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 A329152 a(n) = Sum_{i=1..n-1} Sum_{j=1..i-1} [1 == i*j (mod n)], where [] is the Iverson bracket. 2
 0, 0, 0, 0, 1, 0, 2, 0, 2, 1, 4, 0, 5, 2, 2, 2, 7, 2, 8, 2, 4, 4, 10, 0, 9, 5, 8, 4, 13, 2, 14, 6, 8, 7, 10, 4, 17, 8, 10, 4, 19, 4, 20, 8, 10, 10, 22, 4, 20, 9, 14, 10, 25, 8, 18, 8, 16, 13, 28, 4, 29, 14, 16, 14, 22, 8, 32, 14, 20, 10, 34, 8, 35, 17, 18, 16, 28, 10, 38, 12 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,7 COMMENTS In other words, a(n) is the number of choices for k, 1 <= k <= n, so that k*n + 1 has the form a*b for 1 < a < b < n. Solutions exist only when 1 <= k <= n - 3. Any odd number greater than 3 has at least one solution, 2*ceiling(n/2). Observations: - Unique values of a(n) exist, 6 = a(32), 24 = a(240), 126 = a(512), 144 = a(1320), ..., and n mod 8 = 0. - If a(n) is an odd prime then a(n) = a(2*n) OR a(n) = a(n/2). For example, 1013 = a(2029) = a(2197) = a(4058) = a(4394). LINKS EXAMPLE a(1)=0 because there is no solution to k*1 + 1 = a*b, 1 < a < b < n, 1 <= k < n. a(5)=1 because 1 == 3*2 (mod 5). a(7)=2 because 1 == 4*2 == 5*3 (mod 7). a(11)=4 because 1 == 4*3 == 6*2 == 8*7 == 9*5 (mod 11). a(13)=5 because 1 == 7*2 == 8*5 == 9*3 == 10*4 = 11*6 (mod 13). a(32)=6 because 1 == 11*3 == 13*5 == 23*7 == 25*9 == 27*19 == 29*21 (mod 32). MATHEMATICA Array[Sum[Sum[Boole[Mod[i j, #] == 1], {j, i - 1}], {i, # - 1}] &, 80] (* Michael De Vlieger, Mar 15 2020 *) PROG (PARI) a(n) = {my(x=0); for (i = 1, n - 1, for (ii = 1, i - 1, if(1 == ((ii*i) % n), x++))); return(x)} for (n = 1, 100, print1(a(n), ", ")) CROSSREFS Sequence in context: A305804 A053470 A308202 * A337563 A278648 A353508 Adjacent sequences:  A329149 A329150 A329151 * A329153 A329154 A329155 KEYWORD nonn AUTHOR Torlach Rush, Feb 25 2020 STATUS approved

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Last modified July 6 14:10 EDT 2022. Contains 355110 sequences. (Running on oeis4.)