@GarethMcCaughan Thanks! I edited the answer. It may be strange that 33 is the offset yes. But also why are the individual offsets adjusted the wrong way (to align them with the natural numbers we'd subtract the offset not add it)
maybe I'm being dim (it's nearly 3am here so I probably am). Consider the sequence with f(n)=n, and suppose one of the patternbot sequences was 4567. OEIS might list that sequence as 0,1,2,3,4,5,6,7,... with offset 0, in which case we'd be starting at position 5. Or it might list it as 3,4,5,6,7,... with offset 3, in which case we'd be starting at position 2. If we add starting position + offset then we get 5 either way, and we will get 5 no matter where OEIS chooses to start enumerating.
There is an offset of 1 here arising from the tendency to count things "first, second, third, ...". If we fixed that up then we would get not 2,?,3,?,3,3,?,... but 1,?,2,?,2,2,?,... which, hmm, probably occurs earlier in A000120.
I am definitely not understanding some things you're saying. This sort of thing is terribly easy to get confused by (even when conceptually trivial) so let me try to be as explicit as I can about how I see this, which seems to me to make good sense
For a typical OEIS sequence, there is some function f mapping integers to integers. OEIS then lists a bunch of successive values of f: f(k), f(k+1), f(k+2), ... where k is the "left offset".
[I am not going to try to think about it in "your" terms until I have finished trying to be clear on "my" terms because otherwise my head will explode and it will take my wife for ever to clean up the mess]
but I want to state what is going on clearly and (unless I am confused) show that nothing is the wrong way around
Now, the patternbot's n'th sequence starts at some point within that sequence, which we can describe in two ways. (1) We can give its index within the OEIS sequence as given, so that "1" means it starts at f(k) where again k is the left offset.
(This is what you did at first when listing the starting points.)
I claim that if we describe the starting points in the second way -- which is to my mind the less arbitrary way -- then those are the things that follow the Hamming-weight sequence.
so now let's see whether I've got it backwards, and I'll look at the same example as you
(but will not read what you wrote because again it will make my head explode)
so for n=20 (aargh there are too many things for "n" to mean; here I am taking it to mean the 20th sequence generated by the bot) we compute hammingweight(20) which is 2, and then we look at the function described by A000020, which is "number of primitive polys of given degree over GF(2)", and then we feed numbers from hammingweight(20) onwards into that function
Here is a list of words.
Mother
Clam
Game
Anagram
Immune
Maximum
Image
In each word, there is an indication on how to read a secret word.
But some words gave a false indication and should be ignored.
Can you find the secret word?
NOTE
If this proves to be harder than expe...
Though reading down throws away some possible orientations; you may have to read left to right each row, with the circle on the left. Or top to bottom, with the circle at the top. And so on.
er. circle=>center.
Is there a reason for some of your letters being lowercase?
@Sid Well, we had five doors to look through, each one being a trap from the movie, and we knew that movie knowledge was important. Since KxcHT wasn't anything, the best idea was to arrange them in movie order.
Look at spiders. See the arrow on the left side pointing down?
Then on the right side of the torch there's the upper right quadrant of a circle. Ignore the rightmost three spiders and the ones on or below the hand.
perl -w -e 'use strict; my @L=qw(x T c K H); my %used; sub try { my($depth,$prev)=@_; return print("https://i.stack.imgur.com/$prev.jpg\n") if ($depth>4); for my $l (@L) { next if $used{$l}++; try($depth+1,"$prev$l"); $used{$l}=0; }} try(0,""); ' | xargs -n1 wget -O/dev/null
Professor Halfbrain has recently made a fascinating discovery on cuboids in three-dimensional space.
Halfbrain's cuboid theorem:
Let $ABCDEFGH$ be a three-dimensional cuboid that has an inscribed sphere $S_i$ (that touches all six faces) as well as a circumscribed sphere $S_c$ (that contain...
if that doesn't count... we have one move and four changes in quick->first, so we only have one more change to work with, so it has to be hazy? But what is a hazy dog
So of all the images I found, i.stack.imgur.com/QtTKB.jpg earns a special place in my heart. That wall art. Man. I think there's a puzzle to be made just from that.
ANyone understood what the cleverly disguised puzzle is all about? I don't understand what we have to do there except that we have to find out some words which have some connections among themselves...
This puzzle can be solved as puzzle on its own, but its solution is also part of Deusovi's Community Metapuzzle.
You wake up feeling rather dazed. Your head is pounding. Whatever just happened did not agree with you in the least. As you get up from the floor and start exploring the room you see ...
@BeastlyGerbil Welcome back :) I pretty much ground through all the possible permutations of the letters we found, and their dotted/undotted alternatives, with no success