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A374717 For n a power of 2, a(n) = n. Otherwise let k = n - 2^j (> 0) where 2^j is the greatest power of 2 not exceeding n, then a(n) = least novel m*a(k); m a term in A033844.

%I A374717 #38 Aug 23 2025 23:14:05
%S A374717 1,2,3,4,7,6,9,8,19,14,21,12,49,18,27,16,53,38,57,28,133,42,63,24,361,
%T A374717 98,147,36,343,54,81,32,131,106,159,76,371,114,171,56,1007,266,399,84,
%U A374717 931,126,189,48,2809,722,1083,196,2527,294,441,72,6859,686,1029,108,2401,162,243,64,311
%N A374717 For n a power of 2, a(n) = n. Otherwise let k = n - 2^j (> 0) where 2^j is the greatest power of 2 not exceeding n, then a(n) = least novel m*a(k); m a term in A033844.
%C A374717 Definition developed from the Doudna recursion A005940, see Example. Conjectured to be a permutation of A318400, numbers whose prime factorization consists only of primes with indices 2^k (terms in A033844).
%C A374717 From _David James Sycamore_, Aug 09 2024: (Start)
%C A374717 The even bisection when divided by 2 returns the sequence. The odd bisection when transformed by replacing all factors prime(2^k) in a(2*n-1) with prime(2^(k-1)) also returns the sequence (similar to properties of A005940). The sequence is fixed on numbers of the form 2^n or 3*2^n (A029744), since by the definition points at 2^n are already named as such, and if n = 3*2^r then the powers of 2 adjacent to n are 2^r < 2^(r+1) < 3*2^r < 2^(r+2), from which, by the definition we find k = 2^(r+2) - 3*2^r = 2^r(3 - 2) = 2^r, which is a fixed point so a(k) = 2^r, and a(n) is least novel m*a(k); m a term in A033844. Since 2^r is already a term the required m is 3, so a(n) = 3*2^r = n (compare with fixed points of A005940). (End)
%C A374717 Definition analogous to the Name of A005940: Let c_i = number of 1's in binary expansion of n-1 that have i 0's to their right, and let p(j) = j-th prime. Then a(n) = Product_i p(2^i)^c_i. - _Michael De Vlieger_, Aug 09 2024
%H A374717 Michael De Vlieger, <a href="/A374717/b374717.txt">Table of n, a(n) for n = 1..16384</a>
%H A374717 Michael De Vlieger, <a href="/A374717/a374717.png">Log log scatterplot of a(n)</a>, n = 1..2^14, showing primes in red, perfect prime powers in gold, squarefree composites in green, and with blue and purple representing numbers neither squarefree nor prime powers, where purple represents powerful numbers that are not prime powers.
%H A374717 Michael De Vlieger, <a href="/A374717/a374717_1.png">Fan style binary tree showing a(n)</a>, n = 1..8192 using essentially the same color function as immediately above, with exception that composite primorials appear in bright green.
%F A374717 a(2^k) = 2^k, a(3*2^k) = 3*2^k.
%F A374717 a(2^k-1) = 3^(k-1), k >= 1; a(2^k+1) = A033844(k+1); k >= 0.
%e A374717 a(1) = 1, a(2) = 2 because both are powers of 2. a(3) = 3 since for n = 3, k = 1, a(1) = 1 and m = 3. a(4) = 4 because 4 is a power of 2
%e A374717 For a(5), k = 1, a(1) = 1 and therefore a(5) = 1*7 since 7 is least term in A033844 not already used.
%e A374717 Whereas the Name defines each individual term recursively, the following procedure describes a recursion for generating the first 2^k terms from the first 2^(k-1) terms: Let S(0) = {1}, and then S(k) = {2*S(k-1), S(k-1)}, where 2*S(k-1) means twice every term in S(k-1). Thus from
%e A374717        S(0) = {1} we obtain:
%e A374717        S(1) = {2,1},
%e A374717        S(2) = {4,2,2,1},
%e A374717        S(3) = {8,4,4,2,4,2,2,1},
%e A374717        S(4) = {16,8,8,4,8,4,4,2,8,4,4,2,4,2,2,1} etc.
%e A374717 Convert these (indices) to primes as follows:
%e A374717        P(0) = {2},
%e A374717        P(1) = {3,2},
%e A374717        P(2) = {7,3,3,2},
%e A374717        P(3) = {19,7,7,3,7,3,3,2},
%e A374717        P(4) = {53,19,19,7,19,7,7,3,19,7,7,3,7,3,3,2}, etc.
%e A374717 Set U(0) = 1 and U(k) = U(k-1)*P(k-1) prepended by U(k-1), thus:
%e A374717        U(0) = {1},
%e A374717        U(1) = {1,2},
%e A374717        U(2) = {1,2,3,4},
%e A374717        U(3) = {1,2,3,4,7,6,9,8},
%e A374717        U(4) = {1,2,3,4,7,6,9,8,19,14,21,12,49,18,27,16}, etc.
%e A374717 Thus U(k) gives the first 2^k terms of the sequence because the primes in P(k) are the greatest prime factors of the corresponding terms.
%e A374717 From _Michael De Vlieger_, Aug 09 2024: (Start)
%e A374717 Using the alternative binary definition:
%e A374717 For n = 9, n-1 = 1000_2; c_3 = 1, hence a(9) = prime(2^3)^1 = 19.
%e A374717 For n = 10, n-1 = 1001_2; c_0 = 1, c_2 = 1; a(10) = prime(2^0)^1 * prime(2^2)^1 = 2*7 = 14.
%e A374717 For n = 11, n-1 = 1010_2; c_1 = 1, c_2 = 1; a(11) = prime(2^1)^1 * prime(2^2)^1 = 3*7 = 21.
%e A374717 For n = 12, n-1 = 1011_2; c_0 = 2, c_1 = 1; a(12) = prime(2^0)^2 * prime(2^1)^1 = 2^2*3 = 12.
%e A374717 For n = 13, n-1 = 1100_2; c_2 = 2; a(13) = prime(2^2)^2 = 7^2 = 49.
%e A374717 For n = 2^k + 2^(k-1) = 3*2^(k-1), n-1 = 2^(k+1) - 2^(k-1) - 1.
%e A374717   c_0 = k-1, c_1 = 1, therefore we have fixed point a(3*2^(k-1)) = 3*2^(k-1). (End)
%t A374717 Block[{a, c, k, m, t, nn}, nn = 2^7; c[_] = False;
%t A374717 Do[Set[{m, k}, {1, n - 2^Floor[Log2[n]]}];
%t A374717   If[k == 0,
%t A374717     Set[{a[n], c[n]}, {n, True}],
%t A374717     While[Set[t, Prime[2^m] a[k]]; c[t], m++];
%t A374717       Set[{a[n], c[t]}, {t, True}]], {n, nn}];
%t A374717 Array[a, nn] ] (* _Michael De Vlieger_, Aug 06 2024 *)
%Y A374717 Cf. A005940, A029744, A033844, A318400.
%K A374717 nonn,changed
%O A374717 1,2
%A A374717 _David James Sycamore_, Jul 17 2024