A236853 a(n) = Number of times n occurs in A234742.

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

Author

Antti Karttunen, Jan 31 2014

Keywords

• nonn

Comments

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.

Cf. also A236845.

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192

Crossrefs

A236844 gives the positions of zeros, A236845 the positions of terms larger than one, A236842 the positions of nonzero terms.

Cf. A236862, A236833.

Formula

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.