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A069283
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a(n) = -1 + number of odd divisors of n.
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18
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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
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0,10
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COMMENTS
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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+k-1) = 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
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REFERENCES
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Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, see exercise 2.30 on p. 65.
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LINKS
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FORMULA
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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)*(1-1/2^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
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EXAMPLE
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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
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MAPLE
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g:=sum(x^(k*(k+1)/2)/(1-x^k), k=2..20): gser:=series(g, x=0, 115): seq(coeff(gser, x, n), n=0..100); # Emeric Deutsch, Mar 04 2006
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MATHEMATICA
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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 *)
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PROG
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(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&n-1).bit_length())-1 if n else 0 # Chai Wah Wu, Jul 16 2022
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CROSSREFS
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Cf. A095808 (sums of ascending and descending consecutive integers).
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KEYWORD
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nonn,easy,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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