1020
N = ----------------------
999999999989999999999
Find the 250th digit to the right of the decimal point.
n=sym(10)^20/sym('999999999989999999999')
digits 1000
vpa(n)
finds, breaking each 100 digits after the decimal,
ans =
0.1000000000010000000001100000000021000000001310000000034100000001651000000050610000002157100000072181
0000028789100001009701000038886110001398562100052871731001927279410072144525102648724661098631771713
6350423782349821954959848643331948308388279331727214741625660426747983819009105498616839038805177495
8866686139976718636358633872503563057361394264446117505701822569321464343198916466001310629003212022
7560334308565635463313216688941697802352730194712468999749477420192207243201671549852208922741723760
7772694465305144182260824136287913385545139957991739312965462533087646364643841926179526065626439102
5824439170914508782734969526998264204052717611567525440319727972014764805160467376023619438565396661
7616775860561821725223234993112794171656351176709676305938753114236097207448299725188719094459335490
6697820740011571562306813536363079706925937610605622455812982162168735444077500336602943738810106937
7247040131160571470688852754838049099019069343745028742536469356769109722310227566579871385388108941
The fiftieth digit on the third line is the 2 centered in ...172721474...
One thing that seems to be apparent is the appearance of
A015446
Generalized Fibonacci numbers: a(n) = a(n-1) + 10*a(n-2)
within groups of 11 digits.
I'm assuming that when the number get large enough they overflow from the 11-digit limit and interact with previous groups of 11 digits, so a given span of 11 is the sum of its own member of A015446 with carries from members further to the right.
In the vicinity in question:
6833172721221 23rd member of A015446 ; last listed member in OEIS
25316256603331 24th member of A015446 = 23d member plus 10 * 22nd member
-------------
6833172721474
so that if this sequence had been known, an analytic solution would have been made.
Edited on February 9, 2025, 10:48 am
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Posted by Charlie
on 2025-02-09 10:41:44 |
computer solution and discussion
