A359751 Numbers m > 1 such that for all k > 1, m can be written as a product of factorials without using k!.
24, 576, 720, 2880, 13824, 17280, 40320, 69120, 241920, 331776, 362880, 414720, 518400, 725760, 967680, 1451520, 1658880, 2073600, 2903040, 3628800, 5806080, 7962624, 8294400, 8709120, 9953280, 12441600, 14515200, 17418240, 23224320, 29030400, 34836480, 39813120, 43545600
Offset: 1
Keywords
Examples
2880 is in the sequence via 2880 = (2!)^2 * 6! = 4!*5! = (2!)^2 * 3! * 5!. The factorials > 1 that are factors in a least one of these products are 2!, 3!, 4!, 5!, 6!. None of these factorials occur as factors in all of these products. For example, 2! no factor in 4!*5!, 3! no factor in 4!*5!, 4! no factor in (2!)^2 * 6!, 5! no factor in (2!)^2 * 6!, 6! no factor in 4!*5!. 24 is in the sequence (even though it is a factorial number) as 24 = 2! * 2! * 3! = 4!. So 24 can be written as a product of factorials in at least two ways (some of the factorials {2!, 3!, 4!}). But none of these factorials is in every factorization. 48 is NOT in the sequence as 48 = 2! * 2! * 2! * 3! = 2! * 4!. So 48 can be written as a product of factorials in at least two ways (some of the factorials {2!, 3!, 4!}). But 2! is a factor of every factorization. 12 is NOT in the sequence even though it can be written as a product of factorials, namely 2! * 3! = 12. As this is the only way to write 12 as a product of factorials, it is impossible to write 12 as a product of factorials without using 2!.
Links
- David A. Corneth, Table of n, a(n) for n = 1..2908 (terms <= 10^24)
Programs
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PARI
is(n) = { if(n == 1, return(0)); my(i, factorials, e, res, v); f = factor(n); if(prime(#f~) != f[#f~, 1], return(0); ); if(f[,2] != vecsort(f[,2],,4), return(0); ); factorials = List(); e = List(); res = List(); for(i = 2, oo, v = valuation(n, i!); if(v > 0, listput(factorials, i!); listput(e, v); , break ) ); forvec(x = vector(#e-1, i, [0, e[i+1]]), c = prod(i = 1, #e-1, factorials[i+1]^x[i]); if(c <= n && denominator(n/c) == 1&& 1 << logint(n/c, 2) == n/c, listput(res, concat([valuation(n/c, 2)], x)) ) ); for(i = 1, #e, p = 1; for(j = 1, #res, p*=res[j][i]; ); if(p != 0, return(0) ) ); 1 }
Comments