A069799 -id:A069799 - cp's OEIS frontend

A293448 Self-inverse permutation of natural numbers: replace (with multiplicity) each prime factor A000040(k) with A000040(min+(max-k)) in the prime factorization of n, where min = A055396(n) and max = A061395(n).

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, 70, 43, 242, 75, 46, 47, 162, 49, 20, 51, 338, 53, 24, 55, 686, 57, 58, 59, 150, 61, 62, 147, 64, 65, 154, 67, 578, 69, 42, 71, 108, 73, 74, 45, 722, 77, 286, 79, 1250, 81, 82, 83

Offset: 1

Antti Karttunen, Nov 09 2017

nonn

Reverse the prime-indices in such a way that the smallest and the greatest prime dividing n (A020639 and A006530) are preserved.

a(n) = n iff n belongs to A236510. - Rémy Sigrist, Nov 22 2017

			For n = 25 = 5^2 = prime(3)^2, thus min = max = 3, and we form a product prime(3+(3-3))^2, thus a(25) = prime(3)^2 = 25.
For n = 42 = 2*3*7 = prime(1)*prime(2)*prime(4), thus min = 1 and max = 4, so we form a product prime(1+(4-1))*prime(1+(4-2))*prime(1+(4-4)), thus a(42) = prime(4)*prime(3)*prime(1) = 7*5*2 = 70.
For n = 126 = 2 * 3^2 * 7 = prime(1) * prime(2)^2 * prime(4), thus min = 1 and max = 4, so we form a product prime(1+(4-1)) * prime(1+(4-2))^2 * prime(1+(4-4)), thus a(126) = prime(4) * prime(3)^2 * prime(1) = 7 * 5^2 * 2 = 350.

Cf. A000720, A055396, A057889, A061395, A236510 (fixed points), A273258.

Differs from A069799 (and some other related permutations) for the first time at n=42.

A293448(n) = { if(1==n,return(n)); my(f=factor(n), mini = primepi(f[1, 1]), maxi = primepi(f[#f~, 1])); for(i=1,#f~,f[i,1] = prime((maxi-primepi(f[i,1]))+mini)); factorback(f); }

For all even squarefree numbers coincides with A273258, that is, for all n, a(A039956(n)) = A273258(A039956(n)).

A182938 If n = Product (p_j^e_j) then a(n) = Product (binomial(p_j, e_j)).

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 0, 3, 10, 11, 3, 13, 14, 15, 0, 17, 6, 19, 5, 21, 22, 23, 0, 10, 26, 1, 7, 29, 30, 31, 0, 33, 34, 35, 3, 37, 38, 39, 0, 41, 42, 43, 11, 15, 46, 47, 0, 21, 20, 51, 13, 53, 2, 55, 0, 57, 58, 59, 15, 61, 62, 21, 0, 65, 66

Offset: 1

Peter Luschny, Jan 16 2011

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Plot of Sum_{k=1..n} a(k) / n^2 for n = 1..1000000

Cf. A000026, A001414, A008473, A008474, A008475, A008476, A008477, A028310, A069799.

Cf. A027748, A124010, A007318, A328745.

a182938 n = product $ zipWith a007318'
   (a027748_row n) (map toInteger $ a124010_row n)
-- Reinhard Zumkeller, Feb 18 2012

A182938 := proc(n) local e,j; e := ifactors(n)[2]:
mul (binomial(e[j][1], e[j][2]), j=1..nops(e)) end:
seq (A182938(n), n=1..100);

a[n_] := Times @@ (Map[Binomial @@ # &, FactorInteger[n], 1]);
Table[a[n], {n, 1, 100}] (* Kellen Myers, Jan 16 2011 *)

a(n)=prod(i=1,#n=factor(n)~,binomial(n[1,i],n[2,i])) \\ M. F. Hasler

for(n=1, 100, print1(direuler(p=2, n, (1 + X)^p)[n], ", ")) \\ Vaclav Kotesovec, Mar 28 2025

a(A185359(n)) = 0. - Reinhard Zumkeller, Feb 18 2012
Dirichlet g.f.: Product_{p prime} (1 + p^(-s))^p. - Ilya Gutkovskiy, Oct 26 2019
Conjecture: Sum_{k=1..n} a(k) ~ c * n^2, where c = 0.33754... - Vaclav Kotesovec, Mar 28 2025

Given terms checked with new PARI code by M. F. Hasler, Jan 16 2011

A225891 Numbers obtained by rotating left the indices in the prime signature of n.

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, 66

Offset: 1

Paul Tek, May 20 2013

nonn

Inverse permutation to A105119.
A072774 gives the fixed points.
The first term which is different from A105119 is a(60).
The first term which is different from A069799 is a(90).

			a(2250) = a(2*3^2*5^3) = 2^2*3^3*5 = 540.

Paul Tek, Table of n, a(n) for n = 1..10000

Cf. A105119, A072774, A069799.

Table[Times @@ ((ar = Transpose[FactorInteger[n]])[[1]]^RotateLeft[ar[[2]]]), {n, 66}] (* Ivan Neretin, Jul 09 2015 *)

{ a(n)=my(f=factor(n),l=#f[,1]);
return(prod(i=1,l,f[i,1]^f[if(i==l,1,i+1),2])) }

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

Antti Karttunen, May 20 2014

nonn

Numbers that are fixed by the permutation A242419, i.e., for which A242419(n) = n. Also, numbers that are fixed by both A069799 and A242415.
Number n is present if its prime factorization n = p_a^e_a * p_b^e_b * p_c^e_c * ... * p_i^e_i * p_j^e_j * p_k^e_k where a < b < c < ... < i < j < k, satisfies the condition, that both the first differences of prime indices (a-0, b-a, c-b, ..., j-i, k-j) and the respective exponents (e_a, e_b, e_c, ... , e_i, e_j, e_k) form a palindrome.
More formally, numbers n whose prime factorization is either of the form p^e (p prime, e >= 0), i.e., one of the terms of A000961, or of the form p_i1^e_i1 * p_i2^e_i2 * p_i3^e_i3 * ... * p_i_{k-1}^e_{i_{k-1}} * p_{i_k}^e_{i_k}, where p_i1 < p_i2 < ... < p_i_{k-1} < p_k are distinct primes (sorted into ascending order) in the prime factorization of n, and e_i1 .. e_{i_k} are their nonzero exponents (here k =  A001221(n) and i_k = A061395(n), the index of the largest prime present), and the indices of primes satisfy the relation that for each index i_1 <= i_j < i_k present, the index i_{k-j} is also present, and the exponents e_{i_j} and e_{i_{(k-j)+1}} are equal.

			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.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Fixed points of A242419. Intersection of A242413 and A242414.
Subsequences: A000961, A002110.
Cf. A243058, A243068, A242421, A088902, A241912.

A321474 Reverse the nonzero digits of n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 21, 31, 41, 51, 61, 71, 81, 91, 20, 12, 22, 32, 42, 52, 62, 72, 82, 92, 30, 13, 23, 33, 43, 53, 63, 73, 83, 93, 40, 14, 24, 34, 44, 54, 64, 74, 84, 94, 50, 15, 25, 35, 45, 55, 65, 75, 85, 95, 60, 16, 26, 36, 46, 56, 66, 76

Offset: 0

Rémy Sigrist, Nov 11 2018

This sequence is a self-inverse permutation of nonnegative integers.
See A321464 for the ternary variant.
This sequence has similarities with A069799: here we reverse nonzero digits, there we reverse nonzero prime exponents.

			For n = 1024:
- 1024 has 3 nonzero digits: 1, 2 and 4,
- so we replace the first nonzero digit by the third, the third by the first (and the second remains in place),
- and we obtain a(1024) = 4021.

Cf. A004086, A069799, A136400, A321464.

a(n, base=10) = my (d=digits(n, base), t=Vecrev(select(sign, d)), i=0); for (j=1, #d, if (d[j], d[j] = t[i++])); fromdigits(d, base)

a(10 * n) = 10 * a(n).

A136400(a(n)) = A136400(n).

A273258 Write the distinct prime divisors p of n in the (PrimePi(p) - 1)-th place, ignoring multiplicity. Decode the resulting number after first reversing the code, ignoring any leading zeros.

Original entry on oeis.org

1, 2, 2, 2, 2, 6, 2, 2, 2, 10, 2, 6, 2, 14, 6, 2, 2, 6, 2, 10, 10, 22, 2, 6, 2, 26, 2, 14, 2, 30, 2, 2, 14, 34, 6, 6, 2, 38, 22, 10, 2, 70, 2, 22, 6, 46, 2, 6, 2, 10, 26, 26, 2, 6, 10, 14, 34, 58, 2, 30, 2, 62, 10, 2, 14, 154, 2, 34, 38, 42, 2, 6, 2, 74, 6, 38, 6, 286, 2, 10, 2, 82, 2, 70, 22, 86

Offset: 1

Michael De Vlieger, Aug 28 2016

Encode n with the function f(n) = noting the distinct prime divisors p of n by writing "1" in the (PrimePi(n) - 1)-th place, e.g, f(6) = f(12) = "11". This function is akin to A054841(n) except we don't note the multiplicity e of p in n, rather merely note "1" if e > 0.
This sequence decodes f(n) by reversing the digits.
If we decode f(n) without reversal, we have A007947(n), since f(n) sets any multiplicity e > 1 of prime divisor p of n to 1.
All terms except a(1) are of the form 2x with x odd. a(1) = 1, since f(1) = "0" and stands unaffected in reversal and decoding, and any zeros to the right of all 1's are lost in reversal. Thus f(15) = "110" reversed becomes "011" -> "11" decoded equals 2 * 3 = 6. Because we lose leading zeros, we always have 1 in position 1, which decoded is interpreted as the factor 2.
a(p) for p prime = 2, since primes are written via f(p) as 1 in the (PrimePi(p)-1)-th place. There is only one 1 in this number (similar to a perfect power of ten decimally) and when it is reversed, the number loses all leading zeros to become "1" -> 2. This also applies to prime powers p^e, since e is rendered as 1 by f(p^e), i.e., f(p^e) = f(p).

			a(3) = 2 since f(3) = "10" reversed becomes "01", loses leading zeros to become "1" -> 2.
a(6) = a(12) = "11" reversed stays the same -> 2 * 3 = 6.
a(15) = "110" reversed becomes "011", loses leading zeros to become "11" -> 6.
a(42) = "1101" reversed becomes "1011" -> 70 (a(70) = 42).

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from indices in prime factorization

Cf. A007947, A019565, A030101, A054841 (analogous encoding algorithm), A069799, A087207, A137502, A276379, A293448 (a bijective variant of this sequence).

Table[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ If[# == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> 1 &, f]]@ FactorInteger@ #] &@ n, {n, 86}]

(define (A273258 n) (A019565 (A030101 (A087207 n)))) ;; Antti Karttunen, Jun 18 2017

a(n) = A019565(A030101(A087207(n))). - Antti Karttunen, Jun 18 2017

For all n, a(A039956(n)) = A293448(A039956(n)). - Antti Karttunen, Nov 21 2017

A085078 The largest number with the prime signature of n! using primes <= n.

Original entry on oeis.org

1, 2, 6, 54, 750, 11250, 360150, 123531450, 3088286250, 64854011250, 77201350992150, 65389544290351050, 32637304517036749530, 2121424793607388719450, 163349709107768931397650

Offset: 1

Amarnath Murthy, Jul 01 2003

nonn

n! is the smallest number with that prime signature. E.g. 720 = 2^4*3^2*5. (Can we name a(n) as the eldest brother of n!?) Subsidiary sequence: Total number of distinct numbers with prime signature that of n! having prime divisors less than or equal to n.
From Reikku Kulon, Sep 18 2008: (Start)
This is n! with prime exponents reversed. Perhaps it should be denoted with an inverted exclamation mark: (inverted-!)n
7! = 5040 = 2^4 * 3^2 * 5^1 * 7^1
(inverted-!)7 = 360150 = 2^1 * 3^1 * 5^2 * 7^4 (End)

			For n=6, 6!= 720 = 2^4*3^2*5, hence a(6) = 5^4*3^2*2 = 11250.
For n=8, 8! = 40320 = 2^7*3^2*5*7, hence a(8) = 7^7*5^2*3*2 = 123531450.

Amiram Eldar, Table of n, a(n) for n = 1..181
Index entries for sequences related to prime signature

Cf. A000142, A069799.

Cf. A000040 [From Reikku Kulon, Sep 18 2008]

a[n_] := Module[{f = FactorInteger[n!], p, e}, p = First /@ f; e = Last /@ f; Times @@ (p^Reverse[e])]; Array[a, 15] (* Amiram Eldar, Dec 30 2020 *)

for (n = 1, 20, f = factor(n!); c = matsize(f)[1]; a = prod(i = 1, c, f[i, 1]^f[c + 1 - i, 2]); print(a)); \\ David Wasserman, Jan 14 2005

a(n) = A069799(A000142(n)). - Amiram Eldar, Dec 30 2020

More terms from David Wasserman, Jan 14 2005

cp's OEIS Frontend

A293448 Self-inverse permutation of natural numbers: replace (with multiplicity) each prime factor A000040(k) with A000040(min+(max-k)) in the prime factorization of n, where min = A055396(n) and max = A061395(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A182938 If n = Product (p_j^e_j) then a(n) = Product (binomial(p_j, e_j)).

Views

Author

Keywords

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

Extensions

A225891 Numbers obtained by rotating left the indices in the prime signature of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A321474 Reverse the nonzero digits of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A273258 Write the distinct prime divisors p of n in the (PrimePi(p) - 1)-th place, ignoring multiplicity. Decode the resulting number after first reversing the code, ignoring any leading zeros.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Scheme

Formula

A085078 The largest number with the prime signature of n! using primes <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula