|
%I #125 Dec 01 2023 06:14:03
%S 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,
%T 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,
%U 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
%N a(n) = -1 + number of odd divisors of n.
%C 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
%C 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
%C Let M be any positive integer. Then a(n) = number of proper divisors of M^n + 1 of the form M^k + 1.
%C 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
%C 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
%C 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
%C 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
%C a(n) only depends on the prime signature of A000265(n). - _David A. Corneth_, May 30 2020, corrected by _Charles R Greathouse IV_, Oct 31 2021
%D Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, see exercise 2.30 on p. 65.
%H Reinhard Zumkeller, <a href="/A069283/b069283.txt">Table of n, a(n) for n = 0..10000</a>
%H Tom M. Apostol, <a href="http://www.jstor.org/stable/3620570">Sums of Consecutive Positive Integers</a>, The Mathematical Gazette, Vol. 87, No. 508, (March 2003), pp. 98-101.
%H Alfred Heiligenbrunner, <a href="http://ah9.at/ahsummen.htm">Sum of adjacent numbers</a> (in German).
%H Henri Picciotto, <a href="http://www.mathedpage.org/teachers/staircases.pdf">Staircases</a>.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Polite_number">Polite Number</a>.
%F a(n) = 0 if and only if n = 2^k.
%F a(n) = A001227(n)-1.
%F 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
%F G.f.: sum(k>=2, x^(k(k + 1)/2)/(1 - x^k) ). - _Emeric Deutsch_, Mar 04 2006
%F 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
%F Dirichlet g.f.: (zeta(s)*(1-1/2^s) - 1)*zeta(s). - _Geoffrey Critzer_, Feb 15 2015
%F a(n) = (A000005(n) - A001511(n))/A001511(n) = A326987(n)/A001511(n), with n > 0 in both formulas. - _Omar E. Pol_, Aug 24 2019
%F G.f.: Sum_{k>=1} x^(3*k) / (1 - x^(2*k)). - _Ilya Gutkovskiy_, May 30 2020
%F From _David A. Corneth_, May 30 2020: (Start)
%F a(2*n) = a(n).
%F a(n) = A001227(A000265(n)) - 1. (End)
%F 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
%e 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.
%e a(15) = 3 because the divisors of 15 are 1, 3, 5, 15, and of these, all four are odd, and -1 + 4 = 3.
%e a(16) = 0 because 16 has only one odd divisor, and -1 + 1 = 0.
%e 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
%e x^3 + x^5 + x^6 + x^7 + 2*x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + ...
%e 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
%p 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
%p A069283 := proc(n)
%p A001227(n)-1 ;
%p end proc: # _R. J. Mathar_, Jun 18 2015
%t 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}]
%t Table[Length[Select[Divisors[k], OddQ[#] &]] - 1, {k, 100}] (* _Ant King_, Nov 20 2010 *)
%t Join[{0}, Times @@@ (#[[All, 2]] & /@ Replace[FactorInteger[Range[2, 50]], {2, a_} -> {2, 0}, Infinity] + 1) - 1] (* _Horst H. Manninger_, Oct 30 2021 *)
%o (Haskell)
%o a069283 0 = 0
%o a069283 n = length $ tail $ a182469_row n
%o -- _Reinhard Zumkeller_, May 01 2012
%o (PARI) {a(n) = if( n<1, 0, sumdiv( n, d, d%2) - 1)} /* _Michael Somos_, Aug 07 2013 */
%o (PARI) a(n) = numdiv(n >> valuation(n, 2)) - 1 \\ _David A. Corneth_, May 30 2020
%o (Magma) [0] cat [-1 + #[d:d in Divisors(n)| IsOdd(d)]:n in [1..100]]; // _Marius A. Burtea_, Aug 24 2019
%o (Python)
%o from sympy import divisor_count
%o def A069283(n): return divisor_count(n>>(~n&n-1).bit_length())-1 if n else 0 # _Chai Wah Wu_, Jul 16 2022
%Y Cf. A000265, A001227, A001620, A062397, A057934, A138591, A182469.
%Y Cf. A095808 (sums of ascending and descending consecutive integers).
%K nonn,easy
%O 0,10
%A _Reinhard Zumkeller_, Mar 13 2002
%E Edited by _Vladeta Jovovic_, Mar 25 2002
|
