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%I #23 Apr 21 2024 22:14:15
%S 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 18,13,19,14,20,21,15,22,16,23,24,17,25,18,26,19,27,28,20,29,21,30,31,
%U 22,32,23,33,24,34,35,25,36,26,37,38,27,39,28,40,41,29,42,30,43,31,44
%N 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 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
%C (a(n))^2*A006337(n) gives a sequence of k^2 and 2*k^2 ordered in ascending order. - _Mikhail Kurkov_, Dec 15 2021 [verification needed]
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Graham-PollakSequence.html">Graham-Pollak Sequence</a>
%F 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
%e -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, ...
%o (PARI) b(n)=sqrt(2)*(n+1)\1-sqrt(2)*n\1 \\ from A006337
%o c(n,m)=if(b(n)==m,1,0)
%o d(n,m)=if(n==1,c(n,m),d(n-1,m)+c(n,m))
%o a(n)=c(n,1)*d(n,1)+c(n,2)*d(n,2) \\ _Mikhail Kurkov_, Dec 15 2021 [verification needed]
%Y Cf. A001521, A003151, A003152, A006337.
%K nonn,changed
%O 1,3
%A _Eric W. Weisstein_, Jan 18 2004
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