A064476 For an integer k with prime factorization p_1*p_2*p_3* ... *p_m let k* = (p_1+1)*(p_2+1)*(p_3+1)* ... *(p_m+1) (A064478); sequence gives k such that k* is divisible by k.

1, 6, 12, 36, 72, 144, 216, 432, 864, 1296, 1728, 2592, 5184, 7776, 10368, 15552, 20736, 31104, 46656, 62208, 93312, 124416, 186624, 248832, 279936, 373248, 559872, 746496, 1119744, 1492992, 1679616, 2239488, 2985984, 3359232, 4478976

Offset: 1

Author

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 06 2001

Keywords

• nonn
• easy
• nice

Comments

Could be generalized by defining x* = (p_1+v)*(p_2+v) .. (p_m+v) where v is any integer.

It is not difficult to show that these numbers have the form 2^i*3^j with j <= i <= 2j. Hence 1 is the only odd term; also if k|k* then k*|k**. The values of i and j are given in A064514 and A064515. - Vladeta Jovovic and N. J. A. Sloane, Oct 07 2001

Examples

			12 is in the sequence because 12 = 2 * 2 * 3, so 12* is 3 * 3 * 4 = 36 and 36 is divisible by 12.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..50 from Harry J. Smith)

Crossrefs

Cf. A064478, A064514, A064515, A064518, A064522, A003959.

Programs

ARIBAS

function p2p3(stop:integer): array; var c,i,j,x: integer; b: boolean; ar: array; begin ar := alloc(array,stop); x := 0; c := 0; b := c < stop; while b do i := x; j := x - i; while b and i >= j do if i <= 2*j then ar[c] := (2^i * 3^j,i,j); inc(c); b := c < stop; end; dec(i); inc(j); end; inc(x); end; return sort(ar, comparefirst); end; function comparefirst(x,y: array): integer; begin return y[0] - x[0]; end; function a064476(maxarg: integer); var j: integer; ar: array; begin ar := p2p3(maxarg); for j := 0 to maxarg - 1 do write(ar[j][0]," "); end; end; a064476(35);

Haskell

a064476 n = a064476_list !! (n-1)
a064476_list = filter (\x -> a003959 x `mod` x == 0) [1..]
-- Reinhard Zumkeller, Feb 28 2013

Mathematica

diQ[n_]:=Divisible[Times@@(#+1&/@Flatten[Table[First[#],{Last[#]}]&/@ FactorInteger[n]]),n]; Select[Range[4500000],diQ] (* Harvey P. Dale, Aug 16 2011 *)
With[{max = 5*10^6}, Select[Flatten[Table[2^i*3^j, {j, 0, Log[6, max]}, {i, j, 2*j}]] // Sort, # <= max &]] (* Amiram Eldar, Mar 29 2025 *)

PARI

ns(n)= { local(f,p=1); f=factor(n); for(i=1, matsize(f)[1], p*=(1 + f[i, 1])^f[i, 2]); return(p) }
{ n=0; for (m=1, 10^9, if (ns(m)%m == 0, write("b064476.txt", n++, " ", m); if (n==100, break)) ) } \\ Harry J. Smith, Sep 15 2009

Python

from sympy import integer_log
def A064476(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(max(0,min((i<<1)+1,(x//3**i).bit_length())-i) for i in range(integer_log(x,3)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 26 2025

Formula

Sum_{n>=1} 1/a(n) = 72/55. - Amiram Eldar, Mar 29 2025

Extensions

More terms from Vladeta Jovovic, Oct 07 2001