1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 2, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 2, 1, 0, 0
Offset: 0
For 3, which is prime in Z, but also irreducible in GF(2)[X] (i.e., it is one of the primes in A091206), we have k = 3 as only solution for A234732(k) = 3, thus a(3)=1.
For 5, which is prime in Z, but factors as 3 X 3 in GF(2)[X] (i.e., it is one of the primes in A091209), there cannot be any k such that A234742(k) = 5, thus a(5)=0.
For 91 = 7*13, both 7 and 13 are irreducible in GF(2)[X], but also the product 91 is (i.e., a term of A014580), this means that both k = 7 X 13 = 35 and k = 91 give such k that A234742(k) = 91, thus a(91)=2.
For 351 = 3*3*3*13, the following subsets of divisors from combinations for which the product of divisors = n, are such that every divisor is a term of A014580: (3*3*3*13), (3*117) and (351), and thus we have 3X3X3X13 = 75, 3X117 = 159 and 351 = 351 (itself in A014580), three different k such that A234741(k) = 351, so a(351) = 3.
(In contrast, the combinations like 9*39 (9X39 = 287) or 13*27 (13X27 = 175) result different A234741(175) = 119 and A234741(287) = 223 values than 351 because neither 9, 39 or 27 are in A014580).
For 1001, which factors as 7*11*13, the following subsets of divisors are such that the product of divisors = n and that every divisor is a term of A014580: (7,11,13), (11,(7*13)), (7,(11*13)), (7*11*13), and when these are multiplied with the carryless multiplication (A048720), we get 7 X 11 X 13 = 381, 11 X 91 = 565, 7 X 143 = 941 and 1001 = 1001, the four different k: 381, 565, 941 and 1001 such that A234742(k) = 1001. Thus a(1001) = 4.
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