

A245579


Number of odd divisors of n multiplied by n.


15



1, 2, 6, 4, 10, 12, 14, 8, 27, 20, 22, 24, 26, 28, 60, 16, 34, 54, 38, 40, 84, 44, 46, 48, 75, 52, 108, 56, 58, 120, 62, 32, 132, 68, 140, 108, 74, 76, 156, 80, 82, 168, 86, 88, 270, 92, 94, 96, 147, 150, 204, 104, 106, 216, 220, 112, 228, 116, 118, 240, 122
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OFFSET

1,2


COMMENTS

Sum of all parts of all partitions of n into consecutive parts.  Omar E. Pol, Apr 23 2017
Sum of all parts of all partitions of n into an odd number of equal parts.  Omar E. Pol, Jun 05 2017
Row sums of A299765.  Omar E. Pol, Jul 23 2018
Row sums of A328362.  Omar E. Pol, Oct 20 2019
Row sums of A285891 and of A328365.  Omar E. Pol, Nov 04 2019
Number of partitions of n into consecutive parts, multiplied by n. Also, number of partitions of n into an odd number of equal parts, multiplied by n.  Omar E. Pol, Nov 05 2019


LINKS

Jens Kruse Andersen, Table of n, a(n) for n = 1..10000


FORMULA

a(n) is multiplicative with a(2^e) = 2^e, a(p^e) = p^e * (e+1) if p>2.
a(n) = n * A001227(n).
G.f.: Sum_{k>0 odd} k * x^k / (1  x^k)^2.
a(n) = n*A000005(n)/A001511(n) = A038040(n)/A001511(n).  Omar E. Pol, Apr 24 2018


EXAMPLE

G.f. = x + 2*x^2 + 6*x^3 + 4*x^4 + 10*x^5 + 12*x^6 + 14*x^7 + 8*x^8 + ...
For n = 10 there are two odd divisors of 10: 1 and 5, so a(10) = 2*10 = 20. On the other hand, for n = 10 there are two partitions of 10 into consecutive integers: [10] and [4, 3, 2, 1], and the sum of all parts of these partitions is 10 + 4 + 3 + 2 + 1 = 20, so a(10) = 20.  Omar E. Pol, Apr 23 2017


MAPLE

seq(n*numtheory:tau(n/2^padic:ordp(n, 2)), n=1..100); # Robert Israel, Apr 26 2017


MATHEMATICA

a[ n_] := If[ n < 1, 0, n Sum[ Mod[d, 2], {d, Divisors @ n}]];
(* Second program: *)
Table[n DivisorSum[n, 1 &, OddQ], {n, 61}] (* Michael De Vlieger, Apr 24 2017 *)


PROG

(PARI) {a(n) = if( n<1, 0, n * sumdiv(n, d, d%2))};
(PARI) {a(n) = if( n<0, 0, polcoeff( sum(k=1, n, if( k%2, k * x^k / (1  x^k)^2), x * O(x^n)), n))};
(PARI) {a(n) = if( n<1, 0, n * numdiv(n / 2^valuation(n, 2)))} \\ Fast when n has many divisors. Jens Kruse Andersen, Jul 26 2014
(Python)
from sympy import divisors
def a(n): return n*len(list(filter(lambda i: i%2==1, divisors(n)))) # Indranil Ghosh, Apr 24 2017


CROSSREFS

Cf. A000005, A001227, A001511, A038040, A285891, A299765, A328362, A328365.
Sequence in context: A119018 A264647 A094748 * A245788 A065879 A065880
Adjacent sequences: A245576 A245577 A245578 * A245580 A245581 A245582


KEYWORD

nonn,mult


AUTHOR

Michael Somos, Jul 26 2014


STATUS

approved



