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.

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