A005041 A self-generating sequence.
1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18
Offset: 0
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- James Propp, Problem 1047, Math. Mag., 52 (1979), 265.
- Jeffrey Shallit, Letter to N. J. A. Sloane, Nov 10 1979. Attached: James Propp, Problem 1047, Math. Mag., 52 (1979), 265. [Annotated scanned copy]
- Aaron Snook, Augmented Integer Linear Recurrences, Thesis, 2012. - From _N. J. A. Sloane_, Dec 19 2012
Crossrefs
Programs
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Haskell
a005041 n = a005041_list !! n a005041_list = 1 : f 1 1 (tail ts) where f y i gs'@((j,a):gs) | i < j = y : f y (i+1) gs' | i == j = a : f a (i+1) gs ts = [(6*k + 3*k*(k-1) `div` 2 + r*(k+2), 3*k+r+1) | k <- [0..], r <- [0,1,2]] -- Reinhard Zumkeller, Mar 16 2012 -
Mathematica
Table[n+1, {n, 0, 20}, {Ceiling[(n+1)/3]+1}] // Flatten (* Jean-François Alcover, Dec 10 2014 *)
Formula
For any k in {0, 1, 2, ...} and r in {0, 1, 2}, we have: if n = 6*k + (3/2)*k*(k-1) + r*(k+2), then a(n) = 3*k + r + 1. E.g., for k=3 and r=1, we have n = 6*3 + (3/2)*3*(3-1) + 1*(3+2) = 32 and so a(32) = 3*3 + 1 + 1 = 11. - Francois Jooste (phukraut(AT)hotmail.com), Mar 12 2002
Extensions
More terms from Samuel Hilliard (sam_spade1977(AT)hotmail.com), Apr 11 2004
Comments