A067280 Number of terms in continued fraction for sqrt(n), excl. 2nd and higher periods.

1, 2, 3, 1, 2, 3, 5, 3, 1, 2, 3, 3, 6, 5, 3, 1, 2, 3, 7, 3, 7, 7, 5, 3, 1, 2, 3, 5, 6, 3, 9, 5, 5, 5, 3, 1, 2, 3, 3, 3, 4, 3, 11, 9, 7, 13, 5, 3, 1, 2, 3, 7, 6, 7, 5, 3, 7, 8, 7, 5, 12, 5, 3, 1, 2, 3, 11, 3, 9, 7, 9, 3, 8, 6, 5, 13, 7, 5, 5, 3, 1, 2, 3, 3, 6, 11, 3, 7, 6, 3, 9, 9, 11, 17, 5, 5, 12, 5

Offset: 1

Author

Frank Ellermann, Feb 23 2002

Keywords

• nonn
• easy

Examples

			a(2)=2: [1,(2)+ ]; a(3)=3: [1,(1,2)+ ]; a(4)=1: [2]; a(5)=2: [2,(4)+ ].
		

References

• H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th edition, 1999, table 1.

Crossrefs

Related sequences: 2 : A040000, ..., 44: A040037, 48: A040041, ..., 51: A040043, 56: A040048, 60: A040052, 63: A040055, ..., 66: A040057. 68: A040059, 72: A040063, 80: A040071.

Related sequences: 45: A010135, ..., 47: A010137, 52: A010138, ..., 55: A010141, 57: A010142, ..., 59: A010144. 61: A010145, 62: A010146. 67: A010147, 69: A010148, ..., 71: A010150.

Cf. A003285.

Programs

Python

from sympy import continued_fraction_periodic
def A067280(n): return len((a := continued_fraction_periodic(0,1,n))[:1]+(a[1] if a[1:] else [])) # Chai Wah Wu, Jun 14 2022

Formula

a(n) = A003285(n) + 1. - Andrey Zabolotskiy, Jun 23 2020

Extensions

Name clarified by Michel Marcus, Jun 22 2020