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%I #19 Apr 13 2014 12:40:22
%S 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,
%T 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,
%U 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
%N a(n) = Number of times n occurs in A234742.
%C First positions where the numbers n=2..10 occur are at: 91, 351, 1001, 3159, 7007, 15561, 24453, 60021, 63063. These factor in Z as: 7*13, 3*3*3*13, 7*11*13, 3*3*3*3*3*13, 7*7*11*13, 3*3*7*13*19, 3*3*11*13*19, 3*3*3*3*3*13*19, 3*3*7*7*11*13.
%C Cf. also A236845.
%H Antti Karttunen, <a href="/A236853/b236853.txt">Table of n, a(n) for n = 0..8192</a>
%F a(n) should have a direct formula computable from the prime factorization of n. See the example section, and comments in A236842 and formula/program code in A236862.
%e 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.
%e 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.
%e 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.
%e 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.
%e (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).
%e 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.
%o (Scheme, with memoizing definec-macro from _Antti Karttunen_'s IntSeq-library)
%o ;; Note that A234742(n) >= n which helps us here. Quadratic, but gives correct results. Macro definec memoizes the computed values for further consumption:
%o (definec (A236853 n) (if (zero? n) 1 (let loop ((k n) (ntimes 0)) (cond ((zero? k) ntimes) ((= (A234742 k) n) (loop (- k 1) (+ ntimes 1))) (else (loop (- k 1) ntimes))))))
%Y A236844 gives the positions of zeros, A236845 the positions of terms larger than one, A236842 the positions of nonzero terms.
%Y Cf. A236862, A236833.
%K nonn
%O 0,92
%A _Antti Karttunen_, Jan 31 2014
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