%I A069283
%S A069283 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,
%T A069283 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,
%U A069283 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 A069283 a(n) = -1 + number of odd divisors of n.
%C A069283 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 (alfred.heiligenbrunner(AT)gmx.at), 
               Jun 07 2004
%C A069283 Comment from Michael Gilleland, Dec 29, 2002: 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.)
%C A069283 Let M be any positive integer. Then a(n) = number of proper divisors 
               of M^n+1 of the form M^k+1.
%C A069283 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 A069283 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 (deutsch(AT)duke.poly.edu), Mar 04 2006
%D A069283 Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed. (Addison-Wesley, 
               1994), see exercise 2.30 on p. 65.
%H A069283 A. Heiligenbrunner, <a href="http://www.heiligenbrunner.at/main/ahsummen.htm">
               Sum of adjacent numbers (in German)</a>.
%F A069283 a(n) = 0 iff n=2^k.
%F A069283 G.f.=sum(x^(k(k+1)/2)/(1-x^k), k=2..infinity). - Emeric Deutsch (deutsch(AT)duke.poly.edu), 
               Mar 04 2006
%e A069283 a(21)=3 as 10^k+1 is a divisor of 10^21+1 for k in {1,3,7}:
%e A069283 10^21+1 = 7 * 7 * 11 * 13 * 127 * 2689 * 459691 * 909091] =
%e A069283 =10000001*99999990000001=(11*909091)*(7*7*13*127*2689*459691)
%e A069283 =1001*999000999000999001=(7*11*13)*(7*127*2689*459691*909091)
%e A069283 =11*90909090909090909091=(11)*(7*7*13*127*2689*459691*909091)
%p A069283 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 (deutsch(AT)duke.poly.edu), Mar 
               04 2006
%t A069283 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}]
%Y A069283 Cf. A001227, A062397, A057934. Equals A001227(n) - 1.
%Y A069283 Sequence in context: A062093 A046214 A115413 this_sequence A033630 A101446 
               A062760
%Y A069283 Adjacent sequences: A069280 A069281 A069282 this_sequence A069284 A069285 
               A069286
%K A069283 nonn
%O A069283 0,10
%A A069283 Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Mar 13 2002
%E A069283 Edited by Vladeta Jovovic (vladeta(AT)eunet.rs), Mar 25 2002