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 1-9 of 9 results.

A056239 If n = Product_{k >= 1} (p_k)^(c_k) where p_k is k-th prime and c_k >= 0 then a(n) = Sum_{k >= 1} k*c_k.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 6, 5, 5, 4, 7, 5, 8, 5, 6, 6, 9, 5, 6, 7, 6, 6, 10, 6, 11, 5, 7, 8, 7, 6, 12, 9, 8, 6, 13, 7, 14, 7, 7, 10, 15, 6, 8, 7, 9, 8, 16, 7, 8, 7, 10, 11, 17, 7, 18, 12, 8, 6, 9, 8, 19, 9, 11, 8, 20, 7, 21, 13, 8, 10, 9, 9, 22, 7, 8, 14, 23, 8, 10, 15, 12, 8, 24, 8, 10
Offset: 1

Views

Author

Leroy Quet, Aug 19 2000

Keywords

Comments

A pseudo-logarithmic function in the sense that a(b*c) = a(b)+a(c) and so a(b^c) = c*a(b) and f(n) = k^a(n) is a multiplicative function. [Cf. A248692 for example.] Essentially a function from the positive integers onto the partitions of the nonnegative integers (1->0, 2->1, 3->2, 4->1+1, 5->3, 6->1+2, etc.) so each value a(n) appears A000041(a(n)) times, first with the a(n)-th prime and last with the a(n)-th power of 2. Produces triangular numbers from primorials. - Henry Bottomley, Nov 22 2001
Michael Nyvang writes (May 08 2006) that the Danish composer Karl Aage Rasmussen discovered this sequence in the 1990's: it has excellent musical properties.
All A000041(a(n)) different n's with the same value a(n) are listed in row a(n) of triangle A215366. - Alois P. Heinz, Aug 09 2012
a(n) is the sum of the parts of the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} (p_j-th prime) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(33) = 7 because the partition with Heinz number 33 = 3 * 11 is [2,5]. - Emeric Deutsch, May 19 2015

Examples

			a(12) = 1*2 + 2*1 = 4, since 12 = 2^2 *3^1 = (p_1)^2 *(p_2)^1.
		

Crossrefs

Programs

  • Haskell
    a056239 n = sum $ zipWith (*) (map a049084 $ a027748_row n) (a124010_row n)
    -- Reinhard Zumkeller, Apr 27 2013
    
  • Maple
    # To get 10000 terms. First make prime table: M:=10000; pl:=array(1..M); for i from 1 to M do pl[i]:=0; od: for i from 1 to M do if ithprime(i) > M then break; fi; pl[ithprime(i)]:=i; od:
    # Decode Maple's amazing syntax for factoring integers: g:=proc(n) local e,p,t1,t2,t3,i,j,k; global pl; t1:=ifactor(n); t2:=nops(t1); if t2 = 2 and whattype(t1) <> `*` then p:=op(1,op(1,t1)); e:=op(2,t1); t3:=pl[p]*e; else
    t3:=0; for i from 1 to t2 do j:=op(i,t1); if nops(j) = 1 then e:=1; p:=op(1,j); else e:=op(2,j); p:=op(1,op(1,j)); fi; t3:=t3+pl[p]*e; od: fi; t3; end; # N. J. A. Sloane, May 10 2006
    A056239 := proc(n) add( numtheory[pi](op(1,p))*op(2,p), p = ifactors(n)[2]) ; end proc: # R. J. Mathar, Apr 20 2010
    # alternative:
    with(numtheory): a := proc (n) local B: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: add(B(n)[i], i = 1 .. nops(B(n))) end proc: seq(a(n), n = 1 .. 130); # Emeric Deutsch, May 19 2015
  • Mathematica
    a[1] = 0; a[2] = 1; a[p_?PrimeQ] := a[p] = PrimePi[p];
    a[n_] := a[n] = Total[#[[2]]*a[#[[1]]] & /@ FactorInteger[n]]; a /@ Range[91] (* Jean-François Alcover, May 19 2011 *)
    Table[Total[FactorInteger[n] /. {p_, c_} /; p > 0 :> PrimePi[p] c], {n, 91}] (* Michael De Vlieger, Jul 12 2017 *)
  • PARI
    A056239(n) = if(1==n,0,my(f=factor(n)); sum(i=1, #f~, f[i,2] * primepi(f[i,1]))); \\ Antti Karttunen, Oct 26 2014, edited Jan 13 2020
    
  • Python
    from sympy import primepi, factorint
    def A056239(n): return sum(primepi(p)*e for p, e in factorint(n).items()) # Chai Wah Wu, Jan 01 2023
  • Scheme
    (require 'factor) ;; Uses the function factor available in Aubrey Jaffer's SLIB Scheme library.
    (define (A056239 n) (apply + (map A049084 (factor n))))
    ;; Antti Karttunen, Oct 26 2014
    

Formula

Totally additive with a(p) = PrimePi(p), where PrimePi(n) = A000720(n).
a(n) = Sum_{k=1..A001221(n)} A049084(A027748(k))*A124010(k). - Reinhard Zumkeller, Apr 27 2013
From Antti Karttunen, Oct 11 2014: (Start)
a(n) = n - A178503(n).
a(n) = A161511(A156552(n)).
a(n) = A227183(A243354(n)).
For all n >= 0:
a(A002110(n)) = A000217(n). [Cf. Henry Bottomley's comment above.]
a(A005940(n+1)) = A161511(n).
a(A243353(n)) = A227183(n).
Also, for all n >= 1:
a(A241909(n)) = A243503(n).
a(A122111(n)) = a(n).
a(A242424(n)) = a(n).
A248692(n) = 2^a(n). (End)
a(n) < A329605(n), a(n) = A001222(A108951(n)), a(A329902(n)) = A112778(n). - Antti Karttunen, Jan 14 2020

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

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

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.

A105560 a(1) = 1, and for n >= 2, a(n) = prime(bigomega(n)), where prime(n) = A000040(n) and bigomega(n) = A001222(n).

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 2, 5, 3, 3, 2, 5, 2, 3, 3, 7, 2, 5, 2, 5, 3, 3, 2, 7, 3, 3, 5, 5, 2, 5, 2, 11, 3, 3, 3, 7, 2, 3, 3, 7, 2, 5, 2, 5, 5, 3, 2, 11, 3, 5, 3, 5, 2, 7, 3, 7, 3, 3, 2, 7, 2, 3, 5, 13, 3, 5, 2, 5, 3, 5, 2, 11, 2, 3, 5, 5, 3, 5, 2, 11, 7, 3, 2, 7, 3, 3, 3, 7, 2, 7, 3, 5, 3, 3, 3, 13, 2, 5, 5, 7
Offset: 1

Views

Author

Cino Hilliard, May 03 2005

Keywords

Comments

From Antti Karttunen, Jul 21 2014: (Start)
a(n) divides A122111(n), A242424(n), A243072(n), A243073(n) because a(n) divides all the terms in column n of A243070.
a(2n-1) divides A243505(n) and a(2n-1)^2 divides A122111(2n-1).
(End)

Crossrefs

Programs

  • Mathematica
    Table[Prime[Sum[FactorInteger[n][[i,2]],{i,1,Length[FactorInteger[n]]}]],{n,2,40}] (* Stefan Steinerberger, May 16 2007 *)
  • PARI
    d(n) = for(x=2,n,print1(prime(bigomega(x))","))
    
  • Python
    from sympy import prime, primefactors
    def a001222(n): return 0 if n==1 else a001222(n/primefactors(n)[0]) + 1
    def a(n): return 1 if n==1 else prime(a001222(n)) # Indranil Ghosh, Jun 15 2017

Formula

a(1) = 1, and for n >= 2, a(n) = A000040(A001222(n)).
From Antti Karttunen, Jul 21 2014: (Start)
a(n) = A008578(1 + A001222(n)).
a(n) = A006530(A122111(n)).
a(n) = A122111(n) / A122111(A064989(n)).
a(2n-1) = A122111(2n-1) / A243505(n).
a(n) = A242424(n) / A064989(n).
(End)

Extensions

a(1) = 1 prepended by Antti Karttunen, Jul 21 2014

A243060 Square array read by antidiagonals: rows are successively recursivized versions of Bulgarian solitaire operation (starting from the usual "first order" version, A243051), as applied to partitions listed in A241918.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 3, 4, 2, 1, 8, 3, 4, 2, 1, 25, 8, 3, 4, 2, 1, 16, 6, 8, 3, 4, 2, 1, 9, 16, 6, 8, 3, 4, 2, 1, 9, 5, 16, 6, 8, 3, 4, 2, 1, 343, 9, 5, 16, 6, 8, 3, 4, 2, 1, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 64, 35, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 14641, 64, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 125, 24, 64, 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), ... .
Each row is a A241909-conjugate of the corresponding row in A243070.
Rows in both arrays converge towards A122111.
First point where row k differs from row k of A243070 seems to be A000040(k+2): primes from five onward: 5, 7, 11, 13, 17, 19, 23, 29, 31, ...
While the first point where row k differs from A122111 seems to begin as 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, ... A007283 (3*2^n) from the term n=1 onward.
The rows in this table preserve A243503.

Examples

			The top left corner of the array:
  1, 2, 4, 3, 8, 25, 16, 9, 9, 343,  32,  10,  64, 14641,   125,   27, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  35,  64,    24,    18,   25, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  10,  64,    24,    18,    7, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  10,  64,    24,    18,    7, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  10,  64,    24,    18,    7, ...
		

Crossrefs

Row 1: A243051, Row 2: A243052, Row 3: A243053.
Rows converge towards A122111.

Programs

  • Scheme
    (define (A243060 n) (A243060bi (A002260 n) (A004736 n)))
    (define (A243060bi row col) (explist->n (ascpart_to_prime-exps (bulgarian-operation-n-th-order (prime-exps_to_ascpart (primefacs->explist col)) row))))
    (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))))))))
    ;; For other required functions and libraries, please see A243051.

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))).
Showing 1-9 of 9 results.