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A135369
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Let n! = p(1)^b(1)...p(r)^b(r) be the prime factorization of n!. Then a(n) = sum_(i=1..r)p(i)+b(i)
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1
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1, 2, 3, 7, 9, 15, 17, 25, 28, 30, 32, 44, 47, 61, 63, 65, 69, 87, 90, 110, 113, 115, 117, 141, 145, 147, 149, 152, 155, 185, 188, 220, 225, 227, 229, 231, 235, 273, 275, 277, 281, 323, 326, 370, 373, 376, 378, 426, 431, 433, 436, 438, 441, 495, 499, 501, 505
(list; graph; refs; listen; history; internal format)
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