Sequence

[Kelvin Voskuijl]

Hello fellow SeqFans,

Currently working on this sequence, but I am struggling to find a good name for it 

https://oeis.org/draft/A397191

This sequence is also the product of  Sum_{k>=1} 1/(k!)^n.

Kelvin Voskuijl

[Amiram Eldar]

How about "Decimal expansion of Product_{m>=1} (Sum_{k>=1} (1/k!)^m)." ?

[Arthur O'Dwyer]

How about

A397191 "Decimal expansion of Product_{m>=1} Sum_{k>=1} 1/(k!)^m = 2.809985..."

?

At the very least, you should put that in the "Comments" field.

Personally I think "decimal expansion of..." sequences are uninteresting, and should be grandfathered in only if there is something special about the particular real number in question; like, "Decimal digits of pi = 3.1415..." is plausible. But there's nothing intrinsically interesting about the digits of the decimal expansion of most numbers, any more than their binary expansion or their name written out in English or whatever. 

If there's something interesting about this particular real number that led you to "work on" it, you should say that in the Comments field.

Also, you shouldn't take as gospel my assumption that the real number is two-point-something; I didn't check; for all I know it's zero-point-two-something or twenty-eight-point-something or... This should also be in the Comments field.

–Arthur

[K. Voskuyl]

Thanks, I've added that name

Op do 25 jun 2026 om 16:11 schreef Amiram Eldar <amiram...@gmail.com>:

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[K. Voskuyl]

I was inspired to do this sequence by  A021002

Op do 25 jun 2026 om 16:13 schreef Arthur O'Dwyer <arthur....@gmail.com>:

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[Sean A. Irvine]

We do need to be mindful that we are supposed to be an integer sequence database. Here is what I use when making acceptance decisions on constants:

1. It appears extensively in the mathematical literature.

or

2. It is the asymptotic limit or has some other direct association with one or more integer sequences.

or

3. It can be usefully and naturally expressed as some kind of sum or product of integers.

For 2 and 3, this means you should be able to cross-reference integer sequences associated with the constant. (Well actually this is true for nearly any new submission!)

The representation as decimal digits is a pragmatic choice made for overall consistency.

As usual, the other editors might have different points of view.

Sean.

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[Neil Sloane]

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

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[Neil Sloane]

There is another reason for accepting a sequence that might otherwise be borderline:  if it has a clever (I'm tempted to say "witty") definition, or if it makes you laugh, that's always a reason for accepting any sequence (within reason)!

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

On Thu, Jun 25, 2026 at 4:12 PM Sean A. Irvine <sai...@gmail.com> wrote:

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[Georg Fischer]

In

"A273050 Numbers k such that (rol(k) XOR ror(k)) = k"

a recurrence with signature (0, 9, 0, -8) is conjectured,
while in

"A273180 Numbers n such that ror(n) + rol(n) is a power of 2"

a recurrence with signature (0,0,17,0,0,-16) is given
without the "conjectured" attribute.

I would be interested in a proof or disproof for both cases.
Maybe a good exercise for math classes?

Regards - Georg

[Gareth McCaughan]

On 25/06/2026 22:37, Georg Fischer wrote:
> In
>
>   "A273050 Numbers k such that (rol(k) XOR ror(k)) = k"
>
> a recurrence with signature (0, 9, 0, -8) is conjectured,
> while in
>
>   "A273180 Numbers n such that ror(n) + rol(n) is a power of 2"
>
> a recurrence with signature (0,0,17,0,0,-16) is given
> without the "conjectured" attribute.
>
> I would be interested in a proof or disproof for both cases.

Let's take a look at the first one. We're looking for n such that rol(n)
XOR n XOR ror(n) is zero. Obviously we want to think of n not as a
binary number but as a mere sequence of bits, with the restriction that
the most significant bit is a 1.

That is, we have bits a0, a1, ..., ak (corresponding to the 1, 2, 4,
..., 2^k places in the binary expansion of n), we require ak = 1, and
the further constraint is that a(i-1) + a(i) + a(i+1) = 0 mod 2, where
all the indices are mod k+1.

Once we have two consecutive bits, all the others are forced, so the
only possible bit-patterns besides all-zero are 110110110... and
101101101..., and so k+1 must be a multiple of 3 to make things line up
nicely as we wrap between 0 and k.

That is, in binary our sequence goes:

0
101 110
101101 110110
101101101 110110110

and so on. Or, to write it a bit differently: 0, 5*1, 6*1, 5*(1+8),
6*(1+8), 5*(1+8+8^2), 6*(1+8+8^2), 5*(1+8+8^2+8^3), etc.

The claimed recurrence relations have "period 2", which means that what
we're really trying to verify is that the sequences

0, 5*1, 5*(1+8), 5*(1+8+8^2), 5*(1+8+8^2+8^3), ... and
0, 6*1, 6*(1+8), 6*(1+8+8^2), 6*(1+8+8^2+8^3), ...

satisfy the recurrence a(n) = 9 a(n-1) - 8 a(n-2). Well, the values in
either of those sequence are constant linear combinations of (8^n) and
(1^n), so we need the recurrence corresponding to the quadratic equation
(t-8)(t-1)=0, which is indeed t^2 - 9t + 8 = 0. So indeed the claimed
recurrence is correct.

The explicit expressions for the odd and even cases conjectured by Colin
Barker definitely have the right shape; I haven't checked that they are
actually correct but if they aren't it's probably because someone's made
a typo. The unified expression for both cases might for all I know be
correct, but it seems like an ugly kludge whose only purpose is to
combine two separate formulae into one. I would suggest a cleaner
unified expression would be a(n) = (8^floor(n/2)-1)/7 * (5 if n even, 6
if n odd) or a(n) = (8^floor(n/2)-1)/7 * (5 + (n rem 2)), and I think
trying to remove the floor / rem 2 stuff obscures rather than enlightens.

--
g

[Robert Dougherty-Bliss]

Gareth explained how the first recurrence works. The second recurrence is
similar, but perhaps slightly more involved.

Of course rol(n) + ror(n) always has its most-significant bit set to 1, meaning
that it is a power of 2 only if all other bits are made 0 in the addition.

If we write n = (a(k), a(k-1), ..., a(0))_2, then the ith bit in rol(n) +
ror(n) is a(i - 1) + a(i + 1) + carry(i), where carry(i) is 1 if a carry
occurred in the addition and 0 otherwise. (This addition is done mod 2.)

Say that n has a few bits to get the idea down. The 0th bit is a(k) + a(1) = 1
+ a(1), so we need a(1) = 1. This implies that there is a carry after the
first addition.

The 1st bit is a(2) + a(0) + 1, so we must have a(0) + a(2) = 1. This means
that a(0) and a(2) have different values. But one of them is 1, so this
addition also produces a carry.

The 2nd bit is a(3) + a(1) + 1, so we must have a(3) + a(1) = 1. Again, a(3)
and a(1) must have different values, and again we produce a carry.

This keeps going. The conclusion is that the bits of n 'alternate' along
parities, i.e., a(k) = 1 + a(k - 2) = a(k - 4) = 1 + a(k - 6) = ... and a(k -
1) = 1 + a(k - 3) = a(k - 5) = ..., with boundary conditions a(k) = a(1) = 1.

If k is even, then 1 and k are in different parity classes, so n is completely
determined by alternating bits. The possibilities are:

    k = 0: 1 (1)
    k = 2: 110 (6)
    k = 4: 10011 (19)
    k = 6: 1100110 (102)
    k = 8: 100110011 (307)

A formula for these numbers is given by

    (2^k + 2^(k - 4) + ... + 2^(k mod 4))
        + (2^1 + 2^5 + 2^9 + ... 2^(k - 1 - 2[k mod 4 = 0])).

If k = 0 (mod 4), then this is (6/5) (2^k - 1).
If k = 2 (mod 4), then this is (2 / 5) (2^(k + 2) - 1).

If k is odd, then 1 and k are in the same parity class. The bits of n will
alternate (k - 1) / 2 times between 1 and k, and since a(k) = a(1) = 1, it
follows that (k - 1) / 2 is even. This also implies a(0) = a(k - 1). The (k -
1)th bit of rol(n) + ror(n) is therefore a(0) + a(k - 1) + 1 = 1. For rol(n) +
ror(n) to be a power of 2, there cannot be a carry in this sum, so a(0) = a(k -
1) = 0. The possibilities are:

    k = 1: 10 (2)
    k = 3: (doesn't exist)
    k = 5: 100110 (38)
    k = 7: (doesn't exist)
    k = 9: 1001100110 (614)

A formula for these numbers is

    (2^k + 2^(k - 4) + ... + 2^1)
        + (2^2 + 2^6 + ... + 2^(k - 3))
    = (2/5) (3 * 2^k - 1).

For every k >= 0, there is at most one integer n with (k + 1) bits such that
rol(n) + ror(n) is a power of 2. The numbers cycle through the above formulas
in order. Each formula satisfies a linear recurrence with constant
coefficients, so the whole sequence does as well.

Robert

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