

A069283


a(n) = 1 + number of odd divisors of n.


18



0, 0, 0, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 3, 0, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 3, 1, 0, 3, 1, 3, 2, 1, 1, 3, 1, 1, 3, 1, 1, 5, 1, 1, 1, 2, 2, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 1, 5, 0, 3, 3, 1, 1, 3, 3, 1, 2, 1, 1, 5, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 3, 1, 1, 5, 3, 1, 3, 1, 3, 1, 1, 2, 5, 2
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OFFSET

0,10


COMMENTS

Number of nontrivial ways to write n as sum of at least 2 consecutive integers. That is, we are not counting the trivial solution n=n. E.g., a(9)=2 because 9 = 4 + 5 and 9 = 2 + 3 + 4. a(8)=0 because there are no integers m and k such that m + (m+1) + ... + (m+k1) = 8 apart from k=1, m=8.  Alfred Heiligenbrunner, Jun 07 2004
Also number of sums of sequences of consecutive positive integers excluding sequences of length 1 (e.g., 9 = 2+3+4 or 4+5 so a(9)=2). (Useful for cribbage players.)  Michael Gilleland, Dec 29 2002
Let M be any positive integer. Then a(n) = number of proper divisors of M^n + 1 of the form M^k + 1.
This sequence gives the distinct differences of triangular numbers Ti giving n : n = Ti  Tj; none if n = 2^k. If factor a = n or a > (n/a  1)/2 : i = n/a + (a  1)/2; j = n/a  (a+1)/2. Else : i = n/2a + (2a  1)/2; j = n/2a  (2a  1)/2. Examples: 7 is prime; 7 = T4  T2 = (1 + 2 + 3 + 4)  (1 + 2) (a = 7; n/a = 1). The odd factors of 35 are 35, 7 and 5; 35 = T18  T16 (a = 35) = T8  T1 (a = 7) = T5  T7 (a = 5). 144 = T20  T11 (a = 9) = T49  T46 (a = 3).  M. Dauchez (mdzzdm(AT)yahoo.fr), Oct 31 2005
Also number of partitions of n into the form 1 + 2 + ...( k  1) + k + k + ... + k for some k >= 2. Example: a(9) = 2 because we have [2, 2, 2, 2, 1] and [3, 3, 2, 1].  Emeric Deutsch, Mar 04 2006
a(n) is the number of nontrivial runsum representations of n, and is also known as the politeness of n.  Ant King, Nov 20 2010
Also number of nonpowers of 2 dividing n, divided by the number of powers of 2 dividing n, n > 0.  Omar E. Pol, Aug 24 2019


REFERENCES

Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., AddisonWesley, 1994, see exercise 2.30 on p. 65.


LINKS



FORMULA

a(n) = 0 if and only if n = 2^k.
a(n) = 1 if and only if n = 2^k * p where k >= 0 and p is an odd prime.  Ant King, Nov 20 2010
G.f.: sum(k>=2, x^(k(k + 1)/2)/(1  x^k) ).  Emeric Deutsch, Mar 04 2006
If n = 2^k p1^b1 p2^b2 ... pr^br, then a(n) = (1 + b1)(1 + b2) ... (1 + br)  1.  Ant King, Nov 20 2010
Dirichlet g.f.: (zeta(s)*(11/2^s)  1)*zeta(s).  Geoffrey Critzer, Feb 15 2015
a(2*n) = a(n).
Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma + log(2)/2  3/2)*n, where gamma is Euler's constant (A001620).  Amiram Eldar, Dec 01 2023


EXAMPLE

a(14) = 1 because the divisors of 14 are 1, 2, 7, 14, and of these, two are odd, 1 and 7, and 1 + 2 = 1.
a(15) = 3 because the divisors of 15 are 1, 3, 5, 15, and of these, all four are odd, and 1 + 4 = 3.
a(16) = 0 because 16 has only one odd divisor, and 1 + 1 = 0.
Using Ant King's formula: a(90) = 5 as 90 = 2^1 * 3^2 * 5^1, so a(90) = (1 + 2) * (1 + 1)  1 = 5.  Giovanni Ciriani, Jan 12 2013
x^3 + x^5 + x^6 + x^7 + 2*x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + ...
a(120) = 3 as the odd divisors of 120 are the odd divisors of 15 as 120 = 15*2^3. 15 has 4 odd divisors so that gives a(120) = 4  1 = 3.  David A. Corneth, May 30 2020


MAPLE

g:=sum(x^(k*(k+1)/2)/(1x^k), k=2..20): gser:=series(g, x=0, 115): seq(coeff(gser, x, n), n=0..100); # Emeric Deutsch, Mar 04 2006


MATHEMATICA

g[n_] := Module[{dL = Divisors[2n], dP}, dP = Transpose[{dL, 2n/dL}]; Select[dP, ((1 < #[[1]] < #[[2]]) && (Mod[ #[[1]]  #[[2]], 2] == 1)) &] ]; Table[Length[g[n]], {n, 1, 100}]
Table[Length[Select[Divisors[k], OddQ[#] &]]  1, {k, 100}] (* Ant King, Nov 20 2010 *)
Join[{0}, Times @@@ (#[[All, 2]] & /@ Replace[FactorInteger[Range[2, 50]], {2, a_} > {2, 0}, Infinity] + 1)  1] (* Horst H. Manninger, Oct 30 2021 *)


PROG

(Haskell)
a069283 0 = 0
a069283 n = length $ tail $ a182469_row n
(PARI) {a(n) = if( n<1, 0, sumdiv( n, d, d%2)  1)} /* Michael Somos, Aug 07 2013 */
(Magma) [0] cat [1 + #[d:d in Divisors(n) IsOdd(d)]:n in [1..100]]; // Marius A. Burtea, Aug 24 2019
(Python)
from sympy import divisor_count
def A069283(n): return divisor_count(n>>(~n&n1).bit_length())1 if n else 0 # Chai Wah Wu, Jul 16 2022


CROSSREFS

Cf. A095808 (sums of ascending and descending consecutive integers).


KEYWORD

nonn,easy,changed


AUTHOR



EXTENSIONS



STATUS

approved



