next number in this infinite sequence?

[4=5]

so I was reading GEB and I thought of a cool sequence
1 3 8 20 48 112 256

OEIS doesn't have it so I was wondering if any of you can find the pattern

[Robin S]

Should the sixth and seventh terms be 112 and 256? If so, then I know how the sequence was generated, though I haven't yet got an explicit formula.

[4=5]

yes, thank you.
hmm it turns out OEIS does have it

[Robin S]

Here's one possible way of obtaining the sequence:

Spoiler:

Given the n^th term, double it and add 2^n-1 to get the n+1^th term.

The sequence does appear in OEIS, here.

[Cauchy]

Spoiler:

So in general, letting s₀ = 1 and s₁ = 3, we get that s_n = (n+2)2^n-1.

[Moonbeam]

I'm no expert at maths (I don't even know what OEIS is lol ) but I found the sequence pretty easy to work out ....

Spoiler:

Starting with the first number = 1 .... double it and add 1 to get the 2nd number = 3.
Double this and add 2 to give 8.
Double this and add 4 to give 20.
Double this and add 8 to give 48.
Double this and add 16 to give 112.
Double this and add 32 to give 256.
Double this and add 64 to give (*taps away on calculator*) 576.

[Hix]

The partial sums of this finite sequence are:
1 4 12 32 80 192 448
so the nth partial sum seems to be:
n*2^n-1

This means the nth partial sum represents

Spoiler:

the number of "1s" used when writing out the first 2ⁿ nonnegative integers in binary, since each of the n binary place values will be "1" exactly half of the time on this range.

Since that is what the partial sums represent, then the nth term a_nin the original sequence must represent the number of "1s" used when writing out only the integers from 2^n-1 to 2ⁿ-1 (i.e. the "n bit long" integers).

Thus we would expect the sequence to satisfy the recurrence (found by Robin S and Moonbeam):
a₁=1
a_n+1=2*a_n+2^n-1
for n>0, since the collection of all n+1 bit long integers is just the collection of all n bit long integers that have a "0" or "1" tacked on to the end.

I'm curious about the connection to GEB; does this sequence have something to do with Hofstadter's "G diagram" or "H diagram"? That's the only thing I could think of.

[4=5]

it has very little to do with GEB, I just thought of it while reading it and trying to figure out what this meant

Spoiler:

the reason I found it interesting was because when you find the difference and then the difference of the difference and then the difference of the difference of he difference it counts up "1,2,3,4,5,6,7" if you use the first digit of each layer after.

[tricky77puzzle]

it has very little to do with GEB, I just thought of it while reading it and trying to figure out what this meant

Spoiler:

the reason I found it interesting was because when you find the difference and then the difference of the difference and then the difference of the difference of he difference it counts up "1,2,3,4,5,6,7" if you use the first digit of each layer after.

What you're talking about is the first, second, and third differences.

And yes, it does do that. So the equation is polynomial, and quartic.

[Robin S]

What? No, it isn't quartic - Cauchy gives the formula above.

[kilik]

Eh, bellow isn't a hint but it does give a method of finding the next term in the sequence... not nearly as useful as the formula though.

Spoiler:

I know this is probably given by the formula's... well definently since they both refer to the same sequence, but first thing I noticed was that to get the next term in the sequence, you take your current term, multiply it by three, and subtract the sum of the previous terms...

[Tez]

look at the the difference between each term in the sequence:

Spoiler:

2*1+0=2
2*2+1=5
2*5+2=12
2*12+4=28
2*28+8=64
2*64+16=144
2*144+32= 320

etc etc..... see the pattern? so we have:
1 (+2) 3 (+5) 8 (+12) 20(+28) 48 (+64) 112 (+144) 256 (+320) 576........

to me that was the most obvious pattern I'm sure there are dozens more

[ARP]

{1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672, 61440, 131072, 278528, 589824, 1245184, 2621440, 5505024, 11534336, 24117248, 50331648, 104857600, 218103808, 452984832, 939524096, 1946157056}

a(n)= (n+1)*2^(n-2)

[Bad Kitty]

a(n)= (n+1)*2^(n-2)

I worked out (and I've not done pure maths in a long time so am feeling rather proud of myself) that a(n) = [(n+1)/2]*2^(n-1), which I proved to be equal to yours (n+1)*2^(n-2), and can be proven to give the same sequence as the OEIC's result (n+2)*2^(n-1). For our two sequences, start at n=1, for OEIC's start at n=0.

I think that's right. This is the first bit of pure maths I've done in ages and it's good to get some practice in. Feel free to point out any errors.

[sam_i_am]

When checking for polynomial, solutions, I found this fun pattern.

Spoiler:

1 3 8 20 48 112 256 576

2 5 12 28 64 144 320

3 7 16 36 80 176

4 9 20 44 96

5 11 24 52

6 13 28

7 15

8

I also see

Spoiler:

f(n) = f(n-1)*2 + 2^(n-1), which also gets 576

[Qaanol]

so I was reading GEB and I thought of a cool sequence
1 3 8 20 48 112 256

OEIS doesn't have it so I was wondering if any of you can find the pattern

The next number is 4.

The number after that is 4.

In fact, the whole sequence after the first seven terms is all 4’s.