cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A122111 Self-inverse permutation of the positive integers induced by partition enumeration in A112798 and partition conjugation.

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 5, 9, 12, 32, 10, 64, 24, 18, 7, 128, 15, 256, 20, 36, 48, 512, 14, 27, 96, 25, 40, 1024, 30, 2048, 11, 72, 192, 54, 21, 4096, 384, 144, 28, 8192, 60, 16384, 80, 50, 768, 32768, 22, 81, 45, 288, 160, 65536, 35, 108, 56, 576, 1536, 131072, 42
Offset: 1

Views

Author

Keywords

Comments

Factor n; replace each prime(i) with i, take the conjugate partition, replace parts i with prime(i) and multiply out.
From Antti Karttunen, May 12-19 2014: (Start)
For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).
Because the partition conjugation doesn't change the partition's total sum, this permutation preserves A056239, i.e., A056239(a(n)) = A056239(n) for all n.
(Similarly, for all n, A001221(a(n)) = A001221(n), because the number of steps in the Ferrers/Young-diagram stays invariant under the conjugation. - Note added Apr 29 2022).
Because this permutation commutes with A241909, in other words, as a(A241909(n)) = A241909(a(n)) for all n, from which follows, because both permutations are self-inverse, that a(n) = A241909(a(A241909(n))), it means that this is also induced when partitions are conjugated in the partition enumeration system A241918. (Not only in A112798.)
(End)
From Antti Karttunen, Jul 31 2014: (Start)
Rows in arrays A243060 and A243070 converge towards this sequence, and also, assuming no surprises at the rate of that convergence, this sequence occurs also as the central diagonal of both.
Each even number is mapped to a unique term of A102750 and vice versa.
Conversely, each odd number (larger than 1) is mapped to a unique term of A070003, and vice versa. The permutation pair A243287-A243288 has the same property. This is also used to induce the permutations A244981-A244984.
Taking the odd bisection and dividing out the largest prime factor results in the permutation A243505.
Shares with A245613 the property that each term of A028260 is mapped to a unique term of A244990 and each term of A026424 is mapped to a unique term of A244991.
Conversely, with A245614 (the inverse of above), shares the property that each term of A244990 is mapped to a unique term of A028260 and each term of A244991 is mapped to a unique term of A026424.
(End)
The Maple program follows the steps described in the first comment. The subprogram C yields the conjugate partition of a given partition. - Emeric Deutsch, May 09 2015
The Heinz number of the partition that is conjugate to the partition with Heinz number n. The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r). Example: a(3) = 4. Indeed, the partition with Heinz number 3 is [2]; its conjugate is [1,1] having Heinz number 4. - Emeric Deutsch, May 19 2015

Crossrefs

Cf. A088902 (fixed points).
Cf. A112798, A241918 (conjugates the partitions listed in these two tables).
Cf. A243060 and A243070. (Limit of rows in these arrays, and also their central diagonal).
Cf. A319988 (parity of this sequence for n > 1), A336124 (a(n) mod 4).
{A000027, A122111, A241909, A241916} form a 4-group.
{A000027, A122111, A153212, A242419} form also a 4-group.
Cf. also array A350066 [A(i, j) = a(a(i)*a(j))].

Programs

  • Maple
    with(numtheory): c := proc (n) local B, C: B := proc (n) local pf: pf := op(2, ifactors(n)): [seq(seq(pi(op(1, op(i, pf))), j = 1 .. op(2, op(i, pf))), i = 1 .. nops(pf))] end proc: C := proc (P) local a: a := proc (j) local c, i: c := 0; for i to nops(P) do if j <= P[i] then c := c+1 else  end if end do: c end proc: [seq(a(k), k = 1 .. max(P))] end proc: mul(ithprime(C(B(n))[q]), q = 1 .. nops(C(B(n)))) end proc: seq(c(n), n = 1 .. 59); # Emeric Deutsch, May 09 2015
    # second Maple program:
    a:= n-> (l-> mul(ithprime(add(`if`(jAlois P. Heinz, Sep 30 2017
  • Mathematica
    A122111[1] = 1; A122111[n_] := Module[{l = #, m = 0}, Times @@ Power @@@ Table[l -= m; l = DeleteCases[l, 0]; {Prime@Length@l, m = Min@l}, Length@Union@l]] &@Catenate[ConstantArray[PrimePi[#1], #2] & @@@ FactorInteger@n]; Array[A122111, 60] (* JungHwan Min, Aug 22 2016 *)
    a[n_] := Function[l, Product[Prime[Sum[If[jJean-François Alcover, Sep 23 2020, after Alois P. Heinz *)
  • PARI
    A122111(n) = if(1==n,n,my(f=factor(n), es=Vecrev(f[,2]),is=concat(apply(primepi,Vecrev(f[,1])),[0]),pri=0,m=1); for(i=1, #es, pri += es[i]; m *= prime(pri)^(is[i]-is[1+i])); (m)); \\ Antti Karttunen, Jul 20 2020
    
  • Python
    from sympy import factorint, prevprime, prime, primefactors
    from operator import mul
    def a001222(n): return 0 if n==1 else a001222(n/primefactors(n)[0]) + 1
    def a064989(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==2 else prevprime(i)**f[i] for i in f])
    def a105560(n): return 1 if n==1 else prime(a001222(n))
    def a(n): return 1 if n==1 else a105560(n)*a(a064989(n))
    [a(n) for n in range(1, 101)] # Indranil Ghosh, Jun 15 2017
  • Scheme
    ;; Uses Antti Karttunen's IntSeq-library.
    (definec (A122111 n) (if (<= n 1) n (* (A000040 (A001222 n)) (A122111 (A064989 n)))))
    ;; Antti Karttunen, May 12 2014
    
  • Scheme
    ;; Uses Antti Karttunen's IntSeq-library.
    (definec (A122111 n) (if (<= n 1) n (* (A000079 (A241917 n)) (A003961 (A122111 (A052126 n))))))
    ;; Antti Karttunen, May 12 2014
    
  • Scheme
    ;; Uses Antti Karttunen's IntSeq-library.
    (definec (A122111 n) (if (<= n 1) n (* (expt (A000040 (A071178 n)) (A241919 n)) (A242378bi (A071178 n) (A122111 (A051119 n))))))
    ;; Antti Karttunen, May 12 2014
    

Formula

From Antti Karttunen, May 12-19 2014: (Start)
a(1) = 1, a(p_i) = 2^i, and for other cases, if n = p_i1 * p_i2 * p_i3 * ... * p_{k-1} * p_k, where p's are primes, not necessarily distinct, sorted into nondescending order so that i1 <= i2 <= i3 <= ... <= i_{k-1} <= ik, then a(n) = 2^(ik-i_{k-1}) * 3^(i_{k-1}-i_{k-2}) * ... * p_{i_{k-1}}^(i2-i1) * p_ik^(i1).
This can be implemented as a recurrence, with base case a(1) = 1,
and then using any of the following three alternative formulas:
a(n) = A105560(n) * a(A064989(n)) = A000040(A001222(n)) * a(A064989(n)). [Cf. the formula for A242424.]
a(n) = A000079(A241917(n)) * A003961(a(A052126(n))).
a(n) = (A000040(A071178(n))^A241919(n)) * A242378(A071178(n), a(A051119(n))). [Here ^ stands for the ordinary exponentiation, and the bivariate function A242378(k,n) changes each prime p(i) in the prime factorization of n to p(i+k), i.e., it's the result of A003961 iterated k times starting from n.]
a(n) = 1 + A075157(A129594(A075158(n-1))). [Follows from the commutativity with A241909, please see the comments section.]
(End)
From Antti Karttunen, Jul 31 2014: (Start)
As a composition of related permutations:
a(n) = A153212(A242419(n)) = A242419(A153212(n)).
a(n) = A241909(A241916(n)) = A241916(A241909(n)).
a(n) = A243505(A048673(n)).
a(n) = A064216(A243506(n)).
Other identities. For all n >= 1, the following holds:
A006530(a(n)) = A105560(n). [The latter sequence gives greatest prime factor of the n-th term].
a(2n)/a(n) = A105560(2n)/A105560(n), which is equal to A003961(A105560(n))/A105560(n) when n > 1.
A243505(n) = A052126(a(2n-1)) = A052126(a(4n-2)).
A066829(n) = A244992(a(n)) and vice versa, A244992(n) = A066829(a(n)).
A243503(a(n)) = A243503(n). [Because partition conjugation does not change the partition size.]
A238690(a(n)) = A238690(n). - per Matthew Vandermast's note in that sequence.
A238745(n) = a(A181819(n)) and a(A238745(n)) = A181819(n). - per Matthew Vandermast's note in A238745.
A181815(n) = a(A181820(n)) and a(A181815(n)) = A181820(n). - per Matthew Vandermast's note in A181815.
(End)
a(n) = A181819(A108951(n)). [Prime shadow of the primorial inflation of n] - Antti Karttunen, Apr 29 2022

A241909 Self-inverse permutation of natural numbers: a(1)=1, a(p_i) = 2^i, and if n = p_i1 * p_i2 * p_i3 * ... * p_{ik-1} * p_ik, where p's are primes, with their indexes are sorted into nondescending order: i1 <= i2 <= i3 <= ... <= i_{k-1} <= ik, then a(n) = 2^(i1-1) * 3^(i2-i1) * 5^(i3-i2) * ... * p_k^(1+(ik-i_{k-1})). Here k = A001222(n) and ik = A061395(n).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 16, 5, 6, 27, 32, 25, 64, 81, 18, 7, 128, 15, 256, 125, 54, 243, 512, 49, 12, 729, 10, 625, 1024, 75, 2048, 11, 162, 2187, 36, 35, 4096, 6561, 486, 343, 8192, 375, 16384, 3125, 50, 19683, 32768, 121, 24, 45, 1458, 15625, 65536, 21, 108, 2401
Offset: 1

Views

Author

Antti Karttunen, May 03 2014, partly inspired by Marc LeBrun's Jan 11 2006 message on SeqFan mailing list

Keywords

Comments

This permutation maps between the partitions as ordered in A112798 and A241918 (the original motivation for this sequence).
For all n > 2, A007814(a(n)) = A055396(n)-1, which implies that this self-inverse permutation maps between primes (A000040) and the powers of two larger than one (A000079(n>=1)), and apart from a(1) & a(2), this also maps each even number to some odd number, and vice versa, which means there are no fixed points after 2.
A122111 commutes with this one, that is, a(n) = A122111(a(A122111(n))).
Conjugates between A243051 and A242424 and other rows of A243060 and A243070.

Examples

			For n = 12 = 2 * 2 * 3 = p_1 * p_1 * p_2, we obtain by the first formula 2^(1-1) * 3^(1-1) * 5^(1+(2-1)) = 5^2 = 25. By the second formula, as n = 2^2 * 3^1, we obtain the same result, p_{1+2} * p_{2+1} = p_3 * p_3 = 25, thus a(12) = 25.
Using the product formula over the terms of row n of table A241918, we see, because 9450 = 2*3*3*3*5*5*7 = p_1^1 * p_2^3 * p_3^2 * p_4^1, that the corresponding row in A241918 is {2,5,7,7}, and multiplying p_2 * p_5 * p_7^2 yields 3 * 11 * 17 * 17 = 9537, thus a(9450) = 9537.
Similarly, for 9537, the corresponding row in A241918 is {1,2,2,2,3,3,4}, and multiplying p_1^1 * p_2^3 * p_3^2 * p_4^1, we obtain 9450 back.
		

Crossrefs

Cf. also A278220 (= A046523(a(n))), A331280 (its rgs_transform), A331299 (= min(n,a(n))).
{A000027, A122111, A241909, A241916} form a 4-group.

Programs

  • Haskell
    a241909 1 = 1
    a241909 n = product $ zipWith (^) a000040_list $ zipWith (-) is (1 : is)
                where is = reverse ((j + 1) : js)
                      (j:js) = reverse $ map a049084 $ a027746_row n
    -- Reinhard Zumkeller, Aug 04 2014
    
  • Mathematica
    Array[If[# == 1, 1, Function[t, Times @@ Prime@ Accumulate[If[Length@ t < 2, {0}, Join[{1}, ConstantArray[0, Length@ t - 2], {-1}]] + ReplacePart[t, Map[#1 -> #2 & @@ # &, #]]]]@ ConstantArray[0, Transpose[#][[1, -1]]] &[FactorInteger[#] /. {p_, e_} /; p > 0 :> {PrimePi@ p, e}]] &, 56] (* Michael De Vlieger, Jan 23 2020 *)
  • PARI
    A241909(n) = if(1==n||isprime(n),2^primepi(n),my(f=factor(n),h=1,i,m=1,p=1,k=1); while(k<=#f~, p = nextprime(1+p); i = primepi(f[k,1]); m *= p^(i-h); h = i; if(f[k,2]>1, f[k,2]--, k++)); (p*m)); \\ Antti Karttunen, Jan 17 2020

Formula

If n is a prime with index i (p_i), then a(n) = 2^i, otherwise when n = p_i1 * p_i2 * p_i3 * ... p_ik, where p_i1, p_i2, p_i3, ..., p_ik are the primes present (not necessarily all distinct) in the prime factorization of n, sorted into nondescending order, a(n) = 2^(i1-1) * 3^(i2-i1) * 5^(i3-i2) * ... * p_k^(1+(ik-i_{k-1})).
Equally, if n = 2^k, then a(n) = p_k, otherwise, when n = 2^e1 * 3^e2 * 5^e3 * ... * p_k^{e_k}, i.e., where e1 ... e_k are the exponents (some of them possibly zero, except the last) of the primes 2, 3, 5, ... in the prime factorization of n, a(n) = p_{1+e1} * p_{1+e1+e2} * p_{1+e1+e2+e3} * ... * p_{e1+e2+e3+...+e_k}.
From the equivalence of the above two formulas (which are inverses of each other) it follows that a(a(n)) = n, i.e., that this permutation is an involution. For a proof, please see the attached notes.
The first formula corresponds to this recurrence:
a(1) = 1, a(p_k) = 2^k for primes with index k, otherwise a(n) = (A000040(A001222(n))^(A241917(n)+1)) * A052126(a(A052126(n))).
And the latter formula with this recurrence:
a(1) = 1, and for n>1, if n = 2^k, a(n) = A000040(k), otherwise a(n) = A000040(A001511(n)) * A242378(A007814(n), a(A064989(n))).
[Here A242378(k,n) changes each prime p(i) in the prime factorization of n to p(i+k), i.e., it's the result of A003961 iterated k times starting from n.]
We also have:
a(1)=1, and for n>1, a(n) = Product_{i=A203623(n-1)+2..A203623(n)+1} A000040(A241918(i)).
For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).
For all n > 1, a(2n-1) = 2*a(A064216(n)).

Extensions

Typos in the name corrected by Antti Karttunen, May 31 2014

A243503 Sums of parts of partitions (i.e., their sizes) as ordered in the table A241918: a(n) = Sum_{i=A203623(n-1)+2..A203623(n)+1} A241918(i).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 3, 3, 6, 5, 6, 6, 8, 5, 4, 7, 5, 8, 9, 7, 10, 9, 8, 4, 12, 4, 12, 10, 8, 11, 5, 9, 14, 6, 7, 12, 16, 11, 12, 13, 11, 14, 15, 7, 18, 15, 10, 5, 7, 13, 18, 16, 6, 8, 16, 15, 20, 17, 11, 18, 22, 10, 6, 10, 14, 19, 21, 17, 10, 20, 9, 21, 24, 6, 24
Offset: 1

Views

Author

Antti Karttunen, Jun 05 2014

Keywords

Comments

Each n occurs A000041(n) times in total.
Where are the first and the last occurrence of each n located?

Crossrefs

Cf. A243504 (the products of parts), A241918, A000041, A227183, A075158, A056239, A241909.
Sum of prime indices of A241916, the even bisection of A358195.
Sums of even-indexed rows of A358172.
A112798 lists prime indices, length A001222, sum A056239, max A061395.

Programs

  • Mathematica
    Table[If[n==1,0,With[{y=Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]},Last[y]*Length[y]+Last[y]-Total[y]+Length[y]-1]],{n,100}] (* Gus Wiseman, Jan 09 2023 *)

Formula

a(n) = Sum_{i=A203623(n-1)+2..A203623(n)+1} A241918(i).
a(n) = A056239(A241909(n)).
a(n) = A227183(A075158(n-1)).
a(A000040(n)) = a(A000079(n)) = n for all n >= 1.
a(A122111(n)) = a(n) for all n.
a(A243051(n)) = a(n) for all n, and likewise for A243052, A243053 and other rows of A243060.
a(n) = A061395(n) * A001222(n) + A061395(n) - A056239(n) + A001222(n) - 1. - Gus Wiseman, Jan 09 2023
a(n) = A326844(2n) + A001222(n). - Gus Wiseman, Jan 09 2023

A243070 Square array read by antidiagonals: rows are successively recursivized versions of Bulgarian solitaire operation (starting from the usual "first order" version, A242424), as applied to the partitions listed in A112798.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 3, 4, 2, 1, 6, 3, 4, 2, 1, 6, 8, 3, 4, 2, 1, 10, 6, 8, 3, 4, 2, 1, 5, 12, 6, 8, 3, 4, 2, 1, 12, 5, 16, 6, 8, 3, 4, 2, 1, 9, 9, 5, 16, 6, 8, 3, 4, 2, 1, 14, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 10, 20, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 22, 10, 24, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 15, 28, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 18, 18, 40, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The array is read by antidiagonals: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ... .
Please see comments and references in A242424 for more information about Bulgarian Solitaire.
Each row is a A241909-conjugate of the corresponding row in A243060.
Rows in both arrays converge towards A122111.
All the terms in column n are multiples of A105560(n).
The rows of this table (i.e., the corresponding functions) preserve A056239.
First point where row k differs from row k of A243060 seems to be A000040(k+2): primes from five onward: 5, 7, 11, 13, 17, 19, 23, 29, 31, ... and these seem to be also the points where that row differs for the first time from A122111.

Examples

			The top left corner of the array is:
  1,  2,  4,  3,  6,  6, 10,  5, 12,  9, 14, 10, 22, 15, 18, ...
  1,  2,  4,  3,  8,  6, 12,  5,  9, 12, 20, 10, 28, 18, 18, ...
  1,  2,  4,  3,  8,  6, 16,  5,  9, 12, 24, 10, 40, 24, 18, ...
  1,  2,  4,  3,  8,  6, 16,  5,  9, 12, 32, 10, 48, 24, 18, ...
  1,  2,  4,  3,  8,  6, 16,  5,  9, 12, 32, 10, 64, 24, 18, ...
		

Crossrefs

Row 1: A242424, Row 2: A243072, Row 3: A243073.
Rows converge towards A122111.

Programs

Formula

A(1,col) = A242424(col), otherwise, when row > 1, A(row,col) = A000040(A001222(col)) * A(row-1, A064989(col)).

A243051 Integer sequence induced by Bulgarian solitaire operation on partition list A241918: a(n) = A241909(A242424(A241909(n))).

Original entry on oeis.org

1, 2, 4, 3, 8, 25, 16, 9, 9, 343, 32, 10, 64, 14641, 125, 27, 128, 15, 256, 98, 2401, 371293, 512, 30, 27, 24137569, 6, 2662, 1024, 147, 2048, 81, 161051, 893871739, 625, 50, 4096, 78310985281, 4826809, 28, 8192, 3993, 16384, 57122, 50, 14507145975869, 32768, 90, 81
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

In "Bulgarian solitaire" a deck of cards or another finite set of objects is divided into one or more piles, and the "Bulgarian operation" is performed by taking one card from each pile, and making a new pile of them, which is added to the remaining set of piles. Essentially, this operation is a function whose domain and range are unordered integer partitions (cf. A000041) and which preserves the total size of a partition (the sum of its parts). This sequence is induced when the operation is implemented on the partitions as ordered by the list A241918.

Examples

			For n = 10, we see that as 10 = 2*5 = p_1^1 * p_2^0 * p_3^1, it encodes a partition [2,2,2]. Applying one step of Bulgarian solitaire (subtract one from each part, and add a new part as large as there were parts in the old partition) to this partition results a new partition [1,1,1,3], which is encoded in the prime factorization of p_1^0 * p_2^0 * p_3^0 * p_4^3 = 7^3 = 343. Thus a(10) = 343.
For n = 46, we see that as 46 = 2*23 = p_1 * p_9 = p_1^1 * p_2^0 * p_3^0 * ... * p_9^1, it encodes a partition [2,2,2,2,2,2,2,2,2]. Applying one step of Bulgarian solitaire to this partition results a new partition [1,1,1,1,1,1,1,1,1,9], which is encoded in the prime factorization of p_1^0 * p_2^0 * ... * p_9^0 * p_10^9 = 29^9 = 14507145975869. Thus a(46) = 14507145975869.
For n = 1875, we see that as 1875 = p_1^0 * p_2^1 * p_3^4, it encodes a partition [1,2,5]. Applying Bulgarian Solitaire, we get a new partition [1,3,4]. This in turn is encoded by p_1^0 * p_2^2 * p_3^2 = 3^2 * 5^2 = 225. Thus a(1875)=225.
		

References

  • Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.

Crossrefs

Row 1 of A243060 (table which gives successive "recursive iterates" of this sequence and converges towards A122111).
Fixed points: A243054.

Formula

a(n) = A241909(A242424(A241909(n))).
a(n) = 1 + A075157(A226062(A075158(n-1))).
A243503(a(n)) = A243503(n) for all n. [Because Bulgarian operation doesn't change the total sum of the partition].

A243052 Integer sequence induced by second order Bulgarian solitaire operation on partition list A241918: a(n) = A241909(A243072(A241909(n))).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 5, 9, 12, 32, 35, 64, 24, 18, 25, 128, 15, 256, 539, 36, 48, 512, 14, 27, 96, 25, 17303, 1024, 175, 2048, 125, 72, 192, 54, 21, 4096, 384, 144, 154, 8192, 3773, 16384, 485537, 245, 768, 32768, 70, 81, 45, 288, 26977283, 65536, 10, 108, 3146, 576, 1536, 131072
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The usual Bulgarian Solitaire operation (the "first order" version, cf. A243051) applied to an unordered integer partition means: subtract one from each part, and add a new part as large as there were parts in the old partition.
The "Second Order Bulgarian Solitaire" operation means that after subtracting one from each part of the old partition (and discarding the parts that diminished to zero), we apply the (first order) Bulgarian operation to the remaining partition before adding a new part as large as there were parts in the original partition.
How the partitions are encoded in this case, please see A241918.

Crossrefs

Second row of A243060.

Programs

  • Scheme
    (define (A243052 n) (explist->n (ascpart_to_prime-exps (bulgarian-operation-n-th-order (prime-exps_to_ascpart (primefacs->explist n)) 2))))
    (define (bulgarian-operation-n-th-order ascpart n) (if (or (zero? n) (null? ascpart)) ascpart (let ((newpart (length ascpart))) (let loop ((newpartition (list)) (ascpart ascpart)) (cond ((null? ascpart) (sort (cons newpart (bulgarian-operation-n-th-order newpartition (- n 1))) <)) (else (loop (if (= 1 (car ascpart)) newpartition (cons (- (car ascpart) 1) newpartition)) (cdr ascpart))))))))
    ;; Other required functions and libraries, please see A243051.

Formula

a(n) = A241909(A243072(A241909(n))).

A243053 Integer sequence induced by third-order Bulgarian solitaire operation on partition list A241918: a(n) = A241909(A243073(A241909(n))).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 5, 9, 12, 32, 10, 64, 24, 18, 7, 128, 15, 256, 20, 36, 48, 512, 55, 27, 96, 25, 40, 1024, 30, 2048, 49, 72, 192, 54, 21, 4096, 384, 144, 637, 8192, 60, 16384, 80, 50, 768, 32768, 22, 81, 45, 288, 160, 65536, 35, 108, 22627, 576, 1536, 131072
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The usual (first-order) Bulgarian Solitaire operation (cf. A243051) applied to an unordered integer partition means: subtract one from each part, and add a new part as large as there were parts in the old partition.
The "Second-Order Bulgarian Operation" means that after subtracting one from each part of the old partition (and discarding the parts that diminished to zero), we apply the (first-order) Bulgarian operation to the remaining partition before adding a new part as large as there were parts in the original partition.
Similarly, in "Third-Order Bulgarian Solitaire Operation", we apply the Second-Order Bulgarian operation to the remaining partition (after we have subtracted one from each part) before adding a new part as large as there were parts in the original partition.
How the partitions are encoded in this case, see A241918.

Crossrefs

Third row of A243060.
Differs from A122111 for the first time at n=24, where a(24) = 55, while A122111(24) = 14.

Programs

  • Scheme
    (define (A243053 n) (explist->n (ascpart_to_prime-exps (bulgarian-operation-n-th-order (prime-exps_to_ascpart (primefacs->explist n)) 3))))
    (define (bulgarian-operation-n-th-order ascpart n) (if (or (zero? n) (null? ascpart)) ascpart (let ((newpart (length ascpart))) (let loop ((newpartition (list)) (ascpart ascpart)) (cond ((null? ascpart) (sort (cons newpart (bulgarian-operation-n-th-order newpartition (- n 1))) <)) (else (loop (if (= 1 (car ascpart)) newpartition (cons (- (car ascpart) 1) newpartition)) (cdr ascpart))))))))
    ;; Other required functions and libraries, please see A243051.

Formula

a(n) = A241909(A243073(A241909(n))).
Showing 1-7 of 7 results.