

A004956


a(n) = ceiling(n*phi), where phi is the golden ratio, A001622.


11



0, 2, 4, 5, 7, 9, 10, 12, 13, 15, 17, 18, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 43, 44, 46, 47, 49, 51, 52, 54, 56, 57, 59, 60, 62, 64, 65, 67, 68, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89
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OFFSET

0,2


COMMENTS

a(0)=0, a(1)=2; for n > 1, a(n) = a(n1) + 2 if n is already in the sequence, a(n) = a(n1) + 1 otherwise.
Integer solutions to the equation x = ceiling(phi*floor(x/phi))  Benoit Cloitre, Feb 14 2004
From Benoit Cloitre, Mar 05 2007: (Start)
The following is an alternative way to obtain this sequence. NP means "term not in parentheses". Write down the natural numbers and mark the least NP, which is 1:
1* 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the first NP (which is 1) and parenthesize it; mark the least NP (which is 2):
(1*) 2* 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the 2nd NP (which is 3) and parenthesize it; mark the next NP (which is 4):
(1*) 2* (3) 4* 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Take the 4th NP (which is 6) and parenthesize it; mark the next NP (which is 5):
(1*) 2* (3) 4* 5* (6) 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
Continuing in this way we obtain
(1*) 2* (3) 4* 5* (6) 7* (8) 9* 10* (11) 12* 13* (14) 15* (16) 17* (18) 19* ...
The starred entries (after the first) give the sequence. (End)
From Rick L. Shepherd, Dec 05 2009: (Start)
An equivalent statement of the sieving process described by Benoit Cloitre on Mar 05 2007:
Begin with the natural numbers N. Repeatedly perform these two steps:
i) Let k = N's least remaining term not yet used in Step ii).
ii) Remove the kth remaining term from N.
The remaining terms of N are the (positive) terms shared by this sequence and A026351.
The terms removed from N (the complement) are A026352's terms (see also A004957).
The PARI program performs this sieving process and prints the positive terms of this sequence. (End)


REFERENCES

Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3, Fall 1998, p. 176. Solution published in Vol. 12, No. 1, Winter 2000, pp. 6162.


LINKS

Christian Krause, Table of n, a(n) for n = 0..10000
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.


PROG

(PARI) {/* paws = Print Absolute values of all elements in vector v With same Sign as sn */
paws(v, sn) = for(m=1, matsize(v)[2], if(sign(v[m])==sign(sn), \
print1(abs(v[m]), ", ")))}
{/* Sieve through lim numbers; make values negative to signify "removed" */
lim=100; v=vector(lim, i, i); i=0; j=0; c=1;
while(i<lim, i++; if(v[i]>0, k=v[i]; c=c;
while(c<k && j<lim, j++; if(v[j]>0, c++)); v[j]=v[j])); paws(v, 1)\
/* Changing "1" to "1" in paws() above prints out the terms of A026352 instead} */} \\ Rick L. Shepherd, Dec 05 2009
(PARI) a(n) = ceil(n*(1 + sqrt(5))/2); \\ Michel Marcus, Apr 13 2021


CROSSREFS

Cf. A001622, A004957, A026352.
Essentially same as A026351.
Sequence in context: A189382 A087063 A047497 * A026351 A184656 A286989
Adjacent sequences: A004953 A004954 A004955 * A004957 A004958 A004959


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



