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=={{header|J}}==
<lang J>
mp =: $:~ :(+/ .*) NB. matrix product
f =: (mp 0 = [: */ 3 5 |/ ])@:i.
assert 233168 -: f 1000 NB. ****************** THIS IS THE ANSWER FOR 1000
</lang>
For the efficient computation with large n, we start with observation that the sum of these multiples with the reversed list follows a pattern.
<lang J>
g =: #~ (0 = [: */ 3 5&(|/))
assert 0 3 5 6 9 10 12 15 18 20 21 24 25 27 30 33 35 36 39 40 42 45 48 -: g i. 50
assert 48 48 47 46 48 46 47 48 48 47 46 48 46 47 48 48 47 46 48 46 47 48 48 -: (+ |.)g i. 50 NB. the pattern
assert (f -: -:@:(+/)@:(+|.)@:g@:i.) 50 NB. half sum of the pattern.
NB. continue...
</lang>
Stealing the idea from the python implementation to use 3 simple patterns rather than 1 complicated pattern,
<lang J>
first =: 0&{
last =: first + skip * <.@:(skip %~ <:@:(1&{) - first)
skip =: 2&{
terms =: >:@:<.@:(skip %~ last - first)
sum_arithmetic_series =: -:@:(terms * first + last) NB. sum_arithmetic_series FIRST LAST SKIP
NB. interval is [FIRST, LAST)
NB. sum_arithmetic_series is more general than required.
(0,.10 10000 10000000000000000000x)(,"1 0"1 _)3 5 15x NB. demonstration: form input vectors for 10, ten thousand, and 1*10^(many)
0 10 3
0 10 5
0 10 15
0 10000 3
0 10000 5
0 10000 15
0 10000000000000000000 3
0 10000000000000000000 5
0 10000000000000000000 15
(0,.10 10000 10000000000000000000x)+`-/"1@:(sum_arithmetic_series"1@:(,"1 0"1 _))3 5 15x
23 23331668 23333333333333333331666666666666666668
</lang>
=== Simple Solution ===
The problem can also be solved with a simple use of inclusion/exclusion; this solution is more in keeping with how one could approach the problem from a more traditional language.
<lang J>
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