A064514 -id:A064514 - cp's OEIS frontend

A064476 For an integer k with prime factorization p_1p_2p_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

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

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

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

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

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

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);

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

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 *)

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

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

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

More terms from Vladeta Jovovic, Oct 07 2001

A064515 Write A064476(n) = 2^i(n)*3^j(n); sequence gives values of j(n).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 3, 4, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 6, 5, 7, 6, 7, 6, 7, 6, 8, 7, 6, 8, 7, 8, 7, 9, 8, 7, 9, 8, 7, 9, 8, 10, 9, 8, 10, 9, 8, 10, 9, 11, 8, 10, 9, 11, 10, 9, 11, 10, 12, 9, 11, 10, 12, 9, 11, 10, 12, 11, 13, 10, 12, 11, 13, 10, 12, 11, 13, 10, 12, 14, 11, 13, 12

Offset: 1

Vladeta Jovovic, Oct 07 2001, Oct 07 2001

Cf. A064476, A064514.

function a064515(maxarg: integer); var j: integer; ar: array; begin ar := p2p3(maxarg); for j := 0 to maxarg - 1 do write(ar[j][2]," "); end; end; a064515(95);  (* For definition of function p2p3 see A064476. *)

from sympy import integer_log
def A064515(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 integer_log((m:=bisection(f,n,n))>>(m-1&~m).bit_length(),3)[0] # Chai Wah Wu, Mar 26 2025

More terms from Klaus Brockhaus, Oct 12 2001

A064518 For an integer n with prime factorization p_1p_2p_3* ... p_m let n = (p_1+1)(p_2+1)(p_3+1)* ... (p_m+1); sequence gives n such that n* is divisible by n, ordered by increasing value of n.

Original entry on oeis.org

1, 12, 36, 144, 432, 1296, 1728, 5184, 15552, 20736, 46656, 62208, 186624, 248832, 559872, 746496, 1679616, 2239488, 2985984, 6718464, 8957952, 20155392, 26873856, 60466176, 35831808, 80621568, 107495424, 241864704, 322486272

Offset: 1

Vladeta Jovovic, Oct 07 2001

nonn

It is not difficult to show that these numbers have the form a(n) = 3^i*4^j with j <= i <= 2j.

Sean A. Irvine, Table of n, a(n) for n = 1..1000 (terms 1..50 from Harry J. Smith)

Every term is also a term of A064476.

Cf. A064514, A064515.

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, s=ns(m); if (s%m == 0, write("b064518.txt", n++, " ", s); if (n==50, break)) ) } \\ Harry J. Smith, Sep 17 2009

a(n) = 3^A064514(n) * 4^A064515(n). - Chai Wah Wu, Mar 26 2025

Title clarified by Sean A. Irvine, Jul 15 2023

A380574 For an integer k with prime factorization p_1p_2p_3* ... p_m let k = (p_1+1)(p_2+1)(p_3+1)* ... (p_m+1); sequence gives k such that k* is divisible by k.

Original entry on oeis.org

1, 12, 36, 144, 432, 1296, 1728, 5184, 15552, 20736, 46656, 62208, 186624, 248832, 559872, 746496, 1679616, 2239488, 2985984, 6718464, 8957952, 20155392, 26873856, 35831808, 60466176, 80621568, 107495424, 241864704, 322486272, 429981696, 725594112, 967458816

Offset: 1

Chai Wah Wu, Mar 26 2025

Terms of A064518 in increasing order.
Numbers of the form 3^i*4^j with j <= i <= 2j.
Subsequence of A064476.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A064518, A064514, A064515, A064476.

With[{max = 10^9}, Select[Flatten[Table[3^i*4^j, {j, 0, Log[12, max]}, {i, j, 2*j}]] // Sort, # <= max &]] (* Amiram Eldar, Mar 29 2025 *)

from sympy import integer_log
def A380574(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((j:=i<<1),integer_log(x>>j,3)[0])-i+1) for i in range(x.bit_length()+1>>1))
    return bisection(f,n,n)

Sum_{n>=1} 1/a(n) = 432/385. - Amiram Eldar, Mar 29 2025

cp's OEIS Frontend

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.

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A064515 Write A064476(n) = 2^i(n)*3^j(n); sequence gives values of j(n).

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A064518 For an integer n with prime factorization p_1*p_2*p_3* ... *p_m let n* = (p_1+1)*(p_2+1)*(p_3+1)* ... *(p_m+1); sequence gives n* such that n* is divisible by n, ordered by increasing value of n.

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Formula

Extensions

A380574 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); sequence gives k* such that k* is divisible by k.

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Comments

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Programs

Mathematica

Python

Formula

A064476 For an integer k with prime factorization p_1p_2p_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.

A064518 For an integer n with prime factorization p_1p_2p_3* ... p_m let n = (p_1+1)(p_2+1)(p_3+1)* ... (p_m+1); sequence gives n such that n* is divisible by n, ordered by increasing value of n.

A380574 For an integer k with prime factorization p_1p_2p_3* ... p_m let k = (p_1+1)(p_2+1)(p_3+1)* ... (p_m+1); sequence gives k such that k* is divisible by k.