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.

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A064989 Multiplicative with a(2^e) = 1 and a(p^e) = prevprime(p)^e for odd primes p.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 1, 4, 3, 7, 2, 11, 5, 6, 1, 13, 4, 17, 3, 10, 7, 19, 2, 9, 11, 8, 5, 23, 6, 29, 1, 14, 13, 15, 4, 31, 17, 22, 3, 37, 10, 41, 7, 12, 19, 43, 2, 25, 9, 26, 11, 47, 8, 21, 5, 34, 23, 53, 6, 59, 29, 20, 1, 33, 14, 61, 13, 38, 15, 67, 4, 71, 31, 18, 17, 35, 22, 73, 3, 16
Offset: 1

Views

Author

Vladeta Jovovic, Oct 30 2001

Keywords

Comments

From Antti Karttunen, May 12 2014: (Start)
a(A003961(n)) = n for all n. [This is a left inverse function for the injection A003961.]
Bisections are A064216 (the terms at odd indices) and A064989 itself (the terms at even indices), i.e., a(2n) = a(n) for all n.
(End)
From Antti Karttunen, Dec 18-21 2014: (Start)
When n represents an unordered integer partition via the indices of primes present in its prime factorization (for n >= 2, n corresponds to the partition given as the n-th row of A112798) this operation subtracts one from each part. If n is of the form 2^k (a partition having just k 1's as its parts) the result is an empty partition (which is encoded by 1, having an "empty" factorization).
For all odd numbers n >= 3, a(n) tells which number is located immediately above n in square array A246278. Cf. also A246277.
(End)
Alternatively, if numbers are represented as the multiset of indices of prime factors with multiplicity, this operation subtracts 1 from each element and discards the 0's. - M. F. Hasler, Dec 29 2014

Examples

			a(20) = a(2^2*5) = a(2^2)*a(5) = prevprime(5) = 3.
		

Crossrefs

Cf. A064216 (odd bisection), A003961 (inverse), A151799.
Other sequences whose definition involve or are some other way related with this sequence: A105560, A108951, A118306, A122111, A156552, A163511, A200746, A241909, A243070, A243071, A243072, A243073, A244319, A245605, A245607, A246165, A246266, A246268, A246277, A246278, A246361, A246362, A246371, A246372, A246373, A246374, A246376, A246380, A246675, A246682, A249745, A250470.
Similar prime-shifts towards smaller numbers: A252461, A252462, A252463.

Programs

  • Haskell
    a064989 1 = 1
    a064989 n = product $ map (a008578 . a049084) $ a027746_row n
    -- Reinhard Zumkeller, Apr 09 2012
    (MIT/GNU Scheme, with Aubrey Jaffer's SLIB Scheme library)
    (require 'factor)
    (define (A064989 n) (if (= 1 n) n (apply * (map (lambda (k) (if (zero? k) 1 (A000040 k))) (map -1+ (map A049084 (factor n)))))))
    ;; Antti Karttunen, May 12 2014
    (definec (A064989 n) (if (= 1 n) n (* (A008578 (A055396 n)) (A064989 (A032742 n))))) ;; One based on given recurrence and utilizing memoizing definec-macro.
    (definec (A064989 n) (cond ((= 1 n) n) ((even? n) (A064989 (/ n 2))) (else (A163511 (/ (- (A243071 n) 1) 2))))) ;; Corresponds to one of the alternative formulas, but is very unpractical way to compute this sequence. - Antti Karttunen, Dec 18 2014
    
  • Maple
    q:= proc(p) prevprime(p) end proc: q(2):= 1:
    [seq(mul(q(f[1])^f[2], f = ifactors(n)[2]), n = 1 .. 1000)]; # Robert Israel, Dec 21 2014
  • Mathematica
    Table[Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n, {n, 81}] (* Michael De Vlieger, Jan 04 2016 *)
  • PARI
    { for (n=1, 1000, f=factor(n)~; a=1; j=1; if (n>1 && f[1, 1]==2, j=2); for (i=j, length(f), a*=precprime(f[1, i] - 1)^f[2, i]); write("b064989.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 02 2009
    
  • PARI
    a(n) = {my(f = factor(n)); for (i=1, #f~, if ((p=f[i,1]) % 2, f[i,1] = precprime(p-1), f[i,1] = 1);); factorback(f);} \\ Michel Marcus, Dec 18 2014
    
  • PARI
    A064989(n)=factorback(Mat(apply(t->[max(precprime(t[1]-1),1),t[2]],Vec(factor(n)~))~)) \\ M. F. Hasler, Dec 29 2014
    
  • Python
    from sympy import factorint, prevprime
    from operator import mul
    from functools import reduce
    def a(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])
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 15 2017
    
  • Python
    from math import prod
    from sympy import prevprime, factorint
    def A064989(n): return prod(prevprime(p)**e for p, e in  factorint(n>>(~n&n-1).bit_length()).items()) # Chai Wah Wu, Jan 05 2023

Formula

From Antti Karttunen, Dec 18 2014: (Start)
If n = product A000040(k)^e(k) then a(n) = product A008578(k)^e(k) [where A000040(n) gives the n-th prime, and A008578(n) gives 1 for 1 and otherwise the (n-1)-th prime].
a(1) = 1; for n > 1, a(n) = A008578(A055396(n)) * a(A032742(n)). [Above formula represented as a recurrence. Cf. A252461.]
a(1) = 1; for n > 1, a(n) = A008578(A061395(n)) * a(A052126(n)). [Compare to the formula of A252462.]
This prime-shift operation is used in the definitions of many other sequences, thus it can be expressed in many alternative ways:
a(n) = A200746(n) / n.
a(n) = A242424(n) / A105560(n).
a(n) = A122111(A122111(n)/A105560(n)) = A122111(A052126(A122111(n))). [In A112798-partition context: conjugate, remove the largest part (the largest prime factor), and conjugate again.]
a(1) = 1; for n > 1, a(2n) = a(n), a(2n+1) = A163511((A243071(2n+1)-1) / 2).
a(n) = A249818(A250470(A249817(n))). [A250470 is an analogous operation for "going one step up" in the square array A083221 (A083140).]
(End)
Product_{k=1..n} a(k) = n! / A307035(n). - Vaclav Kotesovec, Mar 21 2019
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((p^2-p)/(p^2-q(p))) = 0.220703928... , where q(p) = prevprime(p) (A151799) if p > 2 and q(2) = 1. - Amiram Eldar, Nov 18 2022

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

A242424 Bulgarian solitaire operation on partition list A112798: a(1) = 1, a(n) = A000040(A001222(n)) * A064989(n).

Original entry on oeis.org

1, 2, 4, 3, 6, 6, 10, 5, 12, 9, 14, 10, 22, 15, 18, 7, 26, 20, 34, 15, 30, 21, 38, 14, 27, 33, 40, 25, 46, 30, 58, 11, 42, 39, 45, 28, 62, 51, 66, 21, 74, 50, 82, 35, 60, 57, 86, 22, 75, 45, 78, 55, 94, 56, 63, 35, 102, 69, 106, 42, 118, 87, 100, 13, 99, 70, 122, 65
Offset: 1

Views

Author

Antti Karttunen, May 13 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 A112798.
Please compare to the definition of A122111, which conjugates the partitions encoded with the same system.
a(n) is even if and only if n is either a prime or a multiple of three.
Conversely, a(n) is odd if and only if n is a nonprime not divisible by three.

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 A243070 (table which gives successive "recursive iterates" of this sequence and converges towards A122111).
Fixed points: A002110 (primorial numbers).

Programs

Formula

a(1) = 1, a(n) = A000040(A001222(n)) * A064989(n) = A105560(n) * A064989(n).
a(n) = A241909(A243051(A241909(n))).
a(n) = A243353(A226062(A243354(n))).
a(A000079(n)) = A000040(n) for all n.
A056239(a(n)) = A056239(n) for all n.

A243505 Permutation of natural numbers, take the odd bisection of A122111 and divide the largest prime factor out: a(n) = A052126(A122111(2n-1)).

Original entry on oeis.org

1, 2, 4, 8, 3, 16, 32, 6, 64, 128, 12, 256, 9, 5, 512, 1024, 24, 18, 2048, 48, 4096, 8192, 10, 16384, 27, 96, 32768, 36, 192, 65536, 131072, 20, 72, 262144, 384, 524288, 1048576, 15, 54, 2097152, 7, 4194304, 144, 768, 8388608, 108, 1536, 288, 16777216, 40, 33554432, 67108864, 30
Offset: 1

Views

Author

Antti Karttunen, Jun 25 2014

Keywords

Crossrefs

Programs

Formula

a(n) = A052126(A122111((2*n)-1)).
a(n) = A122111((2*n)-1) / A105560((2*n)-1).
As a composition of related permutations:
a(n) = A122111(A064216(n)).
a(n) = A241916(A243065(n)).
Other identities:
For all n >= 2, a(n) = A070003(A244984(n)-1) / A105560((2*n)-1).
For all n >= 1, a(A006254(n)) = A000079(n) and a(A007051(n)) = A000040(n).
For all n >= 1, A105560(2n-1) divides a(n).

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)).

A244984 Permutation of natural numbers: a(n) = A243283(A122111((2*n)-1)).

Original entry on oeis.org

1, 2, 3, 5, 4, 9, 14, 6, 23, 37, 10, 58, 8, 7, 90, 143, 15, 13, 225, 24, 355, 563, 12, 894, 17, 38, 1426, 20, 60, 2277, 3643, 19, 31, 5839, 96, 9398, 15155, 16, 27, 24518, 11, 39758, 50, 153, 64607, 42, 242, 80, 105250, 30, 171874, 281237, 26
Offset: 1

Views

Author

Antti Karttunen, Jul 21 2014

Keywords

Crossrefs

Programs

Formula

a(n) = A243283(A122111((2*n)-1)).
a(n) = A243283(A105560((2*n)-1) * A243505(n)).
For all n >= 1, a(A006254(n)) = A244986(n+1).

A334107 a(n) = A329697(A122111(n)).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 2, 2, 0, 2, 0, 1, 2, 1, 0, 2, 3, 1, 2, 1, 0, 2, 0, 2, 2, 1, 3, 3, 0, 1, 2, 2, 0, 2, 0, 1, 2, 1, 0, 2, 4, 3, 2, 1, 0, 3, 3, 2, 2, 1, 0, 3, 0, 1, 2, 2, 3, 2, 0, 1, 2, 3, 0, 3, 0, 1, 3, 1, 4, 2, 0, 2, 4, 1, 0, 3, 3, 1, 2, 2, 0, 3, 4, 1, 2, 1, 3, 2, 0, 4, 2, 4, 0, 2, 0, 2, 3
Offset: 1

Views

Author

Antti Karttunen, Apr 29 2020

Keywords

Crossrefs

Programs

  • Mathematica
    Map[Length@ NestWhileList[# - #/FactorInteger[#][[-1, 1]] &, #, # != 2^IntegerExponent[#, 2] &] - 1 &, Array[Times @@ Table[Prime[LengthWhile[#1, # >= j &] /. 0 -> 1], {j, #2}] & @@ {#, Max[#]} &@ PrimePi@ Flatten[ConstantArray[#1, {#2}] & @@@ FactorInteger@ #] &, 105] ] (* Michael De Vlieger, May 14 2020, after Robert G. Wilson v at A329697 *)
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A122111(n) = if(1==n,n,prime(bigomega(n))*A122111(A064989(n)));
    A329697(n) = if(!bitand(n,n-1),0,1+A329697(n-(n/vecmax(factor(n)[, 1]))));
    A334107(n) = A329697(A122111(n));

Formula

a(n) = A329697(A122111(n)) = A329697(A322865(n)).
a(n) = A329697(A105560(n)) + a(A064989(n)).
For n >= 1, a(A001248(n)) = n, and these seem to be also the first occurrences of each n.

A243072 Second order Bulgarian solitaire operation on partition list A112798: a(1) = 1, a(n) = A000040(A001222(n)) * A242424(A064989(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 12, 5, 9, 12, 20, 10, 28, 18, 18, 7, 44, 15, 52, 20, 27, 30, 68, 14, 36, 42, 25, 30, 76, 30, 92, 11, 45, 66, 54, 21, 116, 78, 63, 28, 124, 45, 148, 50, 50, 102, 164, 22, 81, 60, 99, 70, 172, 35, 90, 42, 117, 114, 188, 42, 212, 138, 75, 13, 126
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The usual Bulgarian Solitaire operation (the "first order" version, cf. A242424) 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.
In this context, where the parts of partitions are encoded with the indices of primes in the prime factorization of n (as in A112798), A064989(n) gives the remaining partition after one has been subtracted from each part; A242424 applies the first order Bulgarian operation to it; and multiplying with A000040(A001222(n)) adds a part as large as there originally were parts.

Crossrefs

Row 2 of A243070. Differs from A122111 for the first time at n=7.

Formula

a(1) = 1, a(n) = A000040(A001222(n)) * A242424(A064989(n)) = A105560(n) * A242424(A064989(n)).
a(n) = A241909(A243052(A241909(n))).

A243073 Third-order Bulgarian solitaire operation on partition list A112798: a(1) = 1, a(n) = A000040(A001222(n)) * A243072(A064989(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 5, 9, 12, 24, 10, 40, 24, 18, 7, 56, 15, 88, 20, 36, 36, 104, 14, 27, 60, 25, 40, 136, 30, 152, 11, 54, 84, 54, 21, 184, 132, 90, 28, 232, 60, 248, 60, 50, 156, 296, 22, 108, 45, 126, 100, 328, 35, 81, 56, 198, 204, 344, 42, 376, 228, 100, 13, 135
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The usual (first-order) Bulgarian Solitaire operation (cf. A242424) 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.
In this context, where the parts of partitions are encoded with the indices of primes in the prime factorization of n (as in A112798), A064989(n) gives the remaining partition after one has been subtracted from each part; A243072 applies the second-order Bulgarian operation to it; and multiplying with A000040(A001222(n)) adds a part as large as there originally were parts.

Crossrefs

Row 3 of A243070. Differs from A122111 for the first time at n=11.

Formula

a(1) = 1, a(n) = A000040(A001222(n)) * A243072(A064989(n)) = A105560(n) * A243072(A064989(n)).
a(n) = A241909(A243053(A241909(n))).

A334108 a(n) = A331410(A122111(n)).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 2, 2, 1, 0, 2, 0, 1, 2, 1, 0, 3, 0, 2, 2, 1, 0, 1, 3, 1, 4, 2, 0, 3, 0, 2, 2, 1, 3, 2, 0, 1, 2, 1, 0, 3, 0, 2, 4, 1, 0, 2, 4, 4, 2, 2, 0, 3, 3, 1, 2, 1, 0, 2, 0, 1, 4, 2, 3, 3, 0, 2, 2, 4, 0, 3, 0, 1, 5, 2, 4, 3, 0, 2, 2, 1, 0, 2, 3, 1, 2, 1, 0, 3, 4, 2, 2, 1, 3, 2, 0, 5, 4, 3, 0, 3, 0, 1, 5
Offset: 1

Views

Author

Antti Karttunen, Apr 29 2020

Keywords

Crossrefs

Cf. A008578 (positions of zeros), A064989, A105560, A122111, A322865, A331410, A334107.

Programs

Formula

a(n) = A331410(A122111(n)) = A331410(A322865(n)).
a(n) = A331410(A105560(n)) + a(A064989(n)).
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