A212173 First integer with same second signature as n (cf. A212172).

1, 1, 1, 4, 1, 1, 1, 8, 4, 1, 1, 4, 1, 1, 1, 16, 1, 4, 1, 4, 1, 1, 1, 8, 4, 1, 8, 4, 1, 1, 1, 32, 1, 1, 1, 36, 1, 1, 1, 8, 1, 1, 1, 4, 4, 1, 1, 16, 4, 4, 1, 4, 1, 8, 1, 8, 1, 1, 1, 4, 1, 1, 4, 64, 1, 1, 1, 4, 1, 1, 1, 72, 1, 1, 4, 4, 1, 1, 1, 16, 16, 1, 1, 4

Offset

1,4

Comments

Two integers have the same second signature iff the same exponents >= 2 occur in the canonical prime factorization of each integer, regardless of the order in which they occur in each factorization.

References

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 844.

Links

Jason Kimberley, Table of n, a(n) for n = 1..10000

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Primefan, The First 2500 Integers Factored (1st of 5 pages)

Formula

a(n) = A046523(A057521(n)) = A057521(A046523(n)).

Example

12 = 2^2*3 has 1 exponent >= 2 in its prime factorization, namely, 2. Hence, its second signature is {2}.  The smallest number with second signature {2} is 4; hence, a(12) = 4.

Maple

f:= proc(n) local E, i;
E:= sort(select(`>`, map(t -> t[2], ifactors(n)[2]), 1), `>`);
mul(ithprime(i)^E[i], i=1..nops(E))
end proc:
map(f, [$1..100]); # Robert Israel, Jul 19 2017

Mathematica

Function[s, Sort[Apply[Join, Map[Function[k, Map[{#, First@ k} &, k]], Values@ s]]][[All, -1]]]@ KeySort@ PositionIndex@ Table[Sort@ DeleteCases[FactorInteger[n][[All, -1]], e_ /; e < 2] /. {} -> {1}, {n, 84}] (* Michael De Vlieger, Jul 19 2017 *)

Prog

(Magma) A212173 := func<n| &*[Integers()| NthPrime(j)^s[j]:j in[1..#s]] where s is Reverse(Sort([pe[2]:pe in Factorisation(n)| pe[2]gt 1]))>; [A212173(n):n in[1..85]]; // Jason Kimberley, Jun 14 2012
(Python)
from sympy import factorint
from operator import mul
def P(n): return sorted(factorint(n).values())
def a046523(n):
    x=1
    while True:
        if P(n)==P(x): return x
        else: x+=1
def a057521(n): return 1 if n==1 else reduce(mul, [1 if e==1 else p**e for p, e in factorint(n).items()])
def a(n): return a046523(a057521(n))
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Jul 19 2017
(PARI) a(n) = {my(sn = vecsort(select(x->(x>=2), factor(n)[, 2]))); for (i=1, n, if (vecsort(select(x->(x>=2), factor(i)[, 2])) == sn, return(i)); ); } \\ Michel Marcus, Jul 19 2017

Crossrefs

Cf. A212172, A046523. All terms belong to A181800.

Sequence in context: A368333 A088440 A300253 * A366993 A274006 A203025

Adjacent sequences: A212170 A212171 A212172 * A212174 A212175 A212176

Keyword

nonn,easy

Author

Matthew Vandermast, Jun 03 2012

Status

approved