cp's OEIS Frontend

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 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 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 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 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 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 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 > 6501, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least 10. Since this sequence relaxes the restriction of zero squares and allows one more square, the number of representations for n > 6501 is at least ten. Then an inspection of n <= 6501 completes the proof.

This sequence is finite and complete. See the von Eitzen Link and the proof in A294675 stating that for n > 7845, the number of ways to write n as a sum of 5 squares (without allowing zero squares) is at least 11. Since this sequence relaxes the restriction of zero squares and allows one more square, the number of representations for n > 7845 is at least 11. Then an inspection of n <= 7845 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 one more square, the number of representations for n > 5408 is at least nine. Then an inspection of n <= 5408 completes the proof.