A091524 a(m) is the multiplier of sqrt(2) in the constant alpha(m) = a(m)*sqrt(2) - b(m), where alpha(m) is the value of the constant determined by the binary bits in the recurrence associated with the Graham-Pollak sequence.

1, 1, 2, 2, 3, 4, 3, 5, 4, 6, 7, 5, 8, 6, 9, 7, 10, 11, 8, 12, 9, 13, 14, 10, 15, 11, 16, 12, 17, 18, 13, 19, 14, 20, 21, 15, 22, 16, 23, 24, 17, 25, 18, 26, 19, 27, 28, 20, 29, 21, 30, 31, 22, 32, 23, 33, 24, 34, 35, 25, 36, 26, 37, 38, 27, 39, 28, 40, 41, 29, 42, 30, 43, 31, 44

Offset: 1

Author

Eric W. Weisstein, Jan 18 2004

Keywords

• nonn

Comments

Each integer appears twice. If one deletes the first occurrence of each positive integer one obtains the sequence of positive integers: 1,2,3,4,5,...; i.e., if we enclose in parentheses the first occurrence of 1,2,3,... giving (1),1,(2),2,(3),(4),3,(5),4,(6),(7),5,(8),6,(9),7,(10),... and remove them, we obtain: 1,2,3,4,5,6,7,... The same property holds if one deletes the second occurrence of each positive integer. - Benoit Cloitre, Oct 13 2007

Examples

			-1+sqrt(2), -1+sqrt(2), -2+2*sqrt(2), -2+2*sqrt(2), -4+3*sqrt(2), ..., so the sequence of multipliers is 1, 1, 2, 2, 3, ...
		

Links

• Eric Weisstein's World of Mathematics, Graham-Pollak Sequence

Crossrefs

Cf. A001521, A003151, A003152, A006337.

Formula

Sequence is completely defined by: a(floor(n*(1+sqrt(2))))=n; a(floor(n*(1+1/sqrt(2))))=n, n>=1 since A003151 and A003152 are Beatty sequences partitioning the integers. - Benoit Cloitre, Oct 13 2007

Conjecture: a(n) = sqrt(A028982(n)/A006337(n)). - Mikhail Kurkov, Apr 25 2024