[seqfan] Re: Pi = cumulative sum k of a(n) digits

[M. F. Hasler]

Eric,
The next three terms are again zero, as can be seen by summing the
first 6952 resp. 69817 resp. 698191 digits of pi.
To get one more, one would have to use 7 million digits of pi.
Certainly possible, but maybe not so interesting...
M.
\p700000
P=digits(Pi\.1^700000);
S=0;for(k=1,#P,if(31415<=S+=P[k],return([S,k,P[k]])))
[31420, 6952, 8]
S=0;for(k=1,#P,if(314159<=S+=P[k],return([S,k,P[k]])))
[314162, 69817, 9]
S=0;for(k=1,#P,if(3141592<=S+=P[k],return([S,k,P[k]])))
[3141596, 698191, 6]

On Tue, Aug 19, 2014 at 9:23 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
> Hello SeqFans,
> I cannot compute enough terms of P to check if this is old hat :-(
> Say k is the cumulative sum of the first n digits of Pi's decimal
> expansion (here http://oeis.org/A000796 );
>
> a(1)=1
> ... because the cumulative sum k of the first « 1 » decimals of Pi is = 3
> a(2)=8
> ... because the cumulative sum k of the first « 8 » decimals of Pi is = 31
>    (3+1+4+1+5+9+2+6=31)
> a(3)=0
> ... because _there is no n_ such that the cumulative sum k such is = 314
> a(4)=0
> ... because _there is no n_ such that the cumulative sum k such is = 3141
> Etc.
> So, up to now (if I'm not wrong): P = 1, 8, 0, 0, ...
>
> Best,
> É.
>
>
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Maximilian