A161179 A double interspersion, R(n,k), by antidiagonals.

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

Author

Clark Kimberling, Jun 05 2009

Keywords

• nonn
• tabl

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)

s(i)

(For a [single] interspersion, see A035513, the Wythoff array.)

Examples

			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

A000045, A035513.

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.)

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