277=9^2+14^2

531

[QUOTE=Kees]277=9^2+14^2

531[/QUOTE]

531 is the smallest number with the property that its first 4 multiples contain the digit 1.

Kees please let us know what you had in mind for 531

8753

Mally :coffee:

[QUOTE=mfgoode]:bow: Hats off to you Kees you are a genius!

However my small number 47 has to do with the Fib. and Lucas series. I will give it in my next post if not answered by then.
Mally :coffee:[/QUOTE]

:popcorn: If you examine the Fib.(Un) and Lucas(Vn) series, you will notice that for the primes n= 5,7,11,13,17 and 47 both Un and Vn are primes.

If you continue in the list, nothing more happens until-surprise- for n=148091
Un and Vn are then probable primes.

Says Paulo Ribenboim "When (if) it will be shown that these numbers are indeed primes, we will acquire the feeling that, perhaps there exist also infinitely many primes, n for which Un and Vn are both primes. This problem will be a 'hard nut to crack'.
Dont lose your nights of sleep, nuts are hard to digest"

Mally :coffee:

[QUOTE=mfgoode]
Kees please let us know what you had in mind for 531
[/QUOTE]

I can't keep quiet any longer!

we have a change of 'policy' in this thread: a few months ago it wasn't required that the characteristic noted was one the poster originally intended: it helps the thread along that once a response has been given we move on to the next number...

what do people think?

Richard

my solution was less pretty but more for the eye: 531=3^5+3^2*2^5

8753=92^2+17^2

I agree that a lively thread is a better thread

87539319 (clue: Ramanujan)

166^3 + 435^3 = 227^3 + 422^3 = 254^3 + 413^3 = 87539319 (It is the smallest such number)

Has 153 been asked before?

I don't know, but it is very special: take a random number which we can divide
by three: say 1133568. Take the sum of the decimal cubes:

1^3+1^3+3^3+3^3+5^3+6^3+8^3=909

repeat procedure till stop and you end at 153

909->1458->702->351->153->id

new number

89

[QUOTE=Kees]I don't know, but it is very special: take a random number which we can divide
by three: say 1133568. Take the sum of the decimal cubes:

1^3+1^3+3^3+3^3+5^3+6^3+8^3=909

repeat procedure till stop and you end at 153

909->1458->702->351->153->id

new number

89[/QUOTE]

Your propery is a consequence of mine somewhat : 1^3 + 5^3 + 3^3 = 153

I'd also like to note that 1 + 2 + 3 ... + 16 + 17 = 153
1! + 2! + 3! + 4! + 5! = 153

34 + 55 = 89

145

Special whole numbers
 

[QUOTE=fetofs]Your propery is a consequence of mine somewhat : 1^3 + 5^3 + 3^3 = 153

I'd also like to note that 1 + 2 + 3 ... + 16 + 17 = 153
1! + 2! + 3! + 4! + 5! = 153

34 + 55 = 89

145[/QUOTE]

I'd also like to note that 1 + 2 + 3 ... + 16 + 17 = 153
1! + 2! + 3! + 4! + 5! = 153

34 + 55 = 89

145[/QUOTE]
:unsure:
34 +55 =89. Besides the sequence of digits in order, I regard this as a trivial
interpretation unless they are regarded as U9 + U10 = U11

89: The 11th. Fibonacci number and the 5th Fib. prime.
The reciprocal is generated by the Fib sequence
1/89 =0.011235....... because 89 = 10^2 -10 - 1 (Fib. quadratic x^2 -x -1 )

145 = 1! +4! +5!

175

Mally :coffee:

[QUOTE="mfgoode"]
34 +55 =89. Besides the sequence of digits in order, I regard this as a trivial
interpretation unless they are regarded as U9 + U10 = U11
[/QUOTE]

Exactly what I meant.

Number 89
 

[QUOTE=fetofs]Exactly what I meant.[/QUOTE]

:smile:
Fetoffs: Thanks for the clarification.
Its good to know that you are back on this thread, after quite some time.
Mally :coffee: