From Newsgroup: rec.puzzles
David Entwistle <
qnivq.ragjvfgyr@ogvagrearg.pbz> writes:
Answers:
a) 1, 11, 21, 1211, 111221, 312211
A005150 - Look and Say sequence: describe the previous term! (method A - initial term is 1).
b) 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3
A005589 - Number of letters in the US English name of n, excluding spaces and hyphens.
c) 5, 16, 24, 72, 48, 120, 72
A069482 - a(n) = prime(n+1)^2 - prime(n)^2.
d) 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, 4
A056992 - Digital roots of square numbers
e) 1, 8, 9, 1, 8, 9, 1
A073636 - Period 3: repeat [1, 8, 9] ; Digital root of A000578(n) = n^3
for n >= 1.
I think c), d) and e) are particularly interesting. How the difference between the squares of adjacent primes evolves, that the digital root of
any perfect square will be 1, 4, 7 or 9 and that the digital root of any perfect cube will be either 1, 8 or 9.
I'm annoyed I didn't get (c) given my affinity for prime numbers, I'm
not sure any amount of additional thinking would have got me there.
However, adding "try summing them" to my toolbox might be helpful in the future, I normally don't go beyond taking their differences.
The other two are hardly surprising:
if x=3n, for n>0, then x^2=9n^2 so will have digital root 9
if x=3n+/-1, then x^2=9n^2+/-6n+1, so will have digital root 1, 4, or 7
if x=3n, for n>0, then x^3=9(3n^3), so will have digital root 9
if x=3n+1, then x^3=9(3n^3+3n^2+n)+1 so will have digital root 1
if x=3n+2, then x^3=9(3n^3+6n^2+4n)+8 so will have digital root 8
Phil
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