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If a(22) is nonzero, it has more than 500 2132 digits, too large for a b-file: it is the concatenation of 99999999999999999999999999999998^3 to 100000000000000000000000000000019^3. (End)
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From Robert Israel, Feb 20 2020: (Start)
a(n)=0 if n is divisible by 3, as the sum of three consecutive cubes is divisible by 3. - _Robert Israel_, Feb 20 2020
If a(22) is nonzero, it has more than 500 digits. (End)
a(n)=0 if n is divisible by 3, as the sum of three consecutive cubes is divisible by 3. - Robert Israel, Feb 20 2020
Robert Israel, <a href="/A104375/b104375.txt">Table of n, a(n) for n = 1..21</a>
for k from floor(n/2)*2+1 by 2 do
ccat:= proc(L)
local t, i;
t:= L[1];
for i from 2 to nops(L) do
t:= t*10^(ilog10(L[i])+1)+L[i]
od;
t
end proc:
f:= proc(n) local k, p;
if n mod 3 = 0 then return 0 fi;
for k from floor(n/2)*2+1 do
p:= ccat([seq((k-i)^3, i=n-1..0, -1)]);
if isprime(p) then return p fi
od
end proc:
f(1):= 0:
map(f, [$1..10]); # Robert Israel, Feb 20 2020
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_Shyam Sunder Gupta (guptass(AT)rediffmail.com), _, Apr 17 2005
Smallest prime formed by concatenation of n consecutive cubes, 0 if no such prime exists.
0, 827, 0, 2744337540964913, 49135832685980009261, 0, 16194277163870641658137516777216169745931717351217373979, 64348566539203664467267512696859000696787170778887189057, 0
1,2
a(2)=827 because 827 is the smallest prime formed from concatenation of 2 consecutive cubes i.e. 8 and 27.
base,nonn
Shyam Sunder Gupta (guptass(AT)rediffmail.com), Apr 17 2005
approved