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 1-10 of 16 results. Next

A245454 Self-inverse permutation of nonnegative integers, A075158-conjugate of blue code: a(n) = 1 + A075157(A193231(A075158(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 18, 8, 9, 25, 11, 16, 64, 14, 27, 12, 96, 7, 288, 21, 20, 243, 891, 45, 10, 405, 15, 162, 33750, 30, 78650, 75, 625, 2025, 35, 81, 390390, 224, 875, 250, 41, 375, 16384, 270, 24, 300125, 24576, 150, 125, 54, 6125, 1350, 73728, 50, 108, 350, 594, 140777, 5845851, 98, 221433750, 1446445, 343, 13
Offset: 1

Views

Author

Antti Karttunen, Jul 22 2014

Keywords

Crossrefs

Programs

Formula

a(n) = 1 + A075157(A193231(A075158(n-1))).

A245451 Self-inverse permutation of nonnegative integers, A075158-conjugate of gray code: a(n) = 1 + A075157(A003188(A075158(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 16, 6, 5, 27, 32, 18, 64, 81, 25, 12, 128, 7, 256, 54, 125, 243, 512, 36, 10, 729, 15, 162, 1024, 49, 2048, 24, 625, 2187, 50, 14, 4096, 6561, 3125, 108, 8192, 343, 16384, 486, 75, 19683, 32768, 72, 20, 21, 15625, 1458, 65536, 35, 250, 324, 78125, 59049, 131072, 98, 262144, 177147, 375, 48
Offset: 1

Views

Author

Antti Karttunen, Jul 22 2014

Keywords

Crossrefs

Inverse: A245452.
Similar permutations: A245454, A122111, A241909, A241916.

Programs

Formula

a(n) = 1 + A075157(A003188(A075158(n-1))).

A245452 Self-inverse permutation of nonnegative integers, A075158-conjugate of the inverse of gray code: a(n) = 1 + A075157(A006068(A075158(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 9, 8, 18, 5, 6, 25, 75, 16, 150, 36, 27, 7, 735, 12, 1470, 49, 50, 245, 12705, 32, 15, 300, 10, 72, 25410, 125, 195195, 11, 225, 4235, 54, 24, 390390, 2940, 490, 121, 4339335, 100, 8678670, 847, 81, 65065, 92147055, 64, 30, 35, 2205, 600, 184294110, 20, 147, 144, 8470, 50820, 2565568005, 343, 5131136010, 1446445, 98, 13
Offset: 1

Views

Author

Antti Karttunen, Jul 22 2014

Keywords

Crossrefs

Inverse: A245451.
Similar permutations: A245454, A122111, A241909, A241916.

Programs

Formula

a(n) = 1 + A075157(A006068(A075158(n-1))).

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

A075157 Run lengths in the binary expansion of n gives the vector of exponents in prime factorization of a(n)+1, with the least significant run corresponding to the exponent of the least prime, 2; with one subtracted from each run length, except for the most significant run of 1's.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 8, 7, 11, 14, 6, 9, 17, 24, 26, 15, 23, 44, 34, 29, 13, 10, 20, 19, 35, 74, 48, 49, 53, 124, 80, 31, 47, 134, 174, 89, 69, 76, 104, 59, 27, 32, 12, 21, 41, 54, 62, 39, 71, 224, 244, 149, 97, 120, 146, 99, 107, 374, 342, 249, 161, 624, 242, 63, 95, 404
Offset: 0

Views

Author

Antti Karttunen, Sep 13 2002

Keywords

Comments

To make this a permutation of nonnegative integers, we subtract one from each run count except for the most significant run, e.g. a(11) = 9, as 11 = 1011 and 9+1 = 10 = 5^1 * 3^(1-1) * 2^(2-1).

Crossrefs

Programs

  • Haskell
    import Data.List (group)
    a075157 0 = 0
    a075157 n = product (zipWith (^) a000040_list rs') - 1 where
       rs' = reverse $ r : map (subtract 1) rs
       (r:rs) = reverse $ map length $ group $ a030308_row n
    -- Reinhard Zumkeller, Aug 04 2014
    
  • PARI
    A005811(n) = hammingweight(bitxor(n, n>>1));  \\ This function from Gheorghe Coserea, Sep 03 2015
    A286468(n) = { my(p=((n+1)%2), i=0, m=1); while(n>0, if(((n%2)==p), m *= prime(i), p = (n%2); i = i+1); n = n\2); m };
    A075157(n) = if(!n,n,(prime(A005811(n))*A286468(n))-1);
    
  • Scheme
    (define (A075157 n) (if (zero? n) n (+ -1 (* (A000040 (A005811 n)) (fold-left (lambda (a r) (* (A003961 a) (A000079 (- r 1)))) 1 (binexp->runcount1list n))))))
    (define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
    ;; Or, using the code of A286468:
    (define (A075157 n) (if (zero? n) n (- (* (A000040 (A005811 n)) (A286468 n)) 1)))

Formula

a(n) = A075159(n+1) - 1.
a(0) = 0; for n >= 1, a(n) = (A000040(A005811(n)) * A286468(n)) - 1.
Other identities. For all n >= 1:
a(A000975(n)) = A006093(n) = A000040(n)-1.

Extensions

Entry revised, PARI-program added and the old incorrect Scheme-program replaced with a new one by Antti Karttunen, May 17 2017

A129594 Involution of nonnegative integers induced by the conjugation of the partition encoded in the run lengths of binary expansion of n.

Original entry on oeis.org

0, 1, 3, 2, 4, 7, 6, 5, 11, 12, 15, 8, 9, 14, 13, 10, 20, 27, 28, 19, 16, 31, 24, 23, 22, 25, 30, 17, 18, 29, 26, 21, 43, 52, 59, 36, 35, 60, 51, 44, 47, 48, 63, 32, 39, 56, 55, 40, 41, 54, 57, 38, 33, 62, 49, 46, 45, 50, 61, 34, 37, 58, 53, 42, 84, 107, 116, 75, 68, 123
Offset: 0

Views

Author

Antti Karttunen, May 01 2007

Keywords

Comments

This sequence is based on the fact that compositions (i.e. ordered partitions) can be mapped 1-to-1 to partitions by taking the partial sums of the list where one is subtracted from each composant except the first. (See table A227189 where the parts for each partition are listed).
The inverse process, from partitions to compositions, occurs by inserting the first (i.e. smallest) element of a partition sorted into ascending order to the front of the list obtained by adding one to the first differences of the elements.
Compositions map bijectively to nonnegative integers by assigning each run of k consecutive 1's (or 0's) in binary expansion of n with summand k in the composition.
The graph of this sequence is quite interesting.

Examples

			a(8) = 11, as 8 is 1000 in binary, mapping to composition 3+1 (we scan the binary expansion from the least to the most significant end), which maps to partition 3+3, whose conjugate-partition is 2+2+2, yielding composition 2+1+1, which maps to binary 1011, 11 in decimal. a(13) = 14, as 13 is 1101 in binary, mapping to composition 1+1+2, which maps to the partition 1+1+2, whose conjugate-partition is 1+3, yielding composition 1+3, which maps to binary 1110, 14 in decimal. a(11) = 8 and a(14) = 13, as taking the conjugate of a partition is a self-inverse operation.
		

Crossrefs

a(n) = A075158(A122111(1+A075157(n)) - 1). See A129595 for another kind of encoding of integer partitions.
Sequences related to partitions encoded in this way:
Cf. A227189 (parts of partitions listed on separate rows of the array).
Cf. A005811 (number of parts in the partition).
Cf. A136480 (for n>= 1, the smallest part).
Cf. A227185 (the largest part).
Cf. A227183 (sum of parts).
Cf. A227184 (product of parts).
Note that this permutation maps between A005811 and A227185 as follows: A005811(n) = A227185(a(n)) and vice versa: A227185(n) = A005811(a(n)). On the other hand, it keeps A227183 fixed, as A227183(n) = A227183(a(n)).
Cf. also A226062.

A226062 a(n) = Bulgarian solitaire operation applied to the partition encoded in the runlengths of binary expansion of n.

Original entry on oeis.org

0, 1, 3, 2, 13, 7, 6, 6, 11, 29, 15, 58, 9, 14, 4, 14, 19, 27, 61, 54, 245, 31, 122, 52, 27, 25, 30, 50, 25, 12, 12, 30, 35, 23, 59, 46, 237, 125, 118, 44, 235, 501, 63, 1002, 233, 250, 116, 40, 51, 19, 57, 38, 229, 62, 114, 36, 59, 17, 28, 34, 57, 8, 28, 62
Offset: 0

Views

Author

Antti Karttunen, Jul 06 2013

Keywords

Comments

For this sequence the partitions are encoded in the binary expansion of n in the same way as in A129594.
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. The question originally posed was: on what condition the resulting partitions will eventually reach a fixed point, that is, a collection of piles that will be unchanged by the operation. See Martin Gardner reference and the Wikipedia-page.
A037481 gives the fixed points of this sequence, which are numbers that encode triangular partitions: 1 + 2 + 3 + ... + n.
A227752(n) tells how many times n occurs in this sequence, and A227753 gives the terms that do not occur here.
Of further interest: among each A000041(n) numbers j_i: j1, j2, ..., jk for which A227183(j_i)=n, how many cycles occur and what is the size of the largest one? (Both are 1 when n is in A000217, as then the fixed points are the only cycles.) Cf. A185700, A188160.
Also, A123975 answers how many Garden of Eden partitions there are for the deck of size n in Bulgarian Solitaire, corresponding to values that do not occur as the terms of this sequence.

Examples

			5 has binary expansion "101", whose runlengths are [1,1,1], which are converted to nonordered partition {1+1+1}.
6 has binary expansion "110", whose runlengths are [1,2] (we scan the runs of bits from right to left), which are converted to nonordered partition {1+2}.
7 has binary expansion "111", whose list of runlengths is [3], which is converted to partition {3}.
In "Bulgarian Operation" we subtract one from each part (with 1-parts vanishing), and then add a new part of the same size as there originally were parts, so that the total sum stays same.
Thus starting from a partition encoded by 5, {1,1,1} the operation works as 1-1, 1-1, 1-1 (all three 1's vanish) but appends part 3 as there originally were three parts, thus we get a new partition {3}. Thus a(5)=7.
From the partition {3} -> 3-1 and 1, which gives a new partition {1,2}, so a(7)=6.
For partition {1+2} -> 1-1 and 2-1, thus the first part vanishes, and the second is now 1, to which we add the new part 2, as there were two parts originally, thus {1+2} stays as {1+2}, and we have reached a fixed point, a(6)=6.
		

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

Cf. A037481 (gives the fixed points).
Cf. A227752 (how many times n occurs here).
Cf. A227753 (numbers that do not occur here).
Cf. A129594 (conjugates the partitions encoded with the same system).

Formula

Other identities:
A227183(a(n)) = A227183(n). [This operation doesn't change the total sum of the partition.]
a(n) = A243354(A242424(A243353(n))).
a(n) = A075158(A243051(1+A075157(n))-1).

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

A243354 Permutation of natural numbers which maps between the partitions as encoded in A112798 (prime-index based system, one-based) to A227739 (binary based system, zero-based).

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 15, 5, 4, 14, 31, 13, 63, 30, 12, 10, 127, 9, 255, 29, 28, 62, 511, 26, 8, 126, 11, 61, 1023, 25, 2047, 21, 60, 254, 24, 18, 4095, 510, 124, 58, 8191, 57, 16383, 125, 27, 1022, 32767, 53, 16, 17, 252, 253, 65535, 22, 56, 122, 508, 2046, 131071
Offset: 1

Views

Author

Antti Karttunen, Jun 05 2014

Keywords

Comments

Note the indexing: the domain starts from one, but the range also includes zero.

Crossrefs

Programs

Formula

a(n) = A006068(A156552(n)).
a(n) = A075158(A241909(n)-1). [With A075158's original starting offset].
For all n >= 1, A243353(a(n)) = n.
A056239(n) = A227183(a(n)).
A003963(n) = A227184(a(n)).
A037481(n) = a(A002110(n)).

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].
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