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A358235
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Number of ways n' (the arithmetic derivative of n) can be formed as a sum (x * y') + (x' * y) from two factors x and y of n, with x <= y, so that the said sum does not involve any carries when the addition is done in the primorial base.
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6
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1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 3, 2, 1, 1, 1, 1, 1, 4, 1, 3, 1, 2, 2, 2, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 1, 1, 3, 2, 1, 1, 1, 1, 2
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OFFSET
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1,4
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COMMENTS
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Number of ways n can be factored as x*y, with 1 <= x <= y, so that the sum (x * y') + (x' * y) is carry-free when the addition is done in the primorial base. Here u' stands for A003415(u), the arithmetic derivative of u.
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LINKS
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FORMULA
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a(n) = Sum_{d|n} [d <= (n/d) and A329041(d*A003415(n/d), A003415(d)*(n/d)) == 1], where [ ] is the Iverson bracket, and the dyadic function A329041 returns 1 only when its two arguments do not generate any carries when added together in the primorial base.
For all n >= 1, a(n) <= A038548(n). [See A358673 for the indices where the equality is attained].
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EXAMPLE
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a(6) = 2 because 6 has only two factor pairs, {1, 6} and {2, 3}, and for both of those pairs the criteria is satisfied, as we have A329041(1*A003415(6), A003415(1)*6) = A329041(5, 0) = 1 and A329041(2*A003415(3), A003415(2)*3) = A329041(2, 3) = 1. In the latter case the primorial base expansions of 2 and 3 are "10" and "11" (see A049345), which can be added together cleanly (i.e., without carries) to obtain "21" = A049345(2+3).
a(24) = 3 because 24 can be factored into two factors in four possible ways: 1*24, 2*12, 3*8 and 4*6, of which all others, except pair {4,6} are carry-free: we have A003415(6)*4 = 20 and A003415(4)*6 = 24, with respective primorial base expansions "310" and "400", which when added together, yield a carry at the third digit position from the right, because 3+4 = 7 > 4 (which is the max. allowed digit in that place), and therefore a(24) = 4-1 = 3.
a(63) = 3 because 63 can be factored into two factors in three possible ways: 1*63, 3*21 and 7*9. The trivial factorization always yields a carryless sum, and in this case also the other two factorizations satisfy the criteria: A329041(3*A003415(21), A003415(3)*21) = A329041(30, 21) = 1 (as 21 is "311" and 30 is "1000" in primorial base), and A329041(7*A003415(9), A003415(7)*9) = A329041(42, 9) = 1 (as 9 is "111", 42 is "1200", and 9+42 = 51 is "1311" in primorial base).
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PROG
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(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k, 2] = (f[k, 2]>=f[k, 1])); factorback(f); };
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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