A290251 -id:A290251 - cp's OEIS frontend

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

1, 2, 4, 3, 8, 9, 6, 5, 16, 27, 18, 25, 12, 15, 10, 7, 32, 81, 54, 125, 36, 75, 50, 49, 24, 45, 30, 35, 20, 21, 14, 11, 64, 243, 162, 625, 108, 375, 250, 343, 72, 225, 150, 245, 100, 147, 98, 121, 48, 135, 90, 175, 60, 105, 70, 77, 40, 63, 42, 55, 28, 33, 22, 13, 128

Offset: 0

Leroy Quet, Jul 29 2009

This is a permutation of the positive integers.
From Antti Karttunen, Jun 20 2014: (Start)
Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.
Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.
Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:
                                     1
                                     |
                  ...................2...................
                 4                                       3
       8......../ \........9                   6......../ \........5
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  16       27         18       25         12       15         10       7
32  81   54  125    36  75   50  49     24  45   30  35     20  21   14 11
etc.
Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.
(End)
Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020
From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)
This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).
Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?
Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)
From Antti Karttunen, Jul 25 2023: (Start)
(In the above question, it is assumed that the starting offset would be 1 instead of 0).
Questions:
Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?
It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.
(End)
If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023
Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.
See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

			For n=3, whose binary representation is "11", we have A000120(3)=2, with A163510(3,1) = A163510(3,2) = 0, thus a(3) = p(2) * p(1)^0 * p(2)^0 = 3*1*1 = 3.
For n=9, "1001" in binary, we have A000120(9)=2, with A163510(9,1) = 0 and A163510(9,2) = 2, thus a(9) = p(2) * p(1)^0 * p(2)^2 = 3*1*9 = 27.
For n=10, "1010" in binary, we have A000120(10)=2, with A163510(10,1) = 1 and A163510(10,2) = 1, thus a(10) = p(2) * p(1)^1 * p(2)^1 = 3*2*3 = 18.
For n=15, "1111" in binary, we have A000120(15)=4, with A163510(15,1) = A163510(15,2) = A163510(15,3) = A163510(15,4) = 0, thus a(15) = p(4) * p(1)^0 * p(2)^0 * p(3)^0 * p(4)^0 = 7*1*1*1*1 = 7.
[1], [2], [1,1], [3], [1,2], [2,1] ... -> [1], [2], [3], [1,2], ... -> [0], [1], [2], [0,1], ... -> 2^0, 2^1, 2^2, 2^0*3^1, ... = 1, 2, 4, 3, ... - _Lorenzo Sauras Altuzarra_, Nov 28 2020

Inverse: A243071.
Cf. A000040, A000120, A000225, A000788, A003961, A007814, A054429, A055396, A064216, A135523, A161992, A163510, A245605, A245612, A246375, A246378, A246681, A161511, A228351, A243499, A243503, A243504, A269854, A280873, A285727, A290251, A293437, A337909.
Cf. A007283 (known positions where a(n)=n), A029747, A029748, A364255 [= gcd(n,a(n))], A364258 [= a(n)-n], A364287 (where a(n) < n), A364292 (where a(n) <= n), A364494 (where n|a(n)), A364496 (where a(n)|n), A364963, A364297.
Cf. A365808 (positions of squares), A365801 (of cubes), A365802 (of fifth powers), A365805 [= A052409(a(n))], A366287, A366391.
Cf. A005940, A332214, A332817, A366275 (variants).

f[n_] := Reverse@ Map[Ceiling[(Length@ # - 1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; {1}~Join~
Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}] (* Michael De Vlieger, Jul 25 2016 *)

from sympy import prime
def A163511(n):
    if n:
        k, c, m = n, 0, 1
        while k:
            c += 1
            m *= prime(c)**(s:=(~k&k-1).bit_length())
            k >>= s+1
        return m*prime(c)
    return 1 # Chai Wah Wu, Jul 17 2023

For n >= 1, a(2n) is even, a(2n+1) is odd. a(2^k) = 2^(k+1), for all k >= 0.
From Antti Karttunen, Jun 20 2014: (Start)
a(0) = 1, a(1) = 2, a(2n) = 2*a(n), a(2n+1) = A003961(a(n)).
As a more general observation about the parity, we have:
For n >= 1, A007814(a(n)) = A135523(n) = A007814(n) + A209229(n). [This permutation preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is incremented by one.]
For n >= 1, A055396(a(n)) = A091090(n) = A007814(n+1) + 1 - A036987(n).
For n >= 1, a(A000225(n)) = A000040(n).
(End)
From Antti Karttunen, Oct 11 2014: (Start)
As a composition of related permutations:
a(n) = A005940(1+A054429(n)).
a(n) = A064216(A245612(n))
a(n) = A246681(A246378(n)).
Also, for all n >= 0, it holds that:
A161511(n) = A243503(a(n)).
A243499(n) = A243504(a(n)).
(End)
More linking identities from Antti Karttunen, Dec 30 2017: (Start)
A046523(a(n)) = A278531(n). [See also A286531.]
A278224(a(n)) = A285713(n). [Another filter-sequence.]
A048675(a(n)) = A135529(n) seems to hold for n >= 1.
A250245(a(n)) = A252755(n).
A252742(a(n)) = A252744(n).
A245611(a(n)) = A253891(n).
A249824(a(n)) = A275716(n).
A292263(a(n)) = A292264(n). [A292944(n) + A292264(n) = n.]
--
A292383(a(n)) = A292274(n).
A292385(a(n)) = A292271(n). [A292271(n) +  A292274(n) = n.]
--
A292941(a(n)) = A292942(n).
A292943(a(n)) = A292944(n).
A292945(a(n)) = A292946(n). [A292942(n) + A292944(n) + A292946(n) = n.]
--
A292253(a(n)) = A292254(n).
A292255(a(n)) = A292256(n). [A292944(n) + A292254(n) + A292256(n) = n.]
--
A279339(a(n)) = A279342(n).
a(A071574(n)) = A269847(n).
a(A279341(n)) = A279338(n).
a(A252756(n)) = A250246(n).
(1+A008836(a(n)))/2 = A059448(n).
(End)
From Antti Karttunen, Jul 26 2023: (Start)
For all n >= 0, a(A007283(n)) = A007283(n).
A001222(a(n)) = A290251(n).
(End)

More terms computed and examples added by Antti Karttunen, Jun 20 2014

A366275 The Cat's tongue permutation: a(n) = A163511(A057889(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 6, 5, 16, 27, 18, 15, 12, 25, 10, 7, 32, 81, 54, 45, 36, 75, 30, 21, 24, 125, 50, 35, 20, 49, 14, 11, 64, 243, 162, 135, 108, 225, 90, 63, 72, 375, 150, 105, 60, 147, 42, 33, 48, 625, 250, 175, 100, 245, 70, 55, 40, 343, 98, 77, 28, 121, 22, 13, 128, 729, 486, 405, 324, 675, 270, 189, 216, 1125

Offset: 0

Antti Karttunen, Oct 06 2023

"Cat's tongue" refers to the look of the scatter plot of this sequence.

Cf. A000040, A000225, A007814, A057889, A163511, A209229, A290251, A366276 (inverse map), A366277 (fixed points of map n -> a(n)), A366278, A366279, A366280, A366281 [= A052409(a(n))], A366282 [= a(n)-n], A366283 [= gcd(n,a(n))].

Cf. also A163511, A253563, A366263 (compare the scatter plots).

A030101(n) = if(n<1,0,subst(Polrev(binary(n)),x,2));
A057889(n) = if(!n,n,A030101(n/(2^valuation(n,2))) * (2^valuation(n, 2)));
A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
A366275(n) = A163511(A057889(n));

from sympy import prime
def A366275(n):
    if n:
        k, c, m = int(bin(n>>(r:=(~n & n-1).bit_length()))[:1:-1],2)<>= s+1
        return m*prime(c)
    return 1 # Chai Wah Wu, Oct 08 2023

For n >= 0, A001222(a(n)) = A290251(n).
For n >= 1, A007814(a(n)) = A135523(n) = A007814(n) + A209229(n). [Like A163511, also this permutation preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is incremented by one.]
For n >= 1, a(2*n) = 2*a(n).
For n >= 1, a(A000225(n)) = A000040(n).

A008687 Number of 1's in 2's complement representation of -n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 3, 2, 3, 2, 2, 1, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3, 2, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 7, 6, 6, 5, 6, 5, 5, 4, 6, 5, 5, 4, 5, 4, 4, 3, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3

Offset: 0

R. H. Hardin

a(A127904(n)) = n and a(m) < n for m < A127904(n). - Reinhard Zumkeller, Feb 05 2007
a(n) = A000120(A010078(n)), n>0; a(n) = A023416(A004754(n-1)), n>1. - Reinhard Zumkeller, Dec 04 2015
Conjecture: a(n)+1 is the length of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020
Terms a(n); n >= 2 can be generated recursively, as follows. Let S(0) = {1}, then for k >=1, let S(k) = {S(k-1)+1, S(k-1)}, where +1 means +1 on every term of S(k-1); see Example. Each step of the recursion gives the next 2^k terms of the sequence. - David James Sycamore, Jul 15 2024

			Using the above recursion for a(n); n >= 2, we have:
 S(0) = {1} so a(2) = 1;
 S(1) = {2,1} so a(3,4) = 2,1;
 S(2) = {3,2,2,1}, so a(5,6,7,8) = 3,2,2,1;
As irregular table the sequence for n >= 2 begins:
  1;
  2,1;
  3,2,2,1;
  4,3,3,2,3,2,2,1;
  5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
  6,5,5,4,5,4,4,3,5,4,3,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
and so on (the k-th row contains 2^k terms; k>=0). - _David James Sycamore_, Jul 15 2024

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Wikipedia, Two's complement

This is Guy Steele's sequence GS(4, 3) (see A135416).
Cf. A000120, A004754, A010078, A023416, A290251.
Cf. A007305, A007306, A061313, A088696.

a008687 n = a008687_list !! n
a008687_list = 0 : 1 : c [1] where c (e:es) = e : c (es ++ [e+1,e])
-- Reinhard Zumkeller, Mar 07 2011

a[0] = 0; a[1] = 1; a[n_] := a[n] = Mod[n,2] + a[Mod[n,2] + Floor[n/2]]; Array[a, 96, 0] (* Jean-François Alcover, Aug 12 2017, after Reinhard Zumkeller *)

a(n) = if(n<2,n, n--; logint(n,2) - hammingweight(n) + 2); \\ Kevin Ryde, Apr 14 2021

a(n) = A023416(n-1) + 1.
a(n) = if n<=1 then n else (n mod 2) + a((n mod 2) + floor(n/2)). - Reinhard Zumkeller, Feb 05 2007
a(n) = if n<2 then n else a(ceiling(n/2)) + n mod 2. - Reinhard Zumkeller, Jul 25 2006
Min{m: a(m)=n} = if n>0 then A083318(n-1) else 0. - Reinhard Zumkeller, Jul 25 2006

A332214 Mersenne-prime fixing variant of permutation A163511: a(n) = A332212(A163511(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 6, 7, 16, 27, 18, 49, 12, 21, 14, 5, 32, 81, 54, 343, 36, 147, 98, 25, 24, 63, 42, 35, 28, 15, 10, 31, 64, 243, 162, 2401, 108, 1029, 686, 125, 72, 441, 294, 175, 196, 75, 50, 961, 48, 189, 126, 245, 84, 105, 70, 155, 56, 45, 30, 217, 20, 93, 62, 11, 128, 729, 486, 16807, 324, 7203, 4802, 625, 216, 3087, 2058, 875

Offset: 0

Antti Karttunen, Feb 09 2020

Any Mersenne prime (A000668) times any power of 2, i.e., sequence A335431, is fixed by this map (note the indexing), including also all even perfect numbers. It is not currently known whether there are any additional fixed points.

Because a(n) has the same prime signature as A163511(n), it implies that applying A046523 and A052409 to this sequence gives the same results as with A163511, namely, sequences A278531 and A365805. - Antti Karttunen, Oct 09 2023

Cf. A163511, A332211, A332212, A332215 (inverse permutation).
Cf. A278531 [= A046523(a(n))], A290251 [= A001222(a(n))], A365805 [= A052409(a(n))], A366372 [= a(n)-n], A366373 [= gcd(n,a(n))], A366374 (numerator of n/a(n)), A366375 (denominator of n/a(n)), A366376.
Cf. A000043, A000668, A000396, A324200, A335431 (conjectured to give all the fixed points).

PARI
```
A332214(n) = A332212(A163511(n));
```

\\ Needs precomputed data for A332211:
v332211 = readvec("b332211_to.txt"); \\ Prepared with gawk ' { print $2 } ' < b332211.txt > b332211_to.txt
A332211(n) = v332211[n];
A332214(n) = if(!n, 1, my(i=1, p=A332211(i), t=1); while(n>1, if(!(n%2), (t*=p), i++; p=A332211(i)); n >>= 1); (t*p)); \\ Antti Karttunen, Oct 09 2023

a(n) = A332212(A163511(n)).

A290253 Triangle read by rows. Row n consists of the parts, ordered nonincreasingly, of the integer partition having viabin number n.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 3, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 3, 2, 3, 3, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 1, 1, 2, 2, 2, 1, 3, 2, 1, 3, 3, 1, 4, 1, 2, 2, 2, 2, 3, 2, 2, 3, 3, 2, 4, 2, 3, 3, 3, 4, 3, 4, 4, 5, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 2, 1, 1, 3, 2, 1, 1, 3, 3, 1, 1, 4, 1, 1, 2, 2, 2, 2, 1, 3, 2, 2, 1, 3, 3, 2, 1, 4, 2, 1, 3, 3, 3, 1, 4, 3, 1, 4, 4, 1, 5, 1, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2

Offset: 0

Emeric Deutsch, Aug 23 2017

The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20.
Number of entries in row n is A290251(n).
In the Maple program the command vitopart(n) yields the integer partition having viabin number n.

			Row 25 is 3,2,2. Indeed, the binary form of 25 is 11001. Consequently, the southeast border of the Ferrers board of the associated partition is EENNEN, where E and N are the steps [1,0] and [0,1], respectively. This leads to the partition [3,2,2].
Triangle begins:
0,
1;
1,1;
2;
1,1,1;
2,1;
2,2;
3;

Alois P. Heinz, Rows n = 0..4095, flattened

Row sums give A161511.
Row lengths give A008687(n+1).
Cf. A253563, A290251.

# (due to W. Edwin Clark)
vitopart := proc (n) local L, i, j, N, p, t; N := 2*n; L := ListTools:-Reverse(convert(N, base, 2)); j := 0; for i to nops(L) do if L[i] = 0 then j := j+1; p[j] := numboccur(L[1 .. i], 1) end if end do; sort([seq(p[t], t = 1 .. j)], `>=`) end proc:
# second Maple program:
T:= proc(n) local m; m:= n; [0]; while m>0 do `if`(1=
      irem(m, 2, 'm'), map(x-> x+1, %), [%[], 0]) od: %[]
    end:
seq(T(n), n=0..50);  # Alois P. Heinz, Aug 23 2017

T[n_] := Module[{L = IntegerDigits[2n, 2], j = 0, p}, Do[If[L[[i]] == 0, j++; p[j] = Count[L[[;;i]], 1]], {i, 1, Length[L]}]; Array[p, j] // Reverse];
Table[T[n], {n, 0, 50}] // Flatten (* Jean-François Alcover, Aug 06 2024, after W. Edwin Clark *)

A342456 A276086 applied to the primorial inflation of Doudna-tree, where A276086(n) is the prime product form of primorial base expansion of n.

Original entry on oeis.org

2, 3, 5, 9, 7, 25, 35, 15, 11, 49, 117649, 625, 717409, 1225, 55, 225, 13, 121, 1771561, 2401, 36226650889, 184877, 1127357, 875, 902613283, 514675673281, 3780549773, 1500625, 83852850675321384784127, 3025, 62004635, 21, 17, 169, 4826809, 14641, 8254129, 143, 2924207, 77, 8223741426987700773289, 59797108943, 546826709

Offset: 0

Antti Karttunen and Michael De Vlieger, Mar 14 2021

nonn

This sequence (which could be viewed as a binary tree, like the underlying A005940 and A329886) is similar to A324289, but unlike its underlying tree A283477 that generates only numbers that are products of distinct primorial numbers (i.e., terms of A129912), here the underlying tree A329886 generates all possible products of primorial numbers, i.e., terms of A025487, but in different order.

Cf. A005940, A025487, A108951, A129912, A276086, A283980, A324886, A342457 [= 2*A246277(a(n))], A342461 [= A001221(a(n))], A342462 [= A001222(a(n))], A342463 [= A342001(a(n))], A342464 [= A051903(a(n))].

Cf. A324289 (a subset of these terms, in different order).

Block[{a, f, r = MixedRadix[Reverse@ Prime@ Range@ 24]}, f[n_] :=
Times @@ MapIndexed[Prime[First[#2]]^#1 &, Reverse@ IntegerDigits[n, r]]; a[0] = 1; a[1] = 2; a[n_] := a[n] = If[EvenQ@ n, (Times @@ Map[Prime[PrimePi@ #1 + 1]^#2 & @@ # &, FactorInteger[#]] - Boole[# == 1])*2^IntegerExponent[#, 2] &[a[n/2]], 2 a[(n - 1)/2]]; Array[f@ a[#] &, 43, 0]] (* Michael De Vlieger, Mar 17 2021 *)

A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A283980(n) = {my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 6, nextprime(p+1))^e)};
A329886(n) = if(n<2,1+n,if(!(n%2),A283980(A329886(n/2)),2*A329886(n\2)));
A342456(n) = A276086(A329886(n));

a(n) = A276086(A329886(n)) = A324886(A005940(1+n)).
For all n >= 0, gcd(a(n), A329886(n)) = 1.
For all n >= 1, A055396(a(n))-1 = A061395(A329886(n)) = A290251(n) = 1+A080791(n).
For all n >= 0, a(2^n) = A000040(2+n).

A366280 Lexicographically earliest infinite sequence such that a(i) = a(j) => A366279(i) = A366279(j) for all i, j >= 0.

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 5, 2, 6, 4, 7, 5, 7, 3, 5, 2, 8, 6, 9, 7, 10, 7, 11, 5, 9, 4, 7, 5, 7, 3, 5, 2, 12, 8, 13, 9, 14, 10, 15, 7, 14, 9, 15, 11, 15, 7, 11, 5, 13, 6, 9, 7, 10, 7, 11, 5, 9, 4, 7, 5, 7, 3, 5, 2, 16, 12, 17, 13, 18, 14, 19, 9, 20, 14, 21, 15, 21, 10, 15, 7, 18, 13, 19, 15, 21, 15, 22, 11, 19, 9, 15, 11, 15

Offset: 0

Antti Karttunen, Oct 06 2023

Restricted growth sequence transform of A366279.

For all i, j >= 0, a(i) = a(j) => A290251(i) = A290251(j).

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A046523, A057889, A163511, A278531, A290251, A366275, A366279.

Cf. also A286531, A286602, A366262.

up_to = 65537;
rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };
A030101(n) = if(n<1,0,subst(Polrev(binary(n)),x,2));
A057889(n) = if(!n,n,A030101(n/(2^valuation(n,2))) * (2^valuation(n, 2)));
A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
A366275(n) = A163511(A057889(n));
A366279(n) = A046523(A366275(n));
v366280 = rgs_transform(vector(1+up_to,n,A366279(n-1)));
A366280(n) = v366280[1+n];

A290252 Number of standard tableaux of the integer partition having viabin number n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 5, 5, 5, 1, 1, 4, 9, 6, 14, 16, 21, 4, 14, 21, 42, 9, 42, 14, 14, 1, 1, 5, 14, 10, 28, 35, 56, 10, 42, 70, 168, 35, 210, 70, 84, 5, 42, 84, 252, 56, 462, 168, 252, 14, 462, 210, 462, 28, 462, 42, 42, 1, 1, 6, 20, 15, 48, 64, 120, 20, 90, 162, 450, 90, 660, 216, 300, 15, 132, 288, 990, 216, 2112

Offset: 0

Emeric Deutsch, Aug 22 2017

The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20.

In the first Maple program (due basically to W. Edwin Clark) (i) the command vitopart(n) yields the partition having viabin number n (for example, vitopart(9) = [2,1,1]), (ii) the command H(p) yields the product of the hooklengths of the partition p (for example, H([2,1,1]) = 8), (iii) the command ST(n) yields the number of standard tableaux corresponding to the partition having viabin number n.

			a(9) = 3; indeed, the partition with viabin number 9 is [2,1,1], the hooklength of the corresponding Ferrers board are 4,1,2,1 and, consequently, the number of standard tableaux is 4!/(4*2) = 3.

Alois P. Heinz, Table of n, a(n) for n = 0..16384

Cf. A290251.

vitopart := proc (n) local L, i, j, N, p, t; N := 2*n; L := ListTools:-Reverse(convert(N, base, 2)); j := 0; for i to nops(L) do if L[i] = 0 then j := j+1; p[j] := numboccur(L[1 .. i], 1) end if end do; sort([seq(p[t], t = 1 .. j)], `>=`) end proc: H := proc (pa) local F, j, p, Q, i, col, a, A; F := proc (x) local i, ct: ct := 0: for i to nops(x) do if 1 < x[i] then ct := ct+1 else  end if end do: ct end proc: for j to nops(pa) do p[1][j] := pa[j] end do: Q[1] := [seq(p[1][j], j = 1 .. nops(pa))]: for i from 2 to pa[1] do for j to F(Q[i-1]) do p[i][j] := Q[i-1][j]-1 end do: Q[i] := [seq(p[i][j], j = 1 .. F(Q[i-1]))] end do: for i to pa[1] do col[i] := [seq(Q[i][j]+nops(Q[i])-j, j = 1 .. nops(Q[i]))] end do: a := proc (i, j) if i <= nops(Q[j]) and j <= pa[1] then Q[j][i]+nops(Q[j])-i else 1 end if end proc: A := matrix(nops(pa), pa[1], a): product(product(A[m, n], n = 1 .. pa[1]), m = 1 .. nops(pa)) end proc: ST := proc (m) options operator, arrow: factorial(add(u, `in`(u, vitopart(m))))/H(vitopart(m)) end proc: seq(ST(q), q = 0 .. 120);
# second Maple program:
a:= proc(n) local l, m; m:= n; [0]; while m>0 do `if`(1=
      irem(m, 2, 'm'), map(x-> x+1, %), [%[], 0]) od: l:=%:
      (h-> add(i, i=l)!/mul(mul(1+l[i]-j+add(`if`(j>l[k]
      , 0, 1), k=i+1..h), j=1..l[i]), i=1..h))(nops(l))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Aug 22 2017

cp's OEIS Frontend

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

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A366275 The Cat's tongue permutation: a(n) = A163511(A057889(n)).

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A008687 Number of 1's in 2's complement representation of -n.

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A332214 Mersenne-prime fixing variant of permutation A163511: a(n) = A332212(A163511(n)).

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A290253 Triangle read by rows. Row n consists of the parts, ordered nonincreasingly, of the integer partition having viabin number n.

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A342456 A276086 applied to the primorial inflation of Doudna-tree, where A276086(n) is the prime product form of primorial base expansion of n.

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A366280 Lexicographically earliest infinite sequence such that a(i) = a(j) => A366279(i) = A366279(j) for all i, j >= 0.

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