A062307 -id:A062307 - cp's OEIS frontend

A008477 If n = Product (p_j^k_j) then a(n) = Product (k_j^p_j).

1, 1, 1, 4, 1, 1, 1, 9, 8, 1, 1, 4, 1, 1, 1, 16, 1, 8, 1, 4, 1, 1, 1, 9, 32, 1, 27, 4, 1, 1, 1, 25, 1, 1, 1, 32, 1, 1, 1, 9, 1, 1, 1, 4, 8, 1, 1, 16, 128, 32, 1, 4, 1, 27, 1, 9, 1, 1, 1, 4, 1, 1, 8, 36, 1, 1, 1, 4, 1, 1, 1, 72, 1

Offset: 1

Olivier Gérard

For any n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... is eventually periodic with period <= 2 [Farrokhi]. - N. J. A. Sloane, Apr 25 2009
a(A005117(n)) = 1; a(A013929(n)) > 1; A010052(a(A122132(n))) = 1. - Reinhard Zumkeller, Feb 17 2012
From Bernard Schott, Mar 26 2021: (Start)
The study of some properties of this sequence was proposed in the 1st problem of Concours Général in 2012 in France (see links).
Terms are precisely the powerful numbers in A001694.
If m is a term, there is a term q such that a(q) = m.
a(a(n)) <= n (see examples). (End)

			For n = 24 = 2^3*3^1, a(24) = 3^2*1^3 = 9, so a(9) = 2^3 = 8 and a(a(24)) = 8 < 24.
For n = 243 = 3^5, a(243) = 5^3 = 125, so a(125) = 3^5 = 243 and a(a(243)) = 243.

M. F. Hasler, Table of n, a(n) for n = 1..10000
Annales Concours Général, Sujet Concours Général 2012 (in French, problems).
Annales Concours Général, Corrigé Concours Général 2012 (in French, solutions).
M. Farrokhi, The Prime Exponentiation of an Integer: Problem 11315, Amer. Math. Monthly, 116 (2009), 470. - from _N. J. A. Sloane_, Apr 25 2009
Index to sequences related to Olympiads and other Mathematical Competitions.

Cf. A001694, A027748, A124010.
Cf. A005117, A013929, A010052, A062307, A122132, A342551.
Cf. A008478 (fixed points).

a008477 n = product $ zipWith (^) (a124010_row n) (a027748_row n)
-- Reinhard Zumkeller, Feb 17 2012

A008477 := proc(n) local e,j; e := ifactors(n)[2]:
mul (e[j][2]^e[j][1], j=1..nops(e)) end:
seq (A008477(n), n=1..60);
# Peter Luschny, Jan 17 2010

Prepend[ Array[ Times @@ Map[ Power @@ RotateLeft[ #1, 1 ]&, FactorInteger[ # ] ]&, 100, 2 ], 1 ]
Table[Times@@(First[#]^Last[#]&/@Transpose[Reverse[ Transpose[ FactorInteger[ n]]]]),{n,80}] (* Harvey P. Dale, Jul 22 2014 *)

A008477(n)=factorback(factor(n)*[0,1;1,0]) \\ M. F. Hasler, May 20 2012

from sympy import factorint, prod
a = lambda n: prod([pk[1]**pk[0] for pk in factorint(n).items()])
print([a(n) for n in range(1,61)]) # Darío Clavijo, Nov 06 2023
(APL, Dyalog dialect) A008477 ← {×/{⍺*⍨≢⍵}⌸factors(⍵)} ⍝ Needs also factors function from https://dfns.dyalog.com/c_factors.htm - Antti Karttunen, Feb 16 2024

Multiplicative with a(p^e) = e^p. - David W. Wilson, Aug 01 2001

A342973 Let f = A008477; nonsquarefree numbers m such that f(m) <> m or f(f(m)) <> m.

Original entry on oeis.org

12, 18, 20, 24, 28, 36, 40, 44, 45, 48, 50, 52, 54, 56, 60, 63, 64, 68, 75, 76, 80, 81, 84, 88, 90, 92, 96, 98, 99, 100, 104, 112, 116, 117, 120, 124, 126, 132, 135, 136, 140, 144, 147, 148, 150, 152, 153, 156, 160, 162, 164, 168, 171, 172, 175, 176, 180, 184, 188, 189, 192

Offset: 1

Bernard Schott, Apr 01 2021

nonn

Equivalently, with f = A008477, terms m of this sequence are precisely the nonsquarefree numbers for which the iterated sequence {m, f(m), f(f(m)), f(f(f(m))), ... } is not periodic.
The first sixteen terms are the same as A126706, then a(17) = 64 while A126706(17) = 68.
There exist only these 4 possibilities:
-> for every squarefree number m in A005117, f(m) = 1, and iterated sequence is for example:  (3, 1, 1, 1, 1, ...).
-> For m nonsquarefree fixed point of f in A008478, f(m) = m, iterated sequence has period = 1, as for example: (4, 4, 4, 4, 4, ...).
-> For m nonsquarefree in A062307, f(m) = q and f(q) = m, iterated sequence has period = 2, as for example: (8, 9, 8, 9, 8, 9, ...).
-> For m in this sequence, f(m) = k and m, k belong to an infinite iterated sequence, as for example: (..., 196, 512, 81, 64, ...) (see example).

			196 = 2^2*7^2 => A008477(196) = 2^2*2^7 = 2^9 = 512.
512 = 2^9 => A008477(512) = 9^2 = 81.
81 = 9^2 = 3^4 => A008477(81) = 4^3 = 64.
196, 512, 81, 64 are not terms of (A008478 U A062307), so they belong to this sequence.

Equals A013929 \ {A008478 U A062307}.

Cf. A008477, A342551.

fun[p_, e_] := e^p; f[n_] := Times @@ fun @@@ FactorInteger[n]; Select[Range[200], !SquareFreeQ[#] && f[#] != # && f[f[#]] != # &] (* Amiram Eldar, Apr 01 2021 *)

f(n) = factorback(factor(n)*[0, 1; 1, 0]); \\ A008477
isok(m) = if (!issquarefree(m), my(mm=f(m)); (mm != m) && (f(mm) != m)); \\ Michel Marcus, Apr 02 2021

A114130 Numbers whose factorization involves only prime exponents, all different, and no number appears both as an exponent and as a prime factor.

Original entry on oeis.org

8, 9, 25, 32, 49, 121, 125, 128, 169, 243, 289, 343, 361, 529, 841, 961, 1331, 1369, 1681, 1849, 2048, 2187, 2197, 2209, 2809, 3481, 3721, 4489, 4913, 5041, 5329, 6125, 6241, 6859, 6889, 7921, 8192, 8575, 9409, 10201, 10609, 10976, 11449, 11881, 11907

Offset: 1

Jon Wild, Feb 14 2006

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..560

Subsequence of A062307.

pedQ[n_]:=Module[{fi=Transpose[FactorInteger[n]],a,b},a=fi[[1]];b = fi[[2]]; AllTrue[b,PrimeQ]&&Length[b]==Length[Union[b]]&&Intersection[ a,b]=={}]; Select[Range[12000],pedQ] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Jan 24 2015 *)

A342551 a(n) is the smallest m such that A008477(m) is the n-th powerful number (A001694).

Original entry on oeis.org

1, 4, 9, 8, 16, 32, 27, 25, 64, 128, 81, 72, 512, 1024, 108, 2048, 243, 49, 4096, 8192, 16384, 288, 729, 32768, 125, 225, 200, 131072, 262144, 2187, 524288, 1152, 1048576, 432, 2097152, 4194304, 972, 196, 8388608, 648, 33554432, 4608, 864, 67108864, 19683, 268435456

Offset: 1

Bernard Schott, Mar 27 2021

nonn

As A008477 is not injective and terms A008477(n) are precisely the powerful numbers, this sequence lists the smallest preimage of each powerful number.
There are these three possibilities (see corresponding examples):
-> If A008477(q) = q is a fixed point in A008478 and if q = A001694(u) then a(u) = q.
-> If k and m are in A062307 and satisfy A008477(k) = m  and A008477(m) = k, if m = A001694(s) and k = A001694(t), then a(t) = m and a(s) = k;
-> If A008477(j) = v where v is a powerful number not in {A008478 U A062307} and j is the smallest preimage of v with v = A001694(z) then a(z) = j.

			-> A008477(16) = 16 is a fixed point and 16 is the 5th powerful number, so a(5) = 16.
-> 25 and 32 are in A062307 and satisfy A008477(25) = 32 and A008477(32) = 25, as 25 = A001694(6) and 32 = A001694(8), so a(6) = 32 and a(8) = 25.
-> A008477(81) = A008477(256) = 64 that is the 11th powerful number, since 81 is the smallest preimage of 64, so a(11) = 81.

Cf. A001694, A008477, A008478, A062307.

pwf(n) = my(k=1, nb=1); while (nb != n, k++; if (ispowerful(k), nb++)); k; \\ A001694
f(n) = factorback(factor(n)*[0, 1; 1, 0]); \\ A008477
a(n) = my(k=1, p=pwf(n)); while (f(k) != p, k++); k; \\ Michel Marcus, Mar 28 2021

More terms from Amiram Eldar, Mar 27 2021

A343293 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 36.

Original entry on oeis.org

36, 64, 81, 512, 196, 16384, 1089, 8589934592, 3844, 4611686018427387904, 31329, 191561942608236107294793378393788647952342390272950272, 478864

Offset: 1

Bernard Schott, Apr 11 2021

Equivalently, when g is the reciprocal map of f = A008477 as defined in the Name, the terms of this sequence are the successive terms of the infinite iterated sequence {m, g(m), g(g(m)), g(g(g(m))), ...} that begins with m = a(1) = 36, hence f(a(n)) = a(n-1).
Why choose 36? Because it is the smallest integer for which there exists such an infinite iterated sequence, with g(36) = 64; then f(36) = 32 with the periodic sequence (32, 25, 32, 25, ...) (see A062307). Explanation: 36 is the first nonsquarefree number in A342973 that is also squareful. The nonsquarefree terms < 36: 12, 18, 20, 24, 28 in A342973 are not squareful (A332785), so they have no preimage by f.
When a(n-1) has several preimages by f, as a(n) is the smallest preimage, this sequence is well defined (see examples).
All the terms are nonsquarefree but also powerful, hence they are in A001694.
a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).
Prime factorizations from a(1) to a(13): 2^2*3^2, 2^6, 3^4, 2^9, 2^2*7^2, 2^14, 3^2*11^2, 2^33, 2^2*31^2, 2^62, 3^2*59^2, 2^177, 2^4*173^2.
It appears that a(2m) = 2^q for some q>1 and a(2m+1) = r^2 for some r>1.
a(14) <= 2^692.

			a(1) = 36; 64 = 2^6 so f(64) = 6^2 = 36, also 192 = 2^6*3^1 and f(192) = 6^2*1^3 = 36 we have f(64) = f(192) = 36; but as 64 < 192, hence g(36) = 64 and a(2) = 64.
a(2) = 64 = f(81) = f(256), but as 81 < 256, g(64) = 81 and a(3) = 81.
a(4) = 512 = f(196) = f(400), but as 196 < 400, g(512) = 196 and a(5) = 196.

Annales Concours Général, Sujet Concours Général 2012 (in French, problems).
Annales Concours Général, Corrigé Concours Général 2012 (in French, solutions).

Cf. A001694, A008477, A062307, A332785, A342973.

a(10)-a(13) from Bert Dobbelaere, Apr 13 2021

A343294 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 100.

Original entry on oeis.org

100, 1024, 625, 33554432, 2116, 70368744177664

Offset: 1

Bernard Schott, Apr 12 2021

Equivalently, when g is the reciprocal map of f = A008477 as defined in the Name, the terms of this sequence are the successive terms of the infinite iterated sequence {m, g(m), g(g(m)), g(g(g(m))), ...} that begins with m = a(1) = 100, hence f(a(n)) = a(n-1).
Why choose 100? Because it is the second integer, after 36, for which there exists a new infinite iterated sequence that begins with g(100) = 1024; then f(100) = 128 with the periodic sequence (128, 49, 128, 49, ...) (see A062307). Explanation: 100 is the 4th nonsquarefree number in A342973 that is also squareful, but the 3 previous such first integers 36, 64, 81 are yet terms of the infinite iterated sequence A343293. Remember that the nonsquarefree terms in A342973 that are not squareful (A332785) have no preimage by f.
When a(n-1) has several preimages by f, as a(n) is the smallest preimage, this sequence is well defined (see examples).
All the terms are nonsquarefree but also powerful, hence they are in A001694.
a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).

			a(1) = 100; 1024 = 2^10 so f(1024) = 10^2 = 100: also 5120 = 2^10*5^1 and f(5120) = 10^2*1^5 = 100; we have f(1024) = f(5120) = 100, but as 1024 < 5120, hence g(100) = 1024 and a(2) = 1024.
a(2) = 1024 = f(625) = f(1250), but as 625 < 1250, g(1024) = 625 and a(3) = 625.

Cf. A001694, A008477, A062307, A332785, A342973, A343293.

A343295 a(n) is the smallest k such that A008477(k) = a(n-1) with a(1) = 144.

Original entry on oeis.org

144, 4096, 1225, 34359738368, 549081

Offset: 1

Bernard Schott, May 10 2021

nonn

The next term is a(6) = 2^741 with 224 digits.
Equivalently, when g is the reciprocal map of f = A008477 as defined in the Name, the terms of this sequence are the successive terms of the infinite iterated sequence {m, g(m), g(g(m)), g(g(g(m))), ...} that begins with m = a(1) = 144, hence f(a(n)) = a(n-1).
Why choose 144? Because it is the third integer, after 36 and 100, for which there exists a new infinite iterated sequence that begins with g(144) = 4096; then f(144) = 128 with the periodic sequence (128, 49, 128, 49, ...) (see A062307). Explanation: 144 is the 5th nonsquarefree number in A342973 that is also squareful; the 3 such first integers 36, 64, 81 are terms of the infinite iterated sequence A343293, while 100 is a term of the infinite iterated sequence A343294.
Remember that the nonsquarefree terms in A342973 that are not squareful (A332785) have no preimage by f.
All the terms are nonsquarefree but also powerful, hence they are in A001694.
a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).
Prime factorizations from a(1) to a(6): 2^4*3^2, 2^12, 5^2*7^2, 2^35, 3^2*13^2*19^2, 2^741.
It appears that a(2m) = 2^q for some q > 1, and a(2m+1) = r^2 for some r > 1.

			a(1) = 144; 4096 = 2^12 so f(4096) = 12^2 = 144: also 12288 = 2^12*3^1 and f(12288) = 12^2*1^3 = 144; we have f(4096) = f(12288) = 144, but as 4096 < 12288, hence g(144) = 4096 and a(2) = 4096.
a(2) = 4096 = f(1225) = f(2450), but as 1225 < 2450, g(4096) = 1225 and a(3) = 1225.

Cf. A001694, A008477, A062307, A332785, A342511, A342973, A343293, A343294.

cp's OEIS Frontend

A008477 If n = Product (p_j^k_j) then a(n) = Product (k_j^p_j).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Formula

A342973 Let f = A008477; nonsquarefree numbers m such that f(m) <> m or f(f(m)) <> m.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

A114130 Numbers whose factorization involves only prime exponents, all different, and no number appears both as an exponent and as a prime factor.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A342551 a(n) is the smallest m such that A008477(m) is the n-th powerful number (A001694).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Extensions

A343293 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 36.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A343294 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 100.

Views

Author

Keywords

Comments

Examples

Crossrefs

A343295 a(n) is the smallest k such that A008477(k) = a(n-1) with a(1) = 144.

Views

Author

Keywords

Comments

Examples

Crossrefs