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A073818
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a(n) = max(prime(i)*(n+1-i) | 1 <= i <= n).
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5
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2, 4, 6, 10, 15, 22, 33, 44, 55, 68, 85, 102, 119, 145, 174, 203, 232, 261, 296, 333, 370, 410, 451, 492, 533, 590, 649, 708, 767, 826, 885, 944, 1005, 1072, 1139, 1207, 1278, 1358, 1455, 1552, 1649, 1746, 1843, 1940, 2037, 2134, 2231, 2328, 2425, 2540, 2667
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OFFSET
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1,1
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COMMENTS
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Smallest m such that m! is divisible by product_{k=0..n} A002110(k).
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
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LINKS
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EXAMPLE
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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.
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PROG
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(PARI) a(n) = {ret = 0; for (i=1, n, ret = max (ret, prime(i)*(n+1-i)); ); return (ret); } \\ Michel Marcus, Jun 16 2013
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CROSSREFS
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KEYWORD
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easy,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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