

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 = ...).


2



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, 14, 62, 30, 30, 62, 6, 6, 254, 14, 2046, 30, 126, 62, 126, 510, 6, 254, 6, 14, 2046, 14, 14, 254, 30, 254, 2046, 254, 30, 254, 4094, 510, 2046, 126, 6, 254, 30, 126, 2046, 14
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OFFSET

1,2


COMMENTS

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.
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).


LINKS

Table of n, a(n) for n=1..64.
Cunningham project factorization tables of 10^k+1


FORMULA

a(n) = tau(10^(2*n)+1)  2.


EXAMPLE

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 10digit number). Alternatively: a(5) = tau(10^(2*5)+1)  2 = tau(101*3541*27961)  2 = 8  2 = 6.


CROSSREFS

Cf. A064942 and A002654 for the derivation of the formula.
Sequence in context: A211512 A139552 A292140 * A081478 A105341 A194676
Adjacent sequences: A064940 A064941 A064942 * A064944 A064945 A064946


KEYWORD

nonn,base


AUTHOR

Ulrich Schimke (ulrschimke(AT)aol.com)


STATUS

approved



