Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #11 Nov 18 2018 16:59:36
%S 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 14,62,30,30,62,6,6,254,14,2046,30,126,62,126,510,6,254,6,14,2046,14,
%U 14,254,30,254,2046,254,30,254,4094,510,2046,126,6,254,30,126,2046,14
%N 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 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 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 <a href="http://homes.cerias.purdue.edu/~ssw/cun/pmain1017">Cunningham project factorization tables of 10^k+1</a>
%F a(n) = tau(10^(2*n)+1) - 2.
%e 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 Cf. A064942 and A002654 for the derivation of the formula.
%K nonn,base
%O 1,2
%A Ulrich Schimke (ulrschimke(AT)aol.com)