[seqfan] Re: A computational challenge from 1973

[Alex Meiburg]

If we group the sum (for squares) by number of digits -- that is,  A = (.01
+ .04 + .09 + .0016 + .0025 + ...) + (.000100 + .000121 + .000144 )...,
Mathematica gives a closed form for each term. Specifically,

f[n_] := 1/3 2^(-5-4 n) 25^(-2-2 n) (10^(1+n)-Ceiling[10^(1/2+n)])
(1+2^(3+2 n) 5^(2+2 n)-3 10^(1+n)-3 Ceiling[10^(1/2+n)]+2^(2+n) 5^(1+n)
Ceiling[10^(1/2+n)]+2 Ceiling[10^(1/2+n)]^2)-1/3 2^(-3-4 n) 5^(-2-4 n)
(-1+10^n-Floor[10^(1/2+n)]) (2^(1+2 n) 5^(2
n)-10^n+Floor[10^(1/2+n)]+2^(1+n) 5^n Floor[10^(1/2+n)]+2
Floor[10^(1/2+n)]^2)

this is found by breaking into those less than 10^(n+1/2) and those more
than 10^(n+1/2), and each sum can be done exactly. The above expression
would be summed over n from 0 onwards. This allows it to converge very
rapidly, yielding (in ~1 second on my computer)

A =
0.1819058902008012156762096779028721234047955026485211521758854214321879901491421189273371395956634796190420836209877147031918045603883217943221957766308226187393585383621162722218465733145947712728914362768939297625807220329993755150893123689842496260086476641169961020828008862830593572110212530631117152305529806492037632632133219059593583028182894120128297404653399814673445759857199248725452332778270108177151273587194246549997630027062718477709037754251457607532299567351237024339989709746488362421925169872670801616144371372507097161319803028219363811375620512512677915544803740657196554696124721097775213404047734600505041085860200873512242914767206681950461568207388112576403619589771428626620412562725797434755213587122866880428136702480851188065728847361715480767330547127979658699973166162970314939749529705726667314373167037314938716342660942289097217132791279915591395912939317201404066753533810115103080147864906643290214333852271864692146387518404231194138382947749162680518961898521092

for the first 1000 digits. The same technique should work for cubes, maybe
possibly those that grow exponentially; but almost certainly not with
factorials. :/

-- Alexander Meiburg

2015-06-16 22:25 GMT-07:00 Neil Sloane <njasloane at gmail.com>:

> Dear Seq Fans, This Spring I started going through
> my file of issues of the newsletter Popular Computing.
> I am now back to 1973.
>
> In Popular Computing, Vol. 1, Number 8, November 1973,
> page PC8-14, Problem 22 asks for the computation
> of various decimal numbers arising from sequences.
>
> Suppose m is a k-digit number. Let c(m) = m/100^k.
> Thus, if m is a 2-digit number, c(m)  = .00m
> Example: m=16, c(m)=.0016 = 16/100^2.
> If m=127, c(m)=.000127 = 127/10^6.
>
> HERE IS THE PROBLEM:
> For a sequence a = a(1), a(2), a(3), ...,
> define f(a) = Sum_{n >= 1} c(a(n)).
>
> For example, take a = 1 4 9 16 25 36 ... the nonzero squares
> Then f(a) is the infinite sum of
> .01
> .04
> .09
> .0016
> .0025
> .0036
> ...
>
> According to Popular Computing, my old friend Hermann P. Robinson
> computed that f(a) = .18190589020080121567...
> Problem 22 asks for more digits, and I'm asking for someone
> to enter this sequence (the sequence of decimal digits,
> with keyword cons, as usual).
>
> Popular Computing and I are also interested in the numbers that arise
> from the cubes, powers of 2, powers of 3, Fibonacci numbers,
> factorials, and subfactorials.
>
> And (this is me) the Catalan numbers, the Motzkin numbers.
>
> If a = primes, f(a)=oo, so we don't consider that one!
>
> Popular Computing gives H P Robinson's values
> for many of these, to 20 decimal
> places, so that will be a good check on your work!
>
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