cp's OEIS Frontend

This sequence is finite and complete. See the von Eitzen Link. For positive integer n, if n > 6501 then the number of ways to write n as a sum of 5 squares is at least 10. So for n > 6501, there are more than eight ways to write n as a sum of 5 squares. For n <= 6501, it has been verified if n is in the sequence by inspection. Hence the sequence is complete.

This sequence is finite and complete. See the von Eitzen Link. For positive integer n, if n > 7845 then the number of ways to write n as a sum of 5 squares is at least 11. So for n > 7845, there are more than nine ways to write n as a sum of 5 squares. For n <= 7845, it has been verified if n is in the sequence by inspection. Hence the sequence is complete.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows one more square, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 5408, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least floor(sqrt(n - 101) / 8) = 9. Since this sequence relaxes the restriction of zero squares and allows two more zero squares, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.