

A273874


Least positive integer k such that k^2 + (k+1)^2 + ... + (k+n2)^2 + (k+n1)^2 is the sum of two nonzero squares. a(n) = 0 if no solution exists.


0



5, 1, 2, 0, 2, 0, 0, 0, 0, 2, 5, 1, 12, 0, 3, 0, 3, 0, 0, 0, 0, 0, 53, 1, 1, 1, 2, 0, 4, 0, 0, 0, 5, 2, 0, 0, 2, 0, 3, 0, 5, 0, 0, 5, 0, 0, 73, 1, 3, 1
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OFFSET

1,1


COMMENTS

Least positive integer k such that sum_{i=0..n1} (k+i)^2 = n*(6*k^2 +6*k*n 6*k +2*n^2 3*n +1)/6 is the sum of two nonzero squares. a(n) = 0 if no k exists for corresponding n.


LINKS

Table of n, a(n) for n=1..50.


EXAMPLE

a(1) = 5 because 5^2 = 3^2 + 4^2.
a(3) = 2 because 2^2 + 3^2 + 4^2 = 2^2 + 5^2.
a(11) = 0 because sum_{i=0..111} (k+i)^2 = 11*(k^2+10*k+35). Since 11 mod 4 = 3, to be expressed as a sum of two squares the term k^2+10*k+35 must be divisible by 11. This happens if k is congruent to 5 or 7 mod 11, but in both cases the sum, once divided by 11^2, is congruent to 3 mod 4, and thus it cannot be written as the sum of two squares. Similar modular arguments can be used for the other terms equal to 0.  Giovanni Resta, Jun 02 2016


CROSSREFS

Cf. A000404, A034705.
Sequence in context: A174919 A156952 A158748 * A086039 A265824 A097413
Adjacent sequences: A273871 A273872 A273873 * A273875 A273876 A273877


KEYWORD

nonn,more


AUTHOR

Altug Alkan, Jun 02 2016


EXTENSIONS

a(7)a(50) from Giovanni Resta, Jun 02 2016


STATUS

approved



