cp's OEIS Frontend

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.

%I A091524 #27 Feb 16 2025 08:32:52
%S A091524 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,
%T A091524 18,13,19,14,20,21,15,22,16,23,24,17,25,18,26,19,27,28,20,29,21,30,31,
%U A091524 22,32,23,33,24,34,35,25,36,26,37,38,27,39,28,40,41,29,42,30,43,31,44
%N 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.
%C A091524 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
%H A091524 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Graham-PollakSequence.html">Graham-Pollak Sequence</a>
%F A091524 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
%F A091524 Conjecture: a(n) = sqrt(A028982(n)/A006337(n)). - _Mikhail Kurkov_, Apr 25 2024
%e A091524 -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, ...
%Y A091524 Cf. A001521, A003151, A003152, A006337.
%K A091524 nonn
%O A091524 1,3
%A A091524 _Eric W. Weisstein_, Jan 18 2004