cp's OEIS Frontend

A064943 Number of integers with 2*n digits that are the sum of the squares of their halves (leading zeros count; 1 does not, to avoid the ambiguity 1 = 0^2 + 1^2 = 00^2 + 01^2 = 000^2 + 001^2 = ...).

%I A064943 #11 Nov 18 2018 16:59:36
%S A064943 0,2,2,2,6,6,14,30,6,14,14,6,6,14,126,14,14,62,6,14,126,14,14,510,126,
%T A064943 14,62,30,30,62,6,6,254,14,2046,30,126,62,126,510,6,254,6,14,2046,14,
%U A064943 14,254,30,254,2046,254,30,254,4094,510,2046,126,6,254,30,126,2046,14
%N A064943 Number of integers with 2*n digits that are the sum of the squares of their halves (leading zeros count; 1 does not, to avoid the ambiguity 1 = 0^2 + 1^2 = 00^2 + 01^2 = 000^2 + 001^2 = ...).
%C A064943 Is there any n > 1 with a(n) = 0? This is equivalent to the question of whether there is any prime of the form 10^(2*n)+1 other than 10^(2*1)+1 = 101. If such a prime exists, n must be a power of 2. Up to now no such prime is known.
%C A064943 68 is the smallest n where a(n) is not a power of two minus 2 (a(68)=22) since (10^136)+1 is the smallest integer among the 10^(2*n)+1 which is not squarefree (10^136+1 = 17^2 * P7 * P11 * P117, so tau(10^136+1) = 24).
%H A064943 <a href="http://homes.cerias.purdue.edu/~ssw/cun/pmain1017">Cunningham project factorization tables of 10^k+1</a>
%F A064943 a(n) = tau(10^(2*n)+1) - 2.
%e A064943 a(5) = 6 because 1765038125 = 17650^2 + 38125^2, 2584043776 = 25840^2+43776^2, 7416043776 = 74160^2+43776^2, 8235038125 = 82350^2+38125^2, 9901009901 = 99010^2+09901^2, 99009901 = 00990^2+09901^2 (the last one counts as a 10-digit number). Alternatively: a(5) = tau(10^(2*5)+1) - 2 = tau(101*3541*27961) - 2 = 8 - 2 = 6.
%Y A064943 Cf. A064942 and A002654 for the derivation of the formula.
%K A064943 nonn,base
%O A064943 1,2
%A A064943 Ulrich Schimke (ulrschimke(AT)aol.com)