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A161179
A double interspersion, R(n,k), by antidiagonals.
3
1, 4, 2, 7, 3, 5, 12, 8, 6, 9, 17, 11, 13, 10, 14, 24, 18, 16, 19, 15, 20, 31, 23, 25, 22, 26, 21, 27, 40, 32, 30, 33, 29, 34, 28, 35, 49, 39, 41, 38, 42, 37, 43, 36, 44, 60, 50, 48, 51, 47, 52, 46, 53, 45, 54, 71, 59, 61, 58, 62, 57, 63, 56, 64, 55, 65, 84, 72, 70, 73, 69, 74
OFFSET
1,2
COMMENTS
(1) Every positive integer occurs exactly once; thus as a sequence, A161179
is a permutation of the positive integers.
(2) The double interspersion property for every pair of rows
s=(s(1),s(2),s(3),...) and t=(t(1),t(2),t(3),...) can be stated for
s(1)<t(1) as follows: there exist indices i and j such that
s(i)<s(i+1)<t(j)<t(j+1)<s(i+2)<s(i+3)<t(j+2)<t(j+3)<...
(For a [single] interspersion, see A035513, the Wythoff array.)
FORMULA
If k is odd and n=1: R(n,k)=(1/2)(k^2+2k-1)
if k is odd and n>1: R(n,k)=(1/2)(n^2+(2k-1)n+k^2-2k-1)
if k is even and n=1: R(n,k)=(1/2)(k^2+2k)
if k is even and n>1: R(n,k)=(1/2)(n^2+(2k-3)n+k^2-2k)
To derive the above, use the easily discerned patterns of
the difference array of R. Quite a different construction
of R takes place in two steps: first, form an array U from the
golden mean, tau, and the Fibonacci numbers A000045, as follows:
if n=1 then U(n,k)=F(k+1)
if n>1, then U(n,k)=F(n+k-1)+F(k+2)-tau*F(k+1).
Then replace each U(n,k) by its rank when all the numbers U(n,k)
are arranged in increasing order. The resulting array is R. (The
rows of U are doubly interspersed and satisfy the Fibonacci
recurrence; no array of rational integers has both of those properties.)
EXAMPLE
Northwest corner:
1....4....7...12...17...24...31...40...49...60
2....3....8...11...18...23...32...39...50...59
5....6...13...16...25...30...41...48...61...70
9...10...19...22...33...38...51...58...73...82
Rows 2 and 4 are doubly interspersed because
3<8<9<10<11<18<19<22<23<32<33<38<...
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Jun 05 2009
STATUS
approved