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A257042
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a(n) = (3*n+7)*n^2.
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1
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0, 10, 52, 144, 304, 550, 900, 1372, 1984, 2754, 3700, 4840, 6192, 7774, 9604, 11700, 14080, 16762, 19764, 23104, 26800, 30870, 35332, 40204, 45504, 51250, 57460, 64152, 71344, 79054, 87300, 96100, 105472, 115434, 126004, 137200, 149040, 161542, 174724
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OFFSET
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0,2
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COMMENTS
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Consider a natural number r such that r has 15 proper divisors and 5 prime factors (note that these prime factors do not have to be distinct). The difference between these two values, say d(r), is in this case 10. Where n is a positive integer, d(r^n)=(3*n+7)*n^2.
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LINKS
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FORMULA
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a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Apr 15 2015
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EXAMPLE
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The smallest integer that satisfies this is 120: it has 15 proper divisors (1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60) and 5 prime factors (2, 2, 2, 3, 5), so d(120)=10. The square of 120, 14400, we would expect to have a difference of 52 between the number of its proper divisors and prime factors, and with respectively 62 and 10, d(120)=52 indeed. Checking this with further integer powers of 120 will continue to generate terms in this sequence.
The integers which satisfy the proper-divisor-prime-factor requirement are those of A189975.
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MAPLE
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MATHEMATICA
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Table[(3 n + 7) n^2, {n, 40}] (* or *) CoefficientList[Series[(10 + 12 x - 4 x^2) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 15 2015 *)
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PROG
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(PARI) lista(nn) = {v = 1; while(!((numdiv(v)-1 == 15) && (bigomega(v) == 5)), v++); for (n=0, nn, vn = v^n; nb = numdiv(vn)-1-bigomega(vn); print1(nb, ", "); ); } \\ Michel Marcus, Apr 16 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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