A064478 -id:A064478 - 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

A064479 Complement of A064478.

Original entry on oeis.org

5, 7, 10, 11, 13, 15, 17, 19, 21, 22, 23, 25, 26, 28, 29, 31, 33, 34, 35, 37, 39, 40, 41, 43, 45, 46, 47, 49, 50, 51, 52, 53, 55, 57, 58, 59, 61, 63, 65, 66, 67, 69, 70, 71, 73, 75, 76, 77, 78, 79, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 97, 99, 100, 101, 103, 105

Offset: 0

N. J. A. Sloane, Oct 06 2001

Sequence includes all odd positive integers except powers of 3.

Cf. A064478, A007617.

More terms and comment from Matthew Conroy, Oct 17 2001

A003959 If n = Product p(k)^e(k) then a(n) = Product (p(k)+1)^e(k), a(1) = 1.

Original entry on oeis.org

1, 3, 4, 9, 6, 12, 8, 27, 16, 18, 12, 36, 14, 24, 24, 81, 18, 48, 20, 54, 32, 36, 24, 108, 36, 42, 64, 72, 30, 72, 32, 243, 48, 54, 48, 144, 38, 60, 56, 162, 42, 96, 44, 108, 96, 72, 48, 324, 64, 108, 72, 126, 54, 192, 72, 216, 80, 90, 60, 216, 62, 96, 128, 729, 84, 144, 68

Offset: 1

Marc LeBrun

Completely multiplicative.

Sum of divisors of n with multiplicity. If n = p^m, the number of ways to make p^k as a divisor of n is C(m,k); and sum(C(m,k)*p^k) = (p+1)^k. The rest follows because the function is multiplicative. - Franklin T. Adams-Watters, Jan 25 2010

Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 1000 terms from T. D. Noe)
Index to divisibility sequences

Apart from initial terms, same as A064478.
Cf. A003958, A027746, A036987, A061142, A163407.
Cf. A063441, A165824, A165825, A165826, A165827.
Cf. A165828, A165829, A165831, A167350, A166590.
Cf. A166642, A166643, A166644, A166645, A166646.
Cf. A166647, A166649, A166650, A166586, A165830.
Cf. A167351, A168065, A168066.

a003959 1 = 1
a003959 n = product $ map (+ 1) $ a027746_row n
-- Reinhard Zumkeller, Apr 09 2012

a:= n-> mul((i[1]+1)^i[2], i=ifactors(n)[2]):
seq(a(n), n=1..80);  # Alois P. Heinz, Sep 13 2017

a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((fi[[All, 1]]+1)^fi[[All, 2]])); a /@ Range[67] (* Jean-François Alcover, Apr 22 2011 *)

a(n)=if(n<1,0,direuler(p=2,n,1/(1-X-p*X))[n]) /* Ralf Stephan */

Multiplicative with a(p^e) = (p+1)^e. - David W. Wilson, Aug 01 2001
Sum_{n>0} a(n)/n^s = Product_{p prime} 1/(1-p^(-s)-p^(1-s)) (conjectured). - Ralf Stephan, Jul 07 2013
This follows from the absolute convergence of the sum (compare with a(n) = n^2) and the Euler product for completely multiplicative functions. Convergence occurs for at least Re(s)>3. - Thomas Anton, Jul 15 2021
Sum_{k=1..n} a(k) ~ c * n^2, where c = A065488/2 = 1/(2*A005596) = 1.3370563627850107544802059152227440187511993141988459926... - Vaclav Kotesovec, Jul 17 2021
From Thomas Scheuerle, Jul 19 2021: (Start)
a(n) = gcd(A166642(n), A166643(n)).
a(n) = A166642(n)/A061142(n).
a(n) = A166643(n)/A165824(n).
a(n) = A166644(n)/A165825(n).
a(n) = A166645(n)/A165826(n).
a(n) = A166646(n)/A165827(n).
a(n) = A166647(n)/A165828(n).
a(n) = A166649(n)/A165830(n).
a(n) = A166650(n)/A165831(n).
a(n) = A167351(n)/A166590(n). (End)
Dirichlet g.f.: zeta(s-1) * Product_{primes p} (1 + 1/(p^s - p - 1)). - Vaclav Kotesovec, Aug 22 2021

Definition reedited (with formula) by Daniel Forgues, Nov 17 2009

A327564 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)^(k_j - 1)).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 9, 4, 1, 1, 3, 1, 1, 1, 27, 1, 4, 1, 3, 1, 1, 1, 9, 6, 1, 16, 3, 1, 1, 1, 81, 1, 1, 1, 12, 1, 1, 1, 9, 1, 1, 1, 3, 4, 1, 1, 27, 8, 6, 1, 3, 1, 16, 1, 9, 1, 1, 1, 3, 1, 1, 4, 243, 1, 1, 1, 3, 1, 1, 1, 36, 1, 1, 6, 3, 1, 1, 1, 27, 64, 1, 1, 3, 1

Offset: 1

Ilya Gutkovskiy, Mar 03 2020

			a(12) = a(2^2 * 3) = (2 + 1)^(2 - 1) * (3 + 1)^(1 - 1) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..16383
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Index entries for sequences computed from exponents in factorization of n.
Index entries for sequences computed from indices in prime factorization.

Cf. A001221, A003557, A003958, A003959, A003968, A005117 (positions of 1's), A007947, A048250, A064478, A064549, A173557, A323363, A325126, A326297, A340368, A348038, A348039, A347960.

a[n_] := Times @@ ((#[[1]] + 1)^(#[[2]] - 1) & /@ FactorInteger[n]); Table[a[n], {n, 1, 85}]

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1]++; f[k,2]--); factorback(f); \\ Michel Marcus, Mar 03 2020

a(1) = 1; a(n) = -Sum_{d|n, dA001221(n/d) * A003557(n/d) * a(d).
a(n) = A003959(n) / A048250(n) = A003968(n) / A007947(n).
a(n) = A348038(n) * A348039(n) = A340368(n) / A173557(n). - Antti Karttunen, Oct 29 2021
Dirichlet g.f.: 1/(zeta(s-1) * Product_{p prime} (1 - 1/p^(s-1) - 1/p^s)). - Amiram Eldar, Dec 07 2023

A326297 If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)^(k_j - 1)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 4, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 6, 2, 4

Offset: 1

Ilya Gutkovskiy, Mar 03 2020

			a(98) = a(2 * 7^2) = (2 - 1)^(1 - 1) * (7 - 1)^(2 - 1) = 6.

Cf. A003557, A003958, A003959, A007947, A023900, A064478, A064549, A122132 (positions of 1's), A125131, A173557, A325126, A327564.

seq(mul((p-1)^(padic[ordp](n,p)-1), p in numtheory[factorset](n)), n =1..100); # Ridouane Oudra, Oct 29 2024

a[n_] := If[n == 1, 1, Times @@ ((#[[1]] - 1)^(#[[2]] - 1) & /@ FactorInteger[n])]; Table[a[n], {n, 1, 100}]

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1]--; f[k,2]--); factorback(f); \\ Michel Marcus, Mar 03 2020

from math import prod
from sympy import factorint
def a(n): return prod((p-1)**(e-1) for p, e in factorint(n).items())
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Aug 30 2021

a(n) = A003958(n) / abs(A023900(n)) = abs(A325126(n)) / A007947(n).
Dirichlet g.f.: Product_{p prime} (1 + 1/(p^s - p + 1)). - Amiram Eldar, Dec 07 2023
a(n) = A003958(n)/A173557(n). - Ridouane Oudra, Oct 29 2024

A304411 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)*k_j).

Original entry on oeis.org

1, 3, 4, 6, 6, 12, 8, 9, 8, 18, 12, 24, 14, 24, 24, 12, 18, 24, 20, 36, 32, 36, 24, 36, 12, 42, 12, 48, 30, 72, 32, 15, 48, 54, 48, 48, 38, 60, 56, 54, 42, 96, 44, 72, 48, 72, 48, 48, 16, 36, 72, 84, 54, 36, 72, 72, 80, 90, 60, 144, 62, 96, 64, 18, 84, 144, 68, 108, 96, 144, 72, 72

Offset: 1

Ilya Gutkovskiy, May 12 2018

			a(24) = a(2^3*3) = (2 + 1)*3 * (3 + 1)*1 = 36.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Ilya Gutkovskiy, Scatter plot of a(n) up to n=50000 .
Index entries for sequences computed from exponents in factorization of n.
Index entries for sequences computed from indices in prime factorization.

Cf. A000005, A000026, A000040, A000203, A003959, A001615, A003961, A005117, A005361, A007947, A008864, A045965, A048250, A064478, A081294 (numbers n such that a(n) is odd), A181797, A304407, A304408, A304409, A304412.

Cf. A072691, A330596.

a[n_] := Times @@ ((#[[1]] + 1) #[[2]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 72}]
Table[Total[Select[Divisors[n], SquareFreeQ]] DivisorSigma[0, n/Last[Select[Divisors[n], SquareFreeQ]]], {n, 72}]

a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); (p+1)*e)} \\ Andrew Howroyd, Jul 24 2018

a(n) = A005361(n)*A048250(n) = A000005(n/A007947(n))*A000203(A007947(n)).
a(p^k) = (p + 1)*k where p is a prime and k > 0.
a(n) = Product_{p|n} (p + 1) if n is a squarefree (A005117).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^2/12) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A072691 * A330596 = 0.6156455744... . - Amiram Eldar, Nov 30 2022

A132705 For an integer n with prime factorization (p_1)(p_2)(p_3)* ... (p_k), a(n) = (p_1+2)(p_2+2)(p_3+2) ... *(p_k+2).

Original entry on oeis.org

2, 3, 4, 5, 16, 7, 20, 9, 64, 25, 28, 13, 80, 15, 36, 35, 256, 19, 100, 21, 112, 45, 52, 25, 320, 49, 60, 125, 144, 31, 140, 33, 128, 65, 76, 63, 400, 49, 84, 75, 448, 43, 180, 45, 208, 175, 100, 49, 1280, 81, 196, 95, 240, 55, 500, 91, 576, 105, 124, 60

Offset: 0

Jonathan Vos Post, Nov 16 2007

a(0)=2 and a(1)=3 by convention. For an integer n with prime factorization prime(i_1)*prime(i_2)*prime(i_3)* ... *prime(i_k), a(n) = A052147(i_1)*A052147(i_2)*A052147(i_3)* ... *A052147(i_k). This sequence is to p+2 as A064478 is to p+1 for primes p.

If a(1) were 1 rather than 3, the sequence would be completely multiplicative with a(p) = p + 2. - Charles R Greathouse IV, Sep 02 2009

Cf. A000040, A003958, A003959, A064476, A064479, A052147, A064478.

from math import prod
from sympy import factorint
def A132705(n): return prod((p+2)**e for p,e in factorint(n).items()) if n!=1 else 3 # Chai Wah Wu, Mar 26 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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A064479 Complement of A064478.

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A003959 If n = Product p(k)^e(k) then a(n) = Product (p(k)+1)^e(k), a(1) = 1.

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A327564 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)^(k_j - 1)).

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A326297 If n = Product (p_j^k_j) then a(n) = Product ((p_j - 1)^(k_j - 1)).

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A304411 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)*k_j).

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A132705 For an integer n with prime factorization (p_1)*(p_2)*(p_3)* ... *(p_k), a(n) = (p_1+2)*(p_2+2)*(p_3+2)* ... *(p_k+2).

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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.

A132705 For an integer n with prime factorization (p_1)(p_2)(p_3)* ... (p_k), a(n) = (p_1+2)(p_2+2)(p_3+2) ... *(p_k+2).