A079352 -id:A079352 - cp's OEIS frontend

A079000 a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is odd".

1, 4, 6, 7, 8, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 23, 25, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 95, 97

Offset: 1

Matthew Vandermast, Feb 01 2003

a(a(n)) = 2n + 3 for n>1.

			a(2) cannot be 2 because 2 is even; it cannot be 3 because that would require 2 to be a member of the sequence. Hence a(2)=4 and the next odd member of the sequence is the fourth member.

Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

N. J. A. Sloane, Table of n, a(n) for n = 1..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.
N. J. A. Sloane, Seven Staggering Sequences.
Index entries for sequences of the a(a(n)) = 2n family

Cf. A079250-A079259, A079313, A079325, A064437, A003605, A079352, A079358.
Cf. also A080596, A080731, A080752, A121053, A080032.
Partial sums give A080566. Differences give A079948.

Digits := 50; A079000 := proc(n) local k,j; if n<=2 then n^2; else k := floor(evalf(log( (n+3)/6 )/log(2)) ); j := n-(9*2^k-3); 12*2^k-3+3*j/2 +abs(j)/2; fi; end;
A002264 := n->floor(n/3): A079944 := n->floor(log[2](4*(n+2)/3))-floor(log[2](n+2)): A000523 := n->floor(log[2](n)): f := n->A079944(A002264(n-4)): g := n->A000523(A002264(n+2)/2): A079000 := proc(n) if n>3 then RETURN(simplify(3*n+3-3*2^g(n)+(-1)^f(n)*(9*2^g(n)-n-3))/2) else if n>0 then RETURN([1,4,6][n]) else RETURN(0) fi fi: end;

a[1] = 1; a[n_] := (k = Floor[Log[2, (n+3)/6]]; j = n-(9*2^k - 3); 12*2^k-3 + 3*j/2 + Abs[j]/2); Table[a[n], {n, 1, 71}] (* Jean-François Alcover, May 21 2012, after Maple *)

a(1) = 1, a(2) = 4, then a(9*2^k-3+j) = 12*2^k-3+3*j/2+|j|/2 for k>=0, -3*2^k <= j <= 3*2^k. Also a(3n) = 3*b(n/3), a(3n+1) = 2*b(n)+b(n+1), a(3n+2) = b(n)+2*b(n+1) for n>=2, where b = A079905. - N. J. A. Sloane and Benoit Cloitre, Feb 20 2003
a(n+1) - 2*a(n) + a(n-1) = 1 for n = 9*2^k - 3, k>=0, = -1 for n = 2 and 3*2^k-3, k>=1 and = 0 otherwise.
a(n) = (3*n + 3 - 3*2^g(n) + (-1)^f(n)*(9*2^g(n) - n - 3))/2 for n>3, f(n) = A079944(A002264(n-4)) and g(n) = A000523(A002264(n+2)/2). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003
Also a(n) = n + 3*2^A000523(A002264(n+2)/2)*(1 - 3*A080584(n-4)) + A080584(n-4)*(n+3) for n>3, where A080584(n)=A079944(A002264(n)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

A356639 Number of integer sequences b with b(1) = 1, b(m) > 0 and b(m+1) - b(m) > 0, of length n which transform under the map S into a nonnegative integer sequence. The transform c = S(b) is defined by c(m) = Product_{k=1..m} b(k) / Product_{k=2..m} (b(k) - b(k-1)).

Original entry on oeis.org

1, 1, 3, 17, 155, 2677, 73327, 3578339, 329652351

Offset: 1

Thomas Scheuerle, Aug 19 2022

This sequence can be calculated by a recursive algorithm:
Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').
The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).
There exist some transformation pairs of infinite sequences in the database:
  A000027 <--> A000142; A000079 <--> A000079; A000045 <--> A001654;
  A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;
  A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;
  A098558 <--> A004277; A010551 <--> A087811; A100538 <--> A329227;
  A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;
  A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;
  A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;
  A279312 <--> A------; A031923 <--> A------.
These transformation pairs are conjectured:
  A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;
  A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;
  A349079 <--> A------.
("A------" means not yet in the database.)
Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.
If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.

			a(4) = 17. The 17 transformation pairs of length 4 are:
  {1, 2, 3, 4}  = S({1, 2, 6, 24}).
  {1, 2, 3, 5}  = S({1, 2, 6, 15}).
  {1, 2, 3, 6}  = S({1, 2, 6, 12}).
  {1, 2, 3, 9}  = S({1, 2, 6, 9}).
  {1, 2, 3, 12} = S({1, 2, 6, 8}).
  {1, 2, 3, 21} = S({1, 2, 6, 7}).
  {1, 2, 4, 5}  = S({1, 2, 4, 20}).
  {1, 2, 4, 6}  = S({1, 2, 4, 12}).
  {1, 2, 4, 8}  = S({1, 2, 4, 8}).
  {1, 2, 4, 12} = S({1, 2, 4, 6}).
  {1, 2, 4, 20} = S({1, 2, 4, 5}).
  {1, 2, 6, 7}  = S({1, 2, 3, 21}).
  {1, 2, 6, 8}  = S({1, 2, 3, 12}).
  {1, 2, 6, 9}  = S({1, 2, 3, 9}).
  {1, 2, 6, 12} = S({1, 2, 3, 6}).
  {1, 2, 6, 15} = S({1, 2, 3, 5}).
  {1, 2, 6, 24} = S({1, 2, 3, 4}).
b(1) = 1 by definition, b(2) = 1+1 as 1 has only 1 as divisor.
a(3) = A000005(b(2)*b(2)) = 3.
The divisors of b(2) are 1,2,4. So b(3) can be b(2)+1, b(2)+2 and b(2)+4.
a(4) = A000005((b(2)+1)*(b(2)+4)) + A000005((b(2)+2)*(b(2)+2)) + A000005((b(2)+4)*(b(2)+1)) = 17.

Cf. A000005.
Cf. A000027, A000045, A000079, A000142, A001405.
Cf. A001654, A004277, A006501, A008233, A019442.
Cf. A010551, A019464, A026549, A031923, A038754.
Cf. A057895, A058295, A062112, A066332, A100071.
Cf. A079352, A082458, A087811, A093968, A098011.
Cf. A098558, A100538, A111286, A166447, A137326.
Cf. A138278, A171647, A205825, A208147, A264557.
Cf. A264635, A308546, A329227, A336496, A349079.
Cf. A349080, A359039.

cp's OEIS Frontend

A079000 a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is odd".

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