A069799 The number obtained by reversing the sequence of nonzero exponents in the prime factorization of n with respect to distinct primes present, as ordered by their indices.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 18, 13, 14, 15, 16, 17, 12, 19, 50, 21, 22, 23, 54, 25, 26, 27, 98, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 250, 41, 42, 43, 242, 75, 46, 47, 162, 49, 20, 51, 338, 53, 24, 55, 686, 57, 58, 59, 150, 61, 62, 147, 64, 65
Offset: 1
Keywords
Examples
a(24) = 54 as 24 = p_1^3 * p_2^1 = 2^3 * 3^1 and 54 = p_1^1 * p_2^3 = 2 * 3^3. For n = 2200, we see that it encodes the partition (1,1,1,3,3,5) in A112798 as 2200 = p_1 * p_1 * p_1 * p_3 * p_3 * p_5 = 2^3 * 5^2 * 11. This in turn corresponds to the following Young diagram in French notation: _ | | | | | |_ _ | | | |_ _ |_ _ _ _ _| Reversing the order of vertical line segment lengths (3,2,1) to (1,2,3), but keeping the order of horizontal line segment lengths as (1,2,2), we get a new Young diagram _ | |_ _ | | | |_ _ | | | | |_ _ _ _ _| which represents the partition (1,3,3,5,5,5), encoded in A112798 by p_1 * p_3^2 * p_5^3 = 2 * 5^2 * 11^3 = 66550, thus a(2200) = 66550.
Links
- Alois P. Heinz (first 1000 terms) & Antti Karttunen, Table of n, a(n) for n = 1..8192
- Wikipedia, Young diagram
- Index entries for sequences that are permutations of the natural numbers
Crossrefs
Programs
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Haskell
a069799 n = product $ zipWith (^) (a027748_row n) (reverse $ a124010_row n) -- Reinhard Zumkeller, Apr 27 2013 (MIT/GNU Scheme, with Aubrey Jaffer's SLIB Scheme library) (require 'factor) (define (A069799 n) (let ((pf (ifactor n))) (apply * (map expt (uniq pf) (reverse (multiplicities pf)))))) (define (ifactor n) (cond ((< n 2) (list)) (else (sort (factor n) <)))) (define (uniq lista) (let loop ((lista lista) (z (list))) (cond ((null? lista) (reverse! z)) ((and (pair? z) (equal? (car z) (car lista))) (loop (cdr lista) z)) (else (loop (cdr lista) (cons (car lista) z)))))) (define (multiplicities lista) (let loop ((mults (list)) (lista lista) (prev #f)) (cond ((not (pair? lista)) (reverse! mults)) ((equal? (car lista) prev) (set-car! mults (+ 1 (car mults))) (loop mults (cdr lista) prev)) (else (loop (cons 1 mults) (cdr lista) (car lista)))))) ;; Antti Karttunen, May 24 2014
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Maple
A069799 := proc(n) local e,j; e := ifactors(n)[2]: mul (e[j][1]^e[nops(e)-j+1][2], j=1..nops(e)) end: seq (A069799(i), i=1..40); # Peter Luschny, Jan 17 2011
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Mathematica
f[n_] := Block[{a = Transpose[ FactorInteger[n]], m = n}, If[ Length[a] == 2, Apply[ Times, a[[1]]^Reverse[a[[2]] ]], m]]; Table[ f[n], {n, 1, 65}]
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PARI
a(n) = {my(f = factor(n)); my(g = f); my(nbf = #f~); for (i=1, nbf, g[i, 1] = f[nbf-i+1, 1];); factorback(g);} \\ Michel Marcus, Jul 02 2015
Formula
If n = p_a^e_a * p_b^e_b * ... * p_j^e_j * p_k^e_k, where p_a < ... < p_k are distinct primes of the prime factorization of n (sorted into ascending order), and e_a, ..., e_k are their nonzero exponents, then a(n) = p_a^e_k * p_b^e_j * ... * p_j^e_b * p_k^e_a.
a(n) = product(A027748(o(n)+1-k)^A124010(k): k=1..o(n)) = product(A027748(k)^A124010(o(n)+1-k): k=1..o(n)), where o(n) = A001221(n). - Reinhard Zumkeller, Apr 27 2013
From Antti Karttunen, Jun 01 2014: (Start)
Can be obtained also by composing/conjugating related permutations:
(End)
Extensions
Edited, corrected and extended by Robert G. Wilson v and Vladeta Jovovic, Apr 15 2002
Definition corrected by Reinhard Zumkeller, Apr 27 2013
Definition again reworded, Comments section edited and Young diagram examples added by Antti Karttunen, May 30 2014
Comments