cp's OEIS Frontend

A181392 Squares in A108571.

%I A181392 #24 Jul 26 2015 06:54:17
%S A181392 1,435343551252544,543345144355225,21343774737727744,
%T A181392 525664562544416656,555621544626466564,645545246266556416,
%U A181392 656542564646552164,666524445565146256,2766717773326766736,8823883385555888521
%N A181392 Squares in A108571.
%C A181392 The last term is 999999999786876856487355368576387875784644 = 999999999893438428238^2. - _Giovanni Resta_, May 19 2013
%C A181392 There are no squares with 43, 44, or 45 digits. Indeed, numbers of 45 digits have sum of digits 1^1+2^2+...+9^9 = 285, which mod 9 is equal to 6. It is easy to verify that no power can be equal to 6 mod 9, hence there are no squares, cubes, etc. of 45 digits. Similarly, the numbers of 44 and 43 digits can only be obtained by omitting the single 1 or the two 2's, so mod 9 they are equal to 5 and 2, respectively. Again, 2 and 5 are not squares or cubes mod 9, but they can be powers with exponents  k = 5, 7, 11, 13, 17, 19, 23, 25,... (numbers not divisible by 2 or 3). Since 10^(44/k) is at most 6.3*10^8 (for k=5) excluding higher powers by generating them is not a tremendous computational effort, which can be further reduced noticing that certain candidates can be excluded based on their last digits. For example, 9993^5 mod 10000 is 3193, which contains a 1. So no number ending in 9993 can be the base for a 5th power of 44 digits (which should lack the 1). Since 4th powers are squares too, they can have at most 42 digits, and since 10^(42/4) is about 3.16*10^10, it is not difficult to ascertain that no 4th powers belong to A108571. - _Giovanni Resta_, Jul 26 2015
%H A181392 Patrick Wieschollek and Giovanni Resta, <a href="/A181392/b181392.txt">Table of n, a(n) for n = 1..1000</a> (first 160 terms from Patrick Wieschollek)
%Y A181392 Cf. A108571, A225886.
%K A181392 base,nonn,fini
%O A181392 1,2
%A A181392 _Patrick Wieschollek_, Oct 17 2010
%E A181392 Edited by _N. J. A. Sloane_, Oct 17 2010