a113047.txt, Kevin Ryde, June 2021 Sequence A113047 is a(n) = C(n) mod 3 for n>=0 = 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, ... where binomial(3n,n) C(n) = -------------- = A001764(n) 2n+1 Paul Barry conjectured that a(n) != 0 iff n = R(j) where 3^j - 1 R(j) = ------- = A003462(j) for j>=0 2 = ternary 111...111 repunit with j many 1s This is true, and further the terms a(n)!=0 are a(n)=1, never a(n)=2, so that a(n) is the characteristic function of the ternary repunits. It's convenient to take C(n) in the following form (per formula by Robert Ferreol in A001764), B(n) C(n) = ---- where B(n) = binomial(3n+1,n) = A045721(n) 3n+1 3n+1 == 1 mod 3 so it has no effect on the remainder mod 3, C(n) == B(n) mod 3 a(n) = 0 when B(n) is divisible by 3. Kummer showed that for a given prime p, the power p^e dividing binomial(x+y,y) is e = number of carries occurring when add x + y by digits in base p B(n) is x = n, y = 2n+1 so that the factors of 3 in B(n) are the number of carries when adding n and 2n+1 in ternary. If n has any ternary digit 2 then n = ... 2 ... ternary 2n+1 = ... [1 or 2] ... Digit 2 in n becomes either 1 or 2 in 2n+1 (depending on carry from below when forming 2n+1, and always 2 at the least significant digit). Digits 2+1 or 2+2 both give a carry and so B(n) has a factor of 3. If n has a 0 digit with a 1 above then v n = ... 1 0 ... ternary 1 above 0 2n+1 = ... 2 d ... has either d=0 or d=1 Digits 1 + 2 give a carry so B(n) has a factor of 3. The 0 in n stops any carry from below when forming 2n+1, so digit 1 in n doubles to 2 in 2n+1. This leaves only repunits. For them, n = 1 1 1 1 ternary repunit n = R(j), j>=0 2n+1 = 1 0 0 0 0 Each digit position is 0+1 so no carries. So B(n) has no factor of 3 and a(n) != 0 as per Paul Barry. To see that a(n) = 1 not 2 when n = R(j), consider the binomial as three factorials (3n+1)! B(n) = ------------ (2n+1)! * n! The product terms 1*2*3*...*n etc in these factorials have various factors of 3 but they cancel since B(n) != 0 mod 3. The remainder B(n) mod 3 is thus determined by the product of the least significant non-0 ternary digits of the product terms. LNZ(i) = lowest non-0 ternary digit of i Any product term with LNZ=1 is already 1 mod 3. Any two LNZ=2 multiply 2*2 == 1 mod 3. The numerator and denominator can be treated together since 2 is its own multiplicative inverse, 2^(-1) == 2 mod 3 So the question is whether an odd or even number of LNZ=2 among all three factorials. Let twos(n) = number of LNZ(i)=2 among i=1..n inclusive = A189672(n) Clark Kimberling in A189672 gives a recurrence for twos(n), based on how many i have least significant digit 2, plus how many i have one or more low 0 digits before LNZ=2, twos(n) = twos(floor(n/3)) + floor((n+1)/3) Expanding repeatedly is the following sum. Each sum term is how many of i = 1..n have LNZ=2 with exactly k low 0s below that 2. / floor(n/3^k) + 1 \ twos(n) = Sum floor| ---------------- | k>=0 \ 3 / For n = R(j), these floors become, written in ternary, n = 1111 2n+1 = 10000 3n+1 = 11111 ternary twos = 111 twos = 1000 twos = 1111 k=0 + 11 + 100 + 111 k=1 + 1 + 10 + 11 ... + 1 + 1 A number in ternary is odd or even according as its number of 1 digits is odd or even. The total 1 digits among twos(n), twos(2n+1), and twos(3n+1) is even since the triangle of 1s in 3n+1 is exactly the combination of the triangle of 1s in n plus the diagonal in 2n+1. twos(n) + twos(2n+1) + twos(3n+1) == 0 mod 2 so B(n) == 1 mod 3 for n = R(j) ------------------------------------------------------------------------------