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 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 (list; graph; refs; listen; history; text; internal format)
 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+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 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 REFERENCES Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, see exercise 2.30 on p. 65. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 Tom M. Apostol, Sums of Consecutive Positive Integers, The Mathematical Gazette, Vol. 87, No. 508, (March 2003), pp. 98-101. Alfred Heiligenbrunner, Sum of adjacent numbers (in German). Henri Picciotto, Staircases. Wikipedia, Polite Number. FORMULA a(n) = 0 if and only if n = 2^k. a(n) = A001227(n)-1. 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(n) = (A000005(n) - A001511(n))/A001511(n) = A326987(n)/A001511(n), with n > 0 in both formulas. - Omar E. Pol, Aug 24 2019 G.f.: Sum_{k>=1} x^(3*k) / (1 - x^(2*k)). - Ilya Gutkovskiy, May 30 2020 From David A. Corneth, May 30 2020: (Start) a(2*n) = a(n). a(n) = A001227(A000265(n)) - 1. (End) 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)/(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 A069283 := proc(n) A001227(n)-1 ; end proc: # R. J. Mathar, Jun 18 2015 MATHEMATICA g[n_] := Module[{dL = Divisors[2n], dP}, dP = Transpose[{dL, 2n/dL}]; Select[dP, ((1 < #[] < #[]) && (Mod[ #[] - #[], 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 -- Reinhard Zumkeller, May 01 2012 (PARI) {a(n) = if( n<1, 0, sumdiv( n, d, d%2) - 1)} /* Michael Somos, Aug 07 2013 */ (PARI) a(n) = numdiv(n >> valuation(n, 2)) - 1 \\ David A. Corneth, May 30 2020 (Magma)  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 CROSSREFS Cf. A000265, A001227, A001620, A062397, A057934, A138591, A182469. Cf. A095808 (sums of ascending and descending consecutive integers). Sequence in context: A115413 A292435 A319094 * A319430 A285337 A328457 Adjacent sequences: A069280 A069281 A069282 * A069284 A069285 A069286 KEYWORD nonn,easy,changed AUTHOR Reinhard Zumkeller, Mar 13 2002 EXTENSIONS Edited by Vladeta Jovovic, Mar 25 2002 STATUS approved

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Last modified December 5 13:14 EST 2023. Contains 367591 sequences. (Running on oeis4.)