Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
A058377
Number of solutions to 1 +- 2 +- 3 +- ... +- n = 0.
21
0, 0, 1, 1, 0, 0, 4, 7, 0, 0, 35, 62, 0, 0, 361, 657, 0, 0, 4110, 7636, 0, 0, 49910, 93846, 0, 0, 632602, 1199892, 0, 0, 8273610, 15796439, 0, 0, 110826888, 212681976, 0, 0, 1512776590, 2915017360, 0, 0, 20965992017, 40536016030, 0, 0, 294245741167
OFFSET
1,7
COMMENTS
Consider the set { 1,2,3,...,n }. Sequence gives number of ways this set can be partitioned into 2 subsets with equal sums. For example, when n = 7, { 1,2,3,4,5,6,7} can be partitioned in 4 ways: {1,6,7} {2,3,4,5}; {2,5,7} {1,3,4,6}; {3,4,7} {1,2,5,6} and {1,2,4,7} {3,5,6}. - sorin (yamba_ro(AT)yahoo.com), Mar 24 2007
The "equal sums" of Sorin's comment are the positive terms of A074378 (Even triangular numbers halved). In the current sequence a(n) <> 0 iff n is the positive index (A014601) of an even triangular number (A014494). - Rick L. Shepherd, Feb 09 2010
a(n) is the number of partitions of n(n-3)/4 into distinct parts not exceeding n-1. - Alon Amit, Oct 18 2017
a(n) is the coefficient of x^(n*(n+1)/4-1) of Product_{k=2..n} (1+x^k). - Jianing Song, Nov 19 2021
LINKS
Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..3342 (terms < 10^1000, first 1000 terms from Alois P. Heinz)
Larry Glasser, A formula for A058377, Jul 29 2019
FORMULA
a(n) is half the coefficient of q^0 in product('(q^(-k)+q^k)', 'k'=1..n) for n >= 1. - Floor van Lamoen, Oct 10 2005
a(4n+1) = a(4n+2) = 0. - Michael Somos, Apr 15 2007
a(n) = [x^n] Product_{k=1..n-1} (x^k + 1/x^k). - Ilya Gutkovskiy, Feb 01 2024
EXAMPLE
1+2-3=0, so a(3)=1;
1-2-3+4=0, so a(4)=1;
1+2-3+4-5-6+7=0, 1+2-3-4+5+6-7=0, 1-2+3+4-5+6-7=0, 1-2-3-4-5+6+7=0, so a(7)=4.
MAPLE
b:= proc(n, i) option remember; local m; m:= i*(i+1)/2;
`if`(n>m, 0, `if`(n=m, 1, b(abs(n-i), i-1) +b(n+i, i-1)))
end:
a:= n-> `if`(irem(n-1, 4)<2, 0, b(n, n-1)):
seq(a(n), n=1..60); # Alois P. Heinz, Oct 30 2011
MATHEMATICA
f[n_, s_] := f[n, s] = Which[n == 0, If[s == 0, 1, 0], Abs[s] > (n*(n + 1))/2, 0, True, f[n - 1, s - n] + f[n - 1, s + n]]; Table[ f[n, 0]/2, {n, 1, 50}]
PROG
(PARI) list(n) = my(poly=vector(n), v=vector(n)); poly[1]=1; for(k=2, n, poly[k]=poly[k-1]*(1+'x^k)); for(k=1, n, if(k%4==1||k%4==2, v[k]=0, v[k]=polcoeff(poly[k], k*(k+1)/4-1))); v \\ Jianing Song, Nov 19 2021
KEYWORD
nonn
AUTHOR
Naohiro Nomoto, Dec 19 2000
EXTENSIONS
More terms from Sascha Kurz, Mar 25 2002
Edited and extended by Robert G. Wilson v, Oct 24 2002
STATUS
approved