A093848 (a(n)) is the earliest monotonic sequence starting with a(1)=1 and satisfying a(n)=length of n-th run of consecutive integers with same parity.
1, 2, 4, 5, 7, 9, 11, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 33, 34, 36, 38, 40, 42, 44, 46, 48, 50, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 122, 124
Offset: 1
Keywords
Examples
Sequence begins: (1),(2,4),(5,7,9,11),(12,14,16,18,20),(21,.... since the number of elements in each run of odd or even integers is 1, 2, 4, 5, ... the sequence itself.
Links
- T. D. Noe, Table of n, a(n) for n=1..1000
Programs
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Maple
A093848 := proc(nmax) local n,par,a,alast,j ; n := 3 ; par := 1 ; a := [1,2,4] ; while nops(a) < nmax do alast := op(-1,a); if type(alast,'even') = type(par,'even') then ; else alast := alast -1 ; end if; for j from 1 to op(n,a) do a := [op(a), alast+2*j] ; end do: par := 1-par ; n := n+1 ; end do: a ; end proc: A093848(120) ; # R. J. Mathar, Jun 22 2010
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Mathematica
t={1,2,4}; Do[t=Join[t,Table[t[[-1]]+2*i-1, {i,t[[n]]}]], {n,3,40}]
Formula
it seems that a(n) = 2n - a*n^b + o(n^b) where a and b are 2 suitable constants. b=0.4.... Does b=2-phi where phi is the golden ratio?
Extensions
Terms starting at a(52) corrected by R. J. Mathar, Jun 22 2010
Comments