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User talk:Peter Polm

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Cheers Peter,

as what comes to your "The sequence for increasing repeating zeroes in increasing Fibonacci numbers":

3,6,12,19,38,42,68,243,384,515,740,1709,5151,11049,45641, ....

and "The sequence for increasing repeating ones":

1,4,10,14,23,42,125,148,272,336,373,484,717,1674,3911,17554, ....

please, go ahead, submit them, and preferably also the program code with which you computed them. — Antti Karttunen 00:41, 2 November 2012 (UTC)


Antti,

code can be found at: http://oeis.org/history?seq=A218076 http://bigintegers.blogspot.nl/2012/10/consecutive-zeros-in-fibonacci-numbers.html There is a faster way to find the terms, a very rough description: As soon as 22 (30,38..) consecutive zerobits are found, only every 2nd (3th,4th..) byte has to be checked for not being zero, and some additional logic if it is.

PP