A076217 a(1)=1, a(n) = a(n-1) + n * sign(n-a(n-1)).
1, 3, 3, 7, 2, 8, 1, 9, 9, 19, 8, 20, 7, 21, 6, 22, 5, 23, 4, 24, 3, 25, 2, 26, 1, 27, 27, 55, 26, 56, 25, 57, 24, 58, 23, 59, 22, 60, 21, 61, 20, 62, 19, 63, 18, 64, 17, 65, 16, 66, 15, 67, 14, 68, 13, 69, 12, 70, 11, 71, 10, 72, 9, 73, 8, 74, 7, 75, 6, 76, 5, 77, 4, 78, 3, 79, 2, 80
Offset: 1
Examples
a(2) = a(1)+sign(2-a(1))*2 = 1 + 2 = 3.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Programs
-
Haskell
a076217 n = a076217_list !! (n-1) a076217_list = 1 : zipWith (+) a076217_list (zipWith (*) [2..] $ map a057427 $ zipWith (-) [2..] a076217_list) -- Reinhard Zumkeller, Apr 21 2013
-
Mathematica
RecurrenceTable[{a[1]==1,a[n]==a[n-1]+n Sign[n-a[n-1]]},a[n],{n,80}] (* Harvey P. Dale, Jun 14 2011 *) -
PARI
alist(N) = my(r, t=0); vector(N, i, t=r=t+i*sign(i-t)); \\ Ruud H.G. van Tol, May 10 2024
Formula
If 3^n>2*m>= 2*3^(n-1); a(3^n-2*m) = m; if 3^n>2*m+1>=2*3^(n-1)+1 a(3^n-2*m-1) = 3^n - m; special case of partial sum: sum(k=1, 3^n, a(k)) = (3/8)*(9^n-1) + (3^(n+1)-1)/2.
Conjecture: a(n) = -a(n-1)+a(n-2)+a(n-3) for n>5. G.f.: -x*(27*x^28 +54*x^27 +27*x^26 +9*x^10 +18*x^9 +9*x^8 +3*x^4 +6*x^3 +5*x^2 +4*x +1) / ((x -1)*(x +1)^2). - Colin Barker, Feb 25 2013
Regarding Barker's conjectured recurrence, it seems to fail at n= powers of 3, and the 2 successive terms. So it holds except for n= 9-11, 27-29, 81-83, 243-245, .... - Bill McEachen, Mar 21 2025
Comments