1-i) | 1 <= i <= n).
%C Smallest m such that m! is divisible by product_{k=0..n} A002110(k).
%C Conjecture: define an operator P(n) which when applied to a sequence s(1),s(2),s(3),... leaves s(1),...,s(n) fixed and 1) sorts in descending order if n is prime; or 2) sorts in ascending order if n is not prime every n consecutive terms thereafter. Apply P(2) to 1,2,3,... to get PS(2), then apply P(3) to PS(2) to get PS(3), then apply P(4) to PS(3), etc. The limit of PS(n) is b(n). Let c(n) = b(A000040(n) + 1), then a(n) = c(n-1) for n > 1 with a single exception at n = 5. - _Mikhail Kurkov_, May 11 2022
%e For n = 5, we take the first 5 primes in ascending order and multiply them by the numbers from 5 to 1 in descending order: 2*5 = 10 3*4 = 12 5*3 = 15 7*2 = 14 11*1 = 11. The largest product is 15, so a(5) = 15.
%o (PARI) a(n) = {ret = 0; for (i=1, n, ret = max (ret, prime(i)*(n+1-i));); return (ret);} \\ _Michel Marcus_, Jun 16 2013
%Y Cf. A073819, A073820.
%K easy,nonn
%O 1,1
%A _David Wasserman_, Aug 13 2002
%E Thanks to _Naohiro Nomoto_ for correcting an editing error, Jun 01 2003
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