A008479 Number of numbers <= n with same prime factors as n.

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 4, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 6, 2, 4, 1, 2, 1, 7, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 3, 2, 1, 1, 1, 5, 4, 1, 1, 2, 1, 1, 1, 3, 1, 3, 1, 2, 1, 1, 1, 9

Offset: 1

Author

Jeffrey Shallit, Olivier Gérard

Keywords

• nonn
• easy
• changed

Comments

For n > 1, a(n) gives the (one-based) index of the row where n is located in arrays A284311 and A285321 or respectively, index of the column where n is in A284457. A285329 gives the other index. - Antti Karttunen, Apr 17 2017

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Paul Erdős and T. Motzkin, Problem 5735, Amer. Math. Monthly, 78 (1971), 680-681. (Incorrect solution!)
• H. N. Shapiro, Problem 5735, Amer. Math. Monthly, 97 (1990), 937.

Crossrefs

Cf. A005117 (positions of ones), A005361, A007947, A008683, A010846, A284311, A284457, A285321, A285328, A285329.

Programs

Maple

N:= 100: # to get a(1)..a(N)
V:= Vector(N):
V[1]:= 1:
for n from 2 to N do
  if V[n] = 0 then
   S:= {n};
   for p in numtheory:-factorset(n) do
     S := S union {seq(seq(s*p^k,k=1..floor(log[p](N/s))),s=S)};
   od:
   S:= sort(convert(S,list));
   for k from 1 to nops(S) do V[S[k]]:= k od:
fi
od:
convert(V,list); # Robert Israel, May 20 2016

Mathematica

PkTbl=Prepend[ Array[ Times @@ First[ Transpose[ FactorInteger[ # ] ] ]&, 100, 2 ], 1 ];1+Array[ Count[ Take[ PkTbl, #-1 ], PkTbl[ [ # ] ] ]&, Length[ PkTbl ] ]
Count[#, k_ /; k == Last@ #] & /@ Function[s, Take[s, #] & /@ Range@ Length@ s]@ Array[Map[First, FactorInteger@ #] &, 120] (* or *)
Table[Sum[(Floor[n^k/k] - Floor[(n^k - 1)/k]) (Floor[k^n/n] - Floor[(k^n - 1)/n]), {k, n}], {n, 120}] (* Michael De Vlieger, May 20 2016 *)

PARI

a(n)=my(f=factor(n)[,1], s); forvec(v=vector(#f, i, [1, logint(n, f[i])]), if(prod(i=1, #f, f[i]^v[i])<=n, s++)); s \\ Charles R Greathouse IV, Oct 19 2017

Scheme

(define (A008479 n) (if (not (zero? (A008683 n))) 1 (+ 1 (A008479 (A285328 n))))) ;; Antti Karttunen, Apr 17 2017

Formula

a(n) = Sum_{k=1..n} (floor(n^k/k)-floor((n^k-1)/k))*(floor(k^n/n)-floor((k^n-1)/n)). - Anthony Browne, May 20 2016

If A008683(n) <> 0 [when n is squarefree, A005117], a(n) = 1, otherwise a(n) = 1+a(A285328(n)). - Antti Karttunen, Apr 17 2017

a(n) <= A010846(n), with equality if and only if n = 1. - Amiram Eldar, May 25 2025

a(m^(k+1)) = A010846(m^k) when m is squarefree. - Flávio V. Fernandes, Aug 20 2025