a019469.txt, Kevin Ryde, May 2021 Sequence A019469 is numbers k for which k does not divide / 2k-4 \ B(k) = binomial | | = A000984(k-2) \ k-2 / Wouter Meeussen conjectured that these k are 3*A096304 and the powers of 2. This is true. Kummer showed that for a given prime p, the power p^e dividing binomial(x+y,y) = (x+y)! / (x! * y!) is e = number of carries occurring when add x + y by digits in base p B(k) here is x = y = k-2 so its prime power is the number of carries when doubling k-2 in base p. For example in base p=5 a carry occurs on doubling digit 3 or 4, and propagates on doubling 2 (but does not begin at 2). digit base 5 ----- ------ 4 always carry: 4+4 = 3 carry 1 3 always carry: 3+3 = 1 carry 1 2 carry does not begin: 2+2 = 4 no carry but does propagate: 1 + 2+2 = 0 carry 1 0, 1 never carry: eg. 1 + 1+1 = 3 no carry The prime factors in B(k) can be compared to the prime factors in k. If k has more of a given p than B(k) does then k does not divide B(k) and k is a term of the sequence. Factors of p in k are low 0s when k is written in base p. In base p=5, if k has one or more low 0s then k = ... w 0 0 0 0 some digit w!=0 k-2 = ... w-1 4 4 4 3 \-----/ Each of the low 0s becomes a low digit 3 or 4 in k-2 and each is a carry on doubling, so there are always enough factors of 5 in B(k) for the factors of 5 in k. The same argument holds for any prime p > 5. For p=2, if k is a power of 2 with one or more low 0s then k = 1 0 0 0 0 low 0s, power of 2 k-2 = 0 1 1 1 0 \-----/ Doubling k-2 in binary causes a carry out of each 1-bit. But k-2 has one fewer 1-bits than k has trailing 0s, so k does not divide B(k). If instead k is not a power of 2 then h w k = 1 ... 1 0 0 0 0 low 0s, not a power of 2 k-2 = 1 ... 0 1 1 1 0 \-----/ k has at least one more 1-bit above position w, such as the most significant bit h. All bits above w are unchanged in k-2, so k-2 has at least as many 1-bits as k has low 0s. For p=3, if k has exactly one low 0 then k = ... w 0 some digit w!=0 k-2 = ... w-1 1 On doubling k-2, a carry begins only at a digit 2. Both w-1 <= 1 and the low 1 shown are not digit 2. The digits of k above w are unchanged in k-2 and if none of them are 2 then k does not divide B(k). If k has two or more low 0s then k = ... w 0 0 0 0 some digit w!=0 k-2 = ... w-1 2 2 2 1 \---/ k-2 has low 2s which are carries on doubling, but one fewer than the low 0s of k. If w=2 so that w-1 = 1 then the carry up from the highest 2 shown will propagate there (1 + 1+1 = 0 carry 1), so that k divides B(k). Or digits above w in k are unchanged in k-2 and if any of them are 2 then that also ensures divisible. But if no digit 2 at or above w then k does not divide B(k). The p=3 cases together show that k does not divide B(k) when k is a multiple of 3, with an arbitrary second least significant digit, but no digit 2 at or above its third least significant digit. So ternary digits of the form high low k = [0 or 1] ... [0 or 1] [0 or 1 or 2] [0] ternary which is 3*A096304 as per Wouter Meeussen. These k are multiples of 3 so are disjoint from the powers of 2 which are also terms. ------------------------------------------------------------------------------