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#265 | |
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Bamboozled!
"𒉺𒌌𒇷𒆷ð’€"
May 2003
Down not across
2·5,393 Posts |
Quote:
In the Bohr formulation of quantum theory, the orbital velocity of the electron in the ground state of the hydrogen atom is \alpha times the speed of light. In a classical (i.e. non-relativistic) quantum theory, the 1s electrons in an atom with Z=137 would be travelling at just a whiff under the speed of light. Needless to say, both of these statements are almost meaningless in terms of modern relativistic quantum mechanics. Paul |
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#266 |
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Nov 2005
101101102 Posts |
OK look at those numbers in binary - they wrap around back to the beginning. I wonder if there's a sort of 'tree' that has all the solutions. It seems that you could create the solutions recursively much like how a serpinski/pascal triangle are related to the golden string.
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#267 | |
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Bronze Medalist
Jan 2004
Mumbai,India
80416 Posts |
Quote:
Paul: In my book ‘Einstein’ edited by Louis de Broglie, N.P. laureate, Louis Armand, French Academy, et.al. there is an essay ‘Einstein the Scientist’ by Roger Nataf, Professor, P. F. des.Sciences. He gives the theory on the absence of Charged particles which is much the same as what you have painstakingly explained in your post. However he says that the probability of the process is the same as in Schrodinger’s theory and is proportional to the square of the elementary charge ‘e’ of the particle, or more precisely to the Dimensional constant 2*pi*e^2/hc which is = 1/137 ( the constant of fine structure. This differs from the formula you have given by a factor of 2 and the absence of epilon_0. The last I take as the permittivity of free space. Dielectric constant 8.8 * 10^ (-12) Farads /meter Could you please elaborate on this discrepancy ?. Mally
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#268 |
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Dec 2005
22·72 Posts |
what is special about
132813776 |
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#269 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22·33·19 Posts |
Quote:
I can only factorise it as 2^4 * 23 * 360907 Mally
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#270 |
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Dec 2005
110001002 Posts |
pretty, but not sufficient. When I posted the number I forgot to check google for a direct link. Gladly enough, I could not find one. Clue will be posted if asked
but one hint never hurts. The hint is 946
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#271 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Quote:
Well Kees your problem number 132813776 is a tough one to crack so how about a clue as offered by you. 946 is a hexagonal pyramidal number. It is also the sum of the first 6 digits +1 [(132 + 813 ) +1] 946 =945 +1 which is one above the first odd abundant number found by Bachet. I find no connection between 946 and your 9 digit number Mally
Last fiddled with by mfgoode on 2006-02-14 at 15:21 |
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#272 | |
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Mar 2005
2·5·17 Posts |
Quote:
the hint may be that 132813775 is itself notable, or contributes to the specialness of 132813776. 132813775 is highly composite since its 5*5*17*47*61*109 but i can't get beyond that. Richard |
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#273 |
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Dec 2005
22×72 Posts |
Surely 946 is more special than just being the neighbour of a special number
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#274 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Quote:
Ha at last!Happy Triangular Numbers: If you iterate the process of summing the squares of the decimal digits of a number and if the process terminates in 1, then the original number is called a Happy number. For example 7 -> 49 -> 97 -> 130 -> 10 -> 1. A Happy Triangular Number is defined as a Triangular number which is also a Happy number. For example, consider a triangular number 946, where 946 -> 133 -> 19 -> 82 -> 68 -> 100 -> 1. So 946 is a Happy triangular Number. Other examples of Happy Triangular Numbers are : Now what is the connetion with 132813776 ? This is not a Happy Triangular number But if one adds the squares of the digits we get 222 =2 *111 and 111 is triangular in base 9 .Mally
Last fiddled with by mfgoode on 2006-02-15 at 16:00 |
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#275 |
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Dec 2005
110001002 Posts |
So close and yet so far. Triangular numbers seem to be a good plan, but in that case you have to look carefully at 946 and think 'why on earth did he choose 946 as a clue'
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