A132666 -id:A132666 - cp's OEIS frontend

A132667 a(1)=1, a(n) = 3*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

1, 3, 2, 6, 5, 4, 12, 11, 10, 9, 8, 7, 21, 20, 19, 18, 17, 16, 15, 14, 13, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 120, 119, 118, 117, 116

Offset: 1

Hieronymus Fischer, Aug 24 2007, Sep 15 2007, Sep 23 2007

nonn

Also: a(1)=1, a(n) = maximal positive integer < a(n-1) not yet in the sequence, if it exists, else a(n) = 3*a(n-1).
Also: a(1)=1, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = 3*a(n-1).
A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

For formulas concerning a general parameter p (with respect to the recurrence rule ... a(n)=p*a(n-1) ...) see A132374. For p=2 to p=10 see A132666 through A132674.

Cf. A087503.

G.f.: g(x) = (x(1-2x)/(1-x) + 3x^2*f'(x^(5/2)) + (5/9)*(f'(x^(1/2)) - 3x - 1))/(1-x) where f(x) = Sum_{k>=0} x^(3^k) and f'(z) = derivative of f(x) at x = z.
a(n) = 4*3^(r/2) - 2 - n if both r and s are even, else a(n) = 7*3^((s-1)/2) - 2 - n, where r = ceiling(2*log_3((2*n+3)/5)), s = ceiling(2*log_3((2*n+3)/3) - 1).
a(n) = (3^floor(1 + (k+1)/2) + 5*3^floor(k/2) - 4)/2 - n, where k=r if r is odd, else k=s (with respect to r and s above; formally, k = ((r+s) - (r-s)*(-1)^r)/2).
a(n) = A087503(m) + A087503(m+1) + 1 - n, where m:=max{ k | A087503(k) a(A087503(n) + 1) = A087503(n+1).
a(A087503(n)) = A087503(n-1) + 1 for n > 0.

A133629 a(1)=1, a(n) = a(n-1) + (p-1)*p^(n/2-1) if n is even, otherwise a(n) = a(n-1) + p^((n-1)/2), where p=5.

Original entry on oeis.org

1, 5, 10, 30, 55, 155, 280, 780, 1405, 3905, 7030, 19530, 35155, 97655, 175780, 488280, 878905, 2441405, 4394530, 12207030, 21972655, 61035155, 109863280, 305175780, 549316405, 1525878905, 2746582030, 7629394530, 13732910155, 38146972655, 68664550780

Offset: 1

Hieronymus Fischer, Sep 19 2007

Partial sums of A133632.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,5,-5).

Sequences with similar recurrence rules: A027383 (p=2), A087503 (p=3), A133628 (p=4).
Related sequences: A132666, A132667, A132668, A132669.
Other related sequences for different p: A016116 (p=2), A038754 (p=3), A084221 (p=4), A133632 (p=5).

a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=5*a[n-2]+5 od: seq(a[n], n=1..29); # Zerinvary Lajos, Mar 17 2008

Vec(x*(1 + 4*x) / ((1 - x) * (1 - 5*x^2)) + O(x^40)) \\ Colin Barker, Nov 25 2016

def A133629(n): return (5+((n&1)<<2))*5**(n>>1)-5>>2 # Chai Wah Wu, Sep 02 2025

a(n) = Sum_{k=1..n} A133632(k).
The following formulas are given for a general natural parameter p > 1 (p=5 for this sequence).
G.f.: x(1+(p-1)x)/((1-px^2)(1-x)).
a(n) = (p/(p-1))*(p^(n/2)-1) if n is even, otherwise a(n)=(p/(p-1))*((2p-1)*p^((n-3)/2)-1).
a(n) = (p/(p-1))*(p^floor(n/2) + p^floor((n-1)/2) - p^floor((n-2)/2)-1).
a(n) = p^floor(n/2) + (p^floor((n+1)/2)-p)/(p-1).
a(n) = A132669(a(n+1)) - 1.
a(n) = A132669(a(n-1)+1) for n > 0.
A132669(a(n)) = a(n-1)+1 for n > 0.
From Colin Barker, Nov 25 2016: (Start)
a(n) = 5*(5^(n/2) - 1)/4 for n even.
a(n) = (9*5^(n/2-1/2) - 5)/4 for n odd.
a(n) = a(n-1) + 5*a(n-2) - 5*a(n-3) for n > 3.
G.f.: x*(1 + 4*x) / ((1 - x) * (1 - 5*x^2)).
(End)

A132668 a(1)=1, a(n) = 4*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

Original entry on oeis.org

1, 4, 3, 2, 8, 7, 6, 5, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49

Offset: 1

Hieronymus Fischer, Aug 24 2007, Sep 15 2007, Sep 23 2007

nonn

Also: a(1)=1, a(n) = maximal positive integer < a(n-1) not yet in the sequence, if it exists, else a(n) = 4*a(n-1).
Also: a(1)=1, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = 4*a(n-1).
A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

For formulas concerning a general parameter p (with respect to the recurrence rule ... a(n) = p*a(n-1) ...) see A132374.
For p=2 to p=10 see A132666 through A132674.
Cf. A133628.

a(n) = (11*4^(r/2) - 5)/3 - n, if both r and s are even, else a(n) = (23*4^((s-1)/2) - 5)/3 - n, where r = ceiling(2*log_4((3n+4)/7)) and s = ceiling(2*log_4((3n+4)/8)).
a(n) = (4^floor(1 + (k+1)/2) + 7*4^floor(k/2) - 5)/3 - n, where k=r, if r is odd, else k=s (with respect to r and s above; formally, k = ((r+s) - (r-s)*(-1)^r)/2).
G.f.: g(x) = (x(1-2x)/(1-x) + 4x^2*f'(x^(7/3)) + (7/16)*(f'(x^(1/3)) - 4x - 1))/(1-x) where f(x) = Sum_{k>=0} x^(4^k) and f'(z) = derivative of f(x) at x = z.
a(n) = A133628(m) + A133628(m+1) + 1 - n, where m:=max{ k | A133628(k) a(A133628(n) + 1) = A133628(n+1).
a(A133628(n)) = A133628(n-1) + 1 for n > 0.

A132669 a(1)=1, a(n) = 5*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

Original entry on oeis.org

1, 5, 4, 3, 2, 10, 9, 8, 7, 6, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 155, 154, 153, 152, 151, 150, 149, 148, 147, 146, 145, 144, 143

Offset: 1

Hieronymus Fischer, Sep 15 2007, Sep 23 2007

nonn

Also: a(1)=1, a(n) = maximal positive integer < a(n-1) not yet in the sequence, if it exists, else a(n) = 5*a(n-1).
Also: a(1)=1, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = 5*a(n-1).
A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

For formulas concerning a general parameter p (with respect to the recurrence rule ... a(n) = p*a(n-1) ...) see A132374.
For p=2 to p=10 see A132666 through A132674.
Cf. A087503.

G.f.: g(x) = (x(1-2x)/(1-x) + 5x^2*f'(x^(9/4)) + (9/25)*(f'(x^(1/4)) - 5x - 1))/(1-x) where f(x) = Sum_{k>=0} x^(5^k) and f'(z) = derivative of f(x) at x = z.
a(n) = (14*5^(r/2) - 6)/4 - n, if both r and s are even, else a(n) = (34*5^((s-1)/2) - 6)/4 - n, where r = ceiling(2*log_5((4n+5)/9)) and s = ceiling(2*log_5((4n+5)/5)) - 1.
a(n) = (5^floor(1 + (k+1)/2) + 9*5^floor(k/2) - 6)/4 - n, where k=r, if r is odd, else k=s (with respect to r and s above; formally, k = ((r+s) - (r-s)*(-1)^r)/2).
a(n) = A133629(m) + A133629(m+1) + 1 - n, where m:=max{ k | A133629(k) < n }.
a(A133629(n) + 1) = A133629(n+1).
a(A133629(n)) = A133629(n-1) + 1 for n > 0.

A132340 a(n+1) = if {a(k):1<=k<=n} is a permutation of [1:n] then 2*a(n) else a(n)-1.

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 10, 9, 8, 7, 14, 13, 12, 11, 22, 21, 20, 19, 18, 17, 16, 15, 30, 29, 28, 27, 26, 25, 24, 23, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85

Offset: 1

Reinhard Zumkeller, Aug 20 2007

nonn

Self-inverse permutation of natural numbers;
a(A052955(n)) = A027383(n);
a(A052955(n)-1)=a(A052955(n))/2; a(A052955(n)+1)=a(A052955(n))-1.
Almost certainly A132666 is a duplicate of this entry. - R. J. Mathar, Jun 12 2008

A182194 a(1)=2, a(n)=a(n-1)^2 if the minimal natural number > 1 not yet in the sequence is greater than a(n-1), else a(n)=a(n-1)-1.

Original entry on oeis.org

2, 4, 3, 9, 8, 7, 6, 5, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54

Offset: 1

Hieronymus Fischer, Apr 30 2012

nonn

A reordering of the natural numbers > 1.

The sequence is quasi self-inverse in that a(a(n-1)-1)=n.

			a(2)=4=a(1)^2, since 3>2=a(1) is the minimal number not yet in the sequence (because of a(1)=2);
a(15)=19=a(14)-1, since the minimal number not yet in the sequence (=10) is <=a(14)=20.
a(10^4)=b(8)+b(7)-10^4-2=877.
a(10^6)=b(10)+b(9)-10^6-2= 103539133.

Hieronymus Fischer, Table of n, a(n) for n = 1..10200

Cf. A020703, A038722, A132666, A132664, A132665, A132674, A210882.

a(n)=a(n-1)-1, if a(n-1)-1 > 1 is not in the set {a(k)| 1<=k<=n-1}, else a(n)=a(n-1)^2.
a(a(n)-1)=n+1.
If we define b(1)=2, b(2)=3, b(k)=b(k-2)^2+1, we get the sequence 2, 3, 5, 10, 26, 101, 677, 10202, 458330, 104080805, …. The b(k) are those terms a(n) of the original sequence for which a(n+1)=a(n)^2.
With these b(k) we obtain for k>1:
a(b(k)-2)=b(k-1),
a(b(k)-1)=b(k-1)^2.
a(b(k))=b(k-1)^2 - 1.
a(n)=b(m)+b(m-1)-n-2, where m is the least index such that b(m)>n+1 (valid for n>=1).

A210882 a(1)=1, a(n)=a(n-1)-1 if a(n-1)-1 > 0 is not in the set {a(k)| 1<=k

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 6, 11, 10, 9, 8, 13, 12, 17, 16, 15, 14, 19, 18, 23, 22, 21, 20, 29, 28, 27, 26, 25, 24, 31, 30, 37, 36, 35, 34, 33, 32, 41, 40, 39, 38, 43, 42, 47, 46, 45, 44, 53, 52, 51, 50, 49, 48, 59, 58, 57, 56, 55, 54, 61, 60, 67, 66, 65, 64, 63, 62, 71

Offset: 1

Hieronymus Fischer, Apr 30 2012

nonn

A reordering of the natural numbers.
The sequence is self-inverse in that a(a(n))=n.
If n is a prime, then a(n+1) is the next prime > n. Hence, the subsequence 2, a(2+1), a(a(2+1)+1), a(a(a(2+1)+1)+1), a(a(a(a(2+1)+1)+1)+1), ... generates the sequence of primes A000040.

			a(4)=5, since 5 is the least prime > a(1), a(2), a(3), and the minimal number not yet in the sequence (=4) is greater than 3=a(3).
a(5)=4, since 4 is not in the set {1,2,3,5}={a(k)| 1<=k<n}.
7=p(4)=a(p(3)+1)=a(a(p(2)+1)+1)= a(a(a(p(1)+1)+1)+1)= a(a(a(2+1)+1)+1).

Hieronymus Fischer, Table of n, a(n) for n = 1..10000

Cf. A020703, A000040, A038722, A132666, A132664, A132665, A132674, A182194.

a(1)=1, a(n)=p (where p is the least prime number > a(k) for 1<=k

a(n)<>n for all n>3.

p(n+1)=a(p(n)+1), where p(n) is the n-th prime.

a(n+1)=p(m+2), if a(n)-1 is the m-th prime, else a(n+1)=a(n)-1, for n>2.

a(n)=p(m)+p(m-1)-n+1, where m is the least index such that p(m)>n-1 (valid for n>2).

A327883 a(0)=a(1)=0; thereafter a(n+1) is the index of the earliest as yet unused occurrence of a(n) if a(n) has occurred before; otherwise a(n+1) = a(a(n)-1). Once an occurrence of a(n) has been used it cannot be used again.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 4, 1, 3, 0, 6, 2, 5, 0, 10, 3, 9, 1, 8, 4, 7, 0, 14, 5, 13, 2, 12, 6, 11, 0, 22, 7, 21, 4, 20, 8, 19, 1, 18, 9, 17, 3, 16, 10, 15, 0, 30, 11, 29, 6, 28, 12, 27, 2, 26, 13, 25, 5, 24, 14, 23, 0, 46, 15, 45, 10, 44, 16, 43, 3, 42, 17, 41, 9, 40, 18, 39, 1, 38, 19, 37, 8

Offset: 0

David James Sycamore, Oct 10 2019

For n > 2, a(n+1) is either the index or repeat of an earlier term (see Name). Hence terms are referred to here as being "index" or "repeat" respectively. Starting from a(3)=1, index and repeat terms occur alternatively throughout the sequence.
a(n) < n for all n >= 0. For k >= 2, numbers 0,1,...,2^k-2 occur as terms in the interval between a(2^k-2) = 0 and a(2^(k+1)-2) = 0; see Formula (e.g., k = 2 --> 0,1,2 occur between a(2) and a(6); k = 3 --> 0,1,..,6 occur between a(6) and a(14)). Thus every integer >= 0 occurs infinitely many times in the sequence.
There appears to be a proper copy subsequence given by a(4*m+2) = a(m-1); m >0 (noticed by Carl J Love). There may be other (independent) copies to be found (e.g. by selecting terms sequentially, one from each of the above mentioned intervals). A family of sequences similar to this one can be described using the same rule as above, with offset k >= 0 and initial terms a(k) = a(k+1) = k. Conjecture: Every such sequence contains a proper copy of itself as a(4*m+2+k) = a(m+k-1).
Index terms > 0 are 1,2,4,3,6,5,10,9,8,7,14,13,... (see A132666). Repeat terms from a(4)=0 are 0,0,1,0,2,0,3,1,4,0,5,2,6,0,7,... (union of the nonnegative integers interleaved with the copy subsequence described above).
For any n >= 0, a k >= 0 exists such that a^k(n)=0 (e.g., n=5 -> k=2; a^2(5)=0).
Replacing a(a(n)-1) with a(a(n)+1) in the Name produces A025480 with 0 prepended.

			a(2) = 0 since a(1) = 0 was last seen as a(0); a(3) = 1 since a(2) = 0 was last seen as a(1); a(4) = 0 since a(3) = 1 has not been seen before, so a(4) = a(a(3)-1) = a(0) = 0; a(327883)=163736.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 0..2^20, showing a(n) = 0 instead as 1/2 for visibility.

Cf. A132666, A025480.

# Code by Carl J Love (via Mapleprimes). (This code is derived from Name. An alternative code (same author), based on a recursion deduced from empirical formulae (see above) produces identical output up to 100000 terms.)
restart:
a:= module()
export
pos:= table([0= 0]), nextpos:= table([0= 1]),
used:= table(sparse, [0= 2]),
ModuleApply:= proc(n::nonnegint)
option remember;
local p:= thisproc(n-1), r:= `if`(used[p] > 1, pos[p], thisproc(p-1));
(pos[r], used[r], nextpos[r]):= (nextpos[r], used[r]+1, n);
r
end proc;
(ModuleApply(0), ModuleApply(1)):= (0,0)
end module :
seq(a(k), k= 0..100);

nn = 1000; c[] := 0; a[0] = a[1] = a[2] = 0; c[0] = 1; Do[If[# == 0, k = a[a[n - 1] - 1], k = #] &[c[a[n - 1]]]; Set[{a[n], c[a[n - 1]]}, {k, n - 1}], {n, 3, nn}]; Array[a, nn + 1, 0] (* _Michael De Vlieger, Jun 25 2025 *)

Conjectured formulae:
a(2^k-2)=0 = a(3*2^k-2) = 0; (k>=0).
a(5*2^k-2) = 1, a(7*2^k-2) = 2, (k>=0).
a(11*2^(2*k)-2) = a(9*2^(2*k+1)-2) = 3 (k>=0).
a(11*2^(2*k+1)-2) = a(9*2^(2k)-2) = 4 (k>=0).
a(2*k+1) = A132666(k); k >= 1.
a(4*k) = k-1; k >= 1.
1st-level copy subsequence: a(k-1) = a(4*k+2), k >= 1.
(m-th)-level (dependent) copy subsequence: a(k) = a(4^m*(k+2) - 2), m >= 1, k >= 0.

A132673 a(1)=1, a(n) = 9*a(n-1) if the minimal positive integer number not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

Original entry on oeis.org

1, 9, 8, 7, 6, 5, 4, 3, 2, 18, 17, 16, 15, 14, 13, 12, 11, 10, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37

Offset: 1

Hieronymus Fischer, Sep 15 2007

nonn

Also: a(1)=1, a(n) = maximal positive number < a(n-1) not yet in the sequence, if it exists, else a(n) = 9*a(n-1).
Also: a(1)=1, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = 9*a(n-1).
A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

For formulas concerning a general parameter p (with respect to the recurrence rule ... a(n) = p*a(n-1) ...) see A132374.

For p=2 to p=10 see A132666 through A132674.

G.f.: g(x) = (x(1-2x)/(1-x) + 9x^2*f'(x^(17/8)) + (17/81)*(f'(x^(1/8)) - 9x - 1)/(1-x) where f(x) = Sum_{k>=0} x^(9^k) and f'(z) = derivative of f(x) at x = z.
a(n) = (26*9^(r/2) - 10)/8 - n if both r and s are even, else a(n) = (107*9^((s-1)/2) - 10)/8 - n, where r = ceiling(2*log_9((8n+9)/17)) and s = ceiling(2*log_9(8n+9)/8)) - 1.
a(n) = (9^floor(1 + (k+1)/2) + 17*9^floor(k/2) - 10)/8 - n, where k=r if r is odd, else k=s (with respect to r and s above; formally, k = ((r+s) - (r-s)*(-1)^r)/2).

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cp's OEIS Frontend

A133632 a(1)=1, a(n) = (p-1)*a(n-1), if n is even, otherwise a(n) = p*a(n-2), where p = 5.

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A132667 a(1)=1, a(n) = 3*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

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A133629 a(1)=1, a(n) = a(n-1) + (p-1)*p^(n/2-1) if n is even, otherwise a(n) = a(n-1) + p^((n-1)/2), where p=5.

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A132668 a(1)=1, a(n) = 4*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

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A132669 a(1)=1, a(n) = 5*a(n-1) if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

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A132340 a(n+1) = if {a(k):1<=k<=n} is a permutation of [1:n] then 2*a(n) else a(n)-1.

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A182194 a(1)=2, a(n)=a(n-1)^2 if the minimal natural number > 1 not yet in the sequence is greater than a(n-1), else a(n)=a(n-1)-1.

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A210882 a(1)=1, a(n)=a(n-1)-1 if a(n-1)-1 > 0 is not in the set {a(k)| 1<=k

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A327883 a(0)=a(1)=0; thereafter a(n+1) is the index of the earliest as yet unused occurrence of a(n) if a(n) has occurred before; otherwise a(n+1) = a(a(n)-1). Once an occurrence of a(n) has been used it cannot be used again.

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A133632 a(1)=1, a(n) = (p-1)a(n-1), if n is even, otherwise a(n) = pa(n-2), where p = 5.