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A328946
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Product of primorials of consecutive integers (second definition A034386).
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0
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1, 1, 2, 12, 72, 2160, 64800, 13608000, 2857680000, 600112800000, 126023688000000, 291114719280000000, 672475001536800000000, 20194424296150104000000000, 606438561613387623120000000000, 18211350005250030322293600000000000, 546886840657658410578476808000000000000
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OFFSET
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0,3
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COMMENTS
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Similar to superprimorials (A006939), but a term of the sequence is a product of primorials of consecutive integers, not consecutive primes. So after 2# each primorial will repeat at least twice in the product. Also similar to superprimorials in that the exponents of the primes decrease linearly, but here it is linearly in p, not in pi(p).
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LINKS
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FORMULA
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a(n) = Product_{k=1..n} A034386(k) = Product_{p prime, p<=n} p^(n-p+1) = Product_{p prime} p^(max(n-p+1,0)) = Product_{p prime,p+k = n+1 and k >= 0} p^k.
a(n) = lcm(n, a(n-1)^2/a(n-2)). - Jon Maiga, Jul 08 2021
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EXAMPLE
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a(7) = 1# * 2# * 3# * 4# * 5# * 6# * 7# = 1*2*(2*3)*(2*3)*(2*3*5)*(2*3*5)*(2*3*5*7) = 2^6 * 3^5 * 5^3 * 7^1. Note that in the prime factorization the sum of each prime and its exponent is constant and equal to 7+1 = 8.
a(23) = 2^22 * 3^21 * 5^19 * 7^17 * 11^13 * 13^11 * 17^7 * 19^5 * 23^1. Here each prime and its exponent add to 24.
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MAPLE
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b:= proc(n) option remember; `if`(n=0, [1$2], (p-> (h->
[h, h*p[2]])(`if`(isprime(n), n, 1)*p[1]))(b(n-1)))
end:
a:= n-> b(n)[2]:
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MATHEMATICA
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b[n_] := b[n] = If[n==0, {1, 1}, Function[p, Function[h, {h, h p[[2]]}][If[ PrimeQ[n], n, 1] p[[1]]]][b[n - 1]]];
a[n_] := b[n][[2]];
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PROG
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(PARI) primo(n) = lcm(primes([2, n])); \\ A034386
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CROSSREFS
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Product of consecutive elements of A034386.
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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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