A242413
Numbers in whose prime factorization the first differences of indices of distinct primes form a palindrome; fixed points of A242415.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 21, 23, 24, 25, 27, 29, 30, 31, 32, 36, 37, 41, 43, 47, 48, 49, 53, 54, 59, 60, 61, 63, 64, 65, 67, 70, 71, 72, 73, 79, 81, 83, 89, 90, 96, 97, 101, 103, 107, 108, 109, 113, 120, 121, 125, 127, 128, 131, 133, 137, 139, 140, 144
Offset: 1
1 is present because it has an empty factorization, so both the sequence of the prime indices and their first differences are empty, and empty sequences are palindromes as well.
12 = 2*2*3 = p_1^2 * p_2 is present, as the first differences (deltas) of prime indices (1-0, 2-1) = (1,1) form a palindrome.
60 = 2*2*3*5 = p_1^2 * p_2 * p_3 is present, as the deltas of prime indices (1-0, 2-1, 3-2) = (1,1,1) form a palindrome.
61 = p_18 is present, as the deltas of prime indices, (18), form a palindrome.
144 = 2^4 * 3^2 = p_1^4 * p_2^2 is present, as the deltas of prime indices (1-0, 2-1) = (1,1) form a palindrome.
Also, any of the cases mentioned in the Example section of A242417 as being present there, are also present in this sequence.
Differs from
A243068 for the first time at n=36, where a(36)=60, while
A243068(36)=61.
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
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. also
A080576,
A036036,
A001221,
A001222,
A061395,
A243503,
A056239,
A052126,
A003961,
A064989,
A071178,
A241917,
A241919,
A242378,
A242424,
A006530,
A105560,
A070003,
A102750,
A066829,
A028260,
A026424,
A244990,
A244991,
A244992,
A108951,
A181815,
A181819,
A181820,
A238745,
A238690,
A242421.
Cf.
A319988 (parity of this sequence for n > 1),
A336124 (a(n) mod 4).
Other related permutations:
A048673-
A064216,
A244981-
A244982,
A244983-
A244984,
A243287-
A243288,
A243505-
A243506,
A245613-
A245614,
A075157,
A075158,
A129594,
A069799,
A242415,
A245451,
A245452,
A245454, also
A336321 &
A336322 (composed with
A225546).
Cf. also array
A350066 [A(i, j) = a(a(i)*a(j))].
-
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
-
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 *)
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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
-
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
-
;; Uses Antti Karttunen's IntSeq-library.
(definec (A122111 n) (if (<= n 1) n (* (A000040 (A001222 n)) (A122111 (A064989 n)))))
;; Antti Karttunen, May 12 2014
-
;; 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
-
;; 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
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.
Original entry on oeis.org
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
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:
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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
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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.
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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
-
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
-
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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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
Definition again reworded, Comments section edited and Young diagram examples added by
Antti Karttunen, May 30 2014
A242378
Square array read by antidiagonals: to obtain A(i,j), replace each prime factor prime(k) in prime factorization of j with prime(k+i).
Original entry on oeis.org
0, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 5, 5, 1, 0, 5, 9, 7, 7, 1, 0, 6, 7, 25, 11, 11, 1, 0, 7, 15, 11, 49, 13, 13, 1, 0, 8, 11, 35, 13, 121, 17, 17, 1, 0, 9, 27, 13, 77, 17, 169, 19, 19, 1, 0, 10, 25, 125, 17, 143, 19, 289, 23, 23, 1, 0, 11, 21, 49, 343, 19, 221, 23, 361, 29, 29, 1, 0
Offset: 0
The top-left corner of the array:
0, 1, 2, 3, 4, 5, 6, 7, 8, ...
0, 1, 3, 5, 9, 7, 15, 11, 27, ...
0, 1, 5, 7, 25, 11, 35, 13, 125, ...
0, 1, 7, 11, 49, 13, 77, 17, 343, ...
0, 1, 11, 13, 121, 17, 143, 19,1331, ...
0, 1, 13, 17, 169, 19, 221, 23,2197, ...
...
A(2,6) = A003961(A003961(6)) = p_{1+2} * p_{2+2} = p_3 * p_4 = 5 * 7 = 35, because 6 = 2*3 = p_1 * p_2.
Taking every second column from column 2 onward gives array
A246278 which is a permutation of natural numbers larger than 1.
Column 2:
A000040 (primes), Column 3:
A065091 (odd primes), Column 4:
A001248 (squares of primes), Column 6:
A006094 (products of two successive primes), Column 8:
A030078 (cubes of primes).
Excluding column 0, a subtable of
A297845.
A242419
Reverse both the exponents and the deltas of the indices of distinct primes present in the prime factorization of n.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 11, 18, 13, 35, 10, 16, 17, 12, 19, 75, 21, 77, 23, 54, 25, 143, 27, 245, 29, 30, 31, 32, 55, 221, 14, 36, 37, 323, 91, 375, 41, 105, 43, 847, 50, 437, 47, 162, 49, 45, 187, 1859, 53, 24, 33, 1715, 247, 667, 59, 150, 61, 899, 147, 64, 65
Offset: 1
For n = 10 = 2*5 = p_1^1 * p_3^1, we get p_(3-1)^1 * p_3^1 = 3 * 5 = 15, thus a(10) = 15.
For n = 20 = 2*2*5 = p_1^2 * p_3^1, we get p_(3-1)^1 * p_3^2 = 3^1 * 5^2 = 3*25 = 75, thus a(20) = 75.
For n = 84 = 2*2*3*7 = p_1^2 * p_2 * p_4, when we reverse the deltas of indices, and reverse also the order of exponents, we get p_(4-2) * p_(4-1) * p_4^2 = 3 * 5 * 7^2 = 735, thus a(84) = 735.
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:
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Reversing the order of "steps", so that each horizontal and vertical line segment centered around a "convex corner" moves as a whole, means that the first stair from the top (one unit wide and three units high) is moved to the last position, the second one (two units wide and two units high) stays in the middle, and the original bottom step (two units wide and one unit high) will be the new topmost step, thus we get the following Young diagram:
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which represents the partition (2,4,4,5,5,5), encoded in A112798 by p_2 * p_4^2 * p_5^3 = 3 * 7^2 * 11^3 = 195657, thus a(2200) = 195657.
Cf.
A112798,
A122111,
A153212,
A000040,
A241919,
A071178,
A242378,
A051119,
A243057,
A105119,
A242420.
A153212
A permutation of the natural numbers: in the prime factorization of n, swap each prime's index difference (from the previous distinct prime that divides n) and the prime's power.
Original entry on oeis.org
1, 2, 4, 3, 8, 6, 16, 5, 9, 18, 32, 15, 64, 54, 12, 7, 128, 10, 256, 75, 36, 162, 512, 35, 27, 486, 25, 375, 1024, 30, 2048, 11, 108, 1458, 24, 21, 4096, 4374, 324, 245, 8192, 150, 16384, 1875, 45, 13122, 32768, 77, 81, 50, 972, 9375, 65536, 14, 72, 1715, 2916, 39366, 131072, 105, 262144, 118098, 225, 13
Offset: 1
Luchezar Belev (l_belev(AT)yahoo.com), Dec 20 2008
For n = 10 we have 10 = 2^1 * 5^1 = p(1)^1 * p(3)^1 then a(10) = p(1)^1 * p(2)^2 = 2^1 * 3^2 = 18.
For n = 18 we have 18 = 2^1 * 3^2 = p(1)^1 * p(2)^2 then a(18) = p(1)^1 * p(3)^1 = 2^1 * 5^1 = 10.
For n = 19 we have 19 = 19^1 = p(8)^1 then a(19) = p(1)^8 = 2^8 = 256.
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:
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Exchanging the order of the horizontal and vertical line segment of each "step", results the following Young diagram:
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which represents the partition (3,5,5,6,6), encoded in A112798 by p_3 * p_5^2 * p_6^2 = 5 * 11^2 * 13^2 = 102245, thus a(2200) = 102245.
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a(n) = {my(f = factor(n)); my(g = f); for (i=1, #f~, if (i==1, g[i,1] = prime(f[i,2]), g[i,1] = prime(f[i,2]+ primepi(g[i-1,1]))); if (i==1, g[i,2] = primepi(f[i,1]), g[i,2] = primepi(f[i,1]) - primepi(f[i-1,1]));); factorback(g);} \\ Michel Marcus, Dec 16 2014
Term a(1)=1 prepended, and also more terms computed by
Antti Karttunen, May 16 2014
A242417
Numbers in whose prime factorization both the first differences of indices of distinct primes and their exponents form a palindrome; fixed points of A242419.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 16, 17, 19, 21, 23, 25, 27, 29, 30, 31, 32, 36, 37, 41, 43, 47, 49, 53, 59, 61, 64, 65, 67, 70, 71, 73, 79, 81, 83, 89, 90, 97, 101, 103, 107, 109, 113, 121, 125, 127, 128, 131, 133, 137, 139, 149, 151, 154, 157, 163, 165, 167, 169
Offset: 1
1 is present because it has an empty factorization, so both sequences are empty, thus palindromes.
3 = p_2^1 is present, as both the sequence of the first differences (deltas) of prime indices (2-0) = (2) and the exponents (1) are palindromes.
6 = p_1^1 * p_2^1 is present, as both the deltas of prime indices (1-0, 2-1) = (1,1) and the exponents (1,1) form a palindrome.
8 = p_1^3 is present, as both the deltas of prime indices (1) and the exponents (3) form a palindrome.
300 = 4*3*25 = p_1^2 * p_2^1 * p_3^2 is present, as both the deltas of prime indices (1-0, 2-1, 3-2) = (1,1,1) 1, 2 and the exponents (2,1,2), form a palindrome.
144 = 2^4 * 3^2 = p_1^4 * p_2^2 is NOT present, as although the deltas of prime indices (1-0, 2-1) = (1,1) are palindrome, the sequence of exponents (4,2) do NOT form a palindrome.
441 = 9*49 = p_2^2 * p_4^2 is present, as both the deltas of prime indices (2-0, 4-2) = (2,2) and the exponents (2,2) form a palindrome.
30030 = 2*3*5*7*11*13 = p_1 * p_2 * p_3 * p_4 * p_5 * p_6 is present, as the exponents are all ones, and the deltas of indices, (6-5,5-4,4-3,3-2,2-1,1-0) = (1,1,1,1,1,1) likewise are all ones, thus both sequences form a palindrome. This is true for all primorial numbers, A002110.
47775 = 3*5*5*7*7*13 = p_2^1 * p_3^2 * p_4^2 * p_6^1 is present, as the deltas of indices (6-4,4-3,3-2,2-0) = (2,1,1,2) and the exponents (1,2,2,1) both form a palindrome.
90000 = 2*2*2*2*3*3*5*5*5*5 = p_1^4 * p_2^2 * p_3^4 is present, as the deltas of indices (3-2,2-1,1-0) = (1,1,1) and the exponents (4,2,4) both form a palindrome.
Showing 1-7 of 7 results.
Comments