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A329888
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a(n) = A329900(A329602(n)); Heinz number of the even bisection (even-indexed parts) of the integer partition with Heinz number n.
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9
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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
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OFFSET
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1,4
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COMMENTS
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Also the product of primes at even positions in the weakly decreasing list (with multiplicity) of prime factors of n. For example, the prime factors of 108 are (3,3,3,2,2), with even bisection (3,2), with product 6, so a(108) = 6.
Proof: A108951(n) gives a number with the same largest prime factor (A006530) and its exponent (A071178) as in n, and with each smaller prime p = 2, 3, 5, 7, ... < A006530(n) having as its exponent the partial sum of the exponents of all prime factors >= p present in n (with primes not present in n having the exponent 0). Then applying A000188 replaces each such "partial sum exponent" k with floor(k/2). Finally, A319626 replaces those halved exponents with their first differences (here the exponent of the largest prime present stays intact, because the next larger prime's exponent is 0 in n). It should be easy to see that if prime q is not present in n (i.e., does not divide it), then neither it is present in a(n). Moreover, if the partial sum exponent of q is odd and only one larger than the partial sum exponent of the next larger prime factor of n, then q will not be present in a(n), while in all other cases q is present in a(n). See also the last example.
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LINKS
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FORMULA
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a(n^2) = n for all n >= 1.
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EXAMPLE
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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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MATHEMATICA
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Table[Times@@Last/@Partition[Reverse[Flatten[Apply[ConstantArray, FactorInteger[n], {1}]]], 2], {n, 100}] (* Gus Wiseman, Oct 13 2021 *)
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PROG
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(PARI) A329888(n) = if(1==n, n, my(f=factor(n), m=1, p=0); forstep(k=#f~, 1, -1, while(f[k, 2], m *= f[k, 1]^(p%2); f[k, 2]--; p++)); (m)); \\ (After Wiseman's new interpretation) - Antti Karttunen, Sep 21 2021
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CROSSREFS
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The sum of prime indices of a(n) is A346700(n).
The odd non-reverse version is A346703.
The non-reverse version is A346704.
A001221 counts distinct prime factors.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A346633 adds up the even bisection of standard compositions.
A346698 adds up the even bisection of prime indices.
Cf. A000097, A035363, A102750, A236913, A253553, A344606, A344617, A344653, A345957, A345958, A345959.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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