1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4, 1, 3, 1, 2, 3, 2, 1, 4, 5, 2, 3, 2, 1, 3, 1, 4, 3, 2, 5, 6, 1, 2, 3, 4, 1, 3, 1, 2, 3, 2, 1, 4, 7, 5, 3, 2, 1, 6, 5, 4, 3, 2, 1, 6, 1, 2, 3, 8, 5, 3, 1, 2, 3, 5, 1, 6, 1, 2, 5, 2, 7, 3, 1, 4, 9, 2, 1, 6, 5, 2, 3, 4, 1, 6, 7, 2, 3, 2, 5, 8, 1, 7, 3, 10, 1, 3, 1, 4, 5
Offset: 1
From _Gus Wiseman_, Aug 15 2021: (Start)
The list of all numbers with image 12 and their corresponding prime factors begins:
144: (3,3,2,2,2,2)
216: (3,3,3,2,2,2)
240: (5,3,2,2,2,2)
288: (3,3,2,2,2,2,2)
336: (7,3,2,2,2,2)
360: (5,3,3,2,2,2)
(End)
The positions from the left are indexed as 1, 2, 3, ..., etc, so e.g., for 240 we pick the second, the fourth and the sixth prime factor, 3, 2 and 2, to obtain a(240) = 3*2*2 = 12. For 288, we similarly pick the second (3), the fourth (2) and the sixth (2) to obtain a(288) = 3*2*2 = 12. - _Antti Karttunen_, Oct 13 2021
Consider n = 11945934 = 2*3*3*3*7*11*13*13*17. Its primorial inflation is A108951(11945934) = 96478365991115908800000 = 2^9 * 3^8 * 5^5 * 7^5 * 11^4 * 13^3 * 17^1. Applying A000188 to this halves each exponent (floored down if the exponent is odd), leaving the factors 2^4 * 3^4 * 5^2 * 7^2 * 11^2 * 13^1 = 2497294800. Then applying A319626 to this number retains the largest prime factor (and its exponent), and subtracts from the exponent of each of the rest of primes the exponent of the next larger prime, so from 2^4 * 3^4 * 5^2 * 7^2 * 11^2 * 13^1 we get 2^(4-4) * 3^(4-2) * 5^(2-2) * 7^(2-2) * 11^(2-1) * 13^1 = 3^2 * 11^1 * 13^1 = 1287 = a(11945934), which is obtained also by selecting every second prime from the list [17, 13, 13, 11, 7, 3, 3, 3, 2] and taking their product. - _Antti Karttunen_, Oct 15 2021
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