This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(1) = 1 because A204830(1) = 120, divisors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 and 1+2+3+4+5+6+8+10+12+15+24+30 = 20+40+60 = 120 = sigma(120)/3. Of course also 1+2+3+5+8+12+15+20+24+30 = 4+6+10+40+60 = 120 = sigma(120)/3 but these three subsets share {120} with the previous ones and therefore they are not disjoint.
a(13) = 2 because A204830(13) = 780, divisors of 780 are 1, 2, 3, 4, 5, 6, 10, 12, 13, 15, 20, 26, 30, 39, 52, 60, 65, 78, 130, 156, 195, 260, 390, 780: 1+3+52+78+260+390 = 2+5+6+10+12+13+15+20+26+30+39+60+65+130+156+195 = 4+780 = sigma(780)/3 and 5+6+10+12+13+15+26+39+52+60+65+130+156+195 = 2+4+20+30+78+260+390 = 1+3+780 = sigma(780)/3.
a(3) = 5 because A204830(3) = 240, divisors of 240 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240 and sigma(240)/3 = 248: it is easy to see that the total number of three disjoint subset is 5 because the only subsets containing 240 that sum to 248 are {1,2,5,240}, {1,3,4,240}, {2,6,240}, {3,5,240} and {8,240}.
with(numtheory): with(combstruct): P:=proc(q,h) local a,b,c,d,f,k,n,r; for n from 1 to q do a:=sigma(n); b:=op(divisors(n)); if a mod h=0 and a>=h*n then k:=0; c:=1/h*a-n; r:=select(m->m<=c,[b]); f:=iterstructs(Combination(r)); while not finished(f) do if c=add(d,d=nextstruct(f)) then k:=k+1; fi; od; lprint(n,k); fi; od; end: P(10^4,3);
Given n = 48, we can partition the divisors thus: 1 + 3 + 4 + 6 + 8 + 16 + 24 = 2 + 12 + 48, therefore 48 is a term (A083206(48) = 5).
From _David A. Corneth_, Dec 04 2024: (Start)
30 is in the sequence. sigma(30) = 72. So we look for distinct divisors of 30 that sum to 72/2 = 36. That set or its complement contains 30. The other divisors in that set containing 30 sum to 36 - 30 = 6. So we look for some distinct proper divisors of 30 that sum to 6. That is from the divisors of {1, 2, 3, 5, 6, 10, 15}. It turns out that both 1+2+3 and 6 satisfy this condition. So 36 is in the sequence.
25 is not in the sequence as sigma(25) = 31 which is odd so the sum of two equal integers cannot be the sum of divisors of 25.
33 is not in the sequence as sigma(33) = 48 < 2*33. So is impossible to have a partition of the set of divisors into two disjoint set the sum of each of them sums to 48/2 = 24 as one of them contains 33 > 24 and any other divisors are nonnegative. (End)
a083207 n = a083207_list !! (n-1) a083207_list = filter (z 0 0 . a027750_row) $ [1..] where z u v [] = u == v z u v (p:ps) = z (u + p) v ps || z u (v + p) ps -- Reinhard Zumkeller, Apr 18 2013
with(numtheory): with(combstruct): is_A083207 := proc(n) local S, R, Found, Comb, a, s; s := sigma(n); if not(modp(s, 2) = 0 and n * 2 <= s) then return false fi; S := s / 2 - n; R := select(m -> m <= S, divisors(n)); Found := false; Comb := iterstructs(Combination(R)): while not finished(Comb) and not Found do Found := add(a, a = nextstruct(Comb)) = S od; Found end: A083207_list := upto -> select(is_A083207, [$1..upto]): A083207_list(272); # Peter Luschny, Dec 14 2009, updated Aug 15 2014
ZumkellerQ[n_] := Module[{d=Divisors[n], t, ds, x}, ds = Plus@@d; If[Mod[ds, 2] > 0, False, t = CoefficientList[Product[1 + x^i, {i, d}], x]; t[[1 + ds/2]] > 0]]; Select[Range[1000], ZumkellerQ] (* T. D. Noe, Mar 31 2010 *)
znQ[n_]:=Length[Select[{#,Complement[Divisors[n],#]}&/@Most[Rest[ Subsets[ Divisors[ n]]]],Total[#[[1]]]==Total[#[[2]]]&]]>0; Select[Range[300],znQ] (* Harvey P. Dale, Dec 26 2022 *)
part(n,v)=if(n<1, return(n==0)); forstep(i=#v,2,-1,if(part(n-v[i],v[1..i-1]), return(1))); n==v[1] is(n)=my(d=divisors(n),s=sum(i=1,#d,d[i])); s%2==0 && part(s/2-n,d[1..#d-1]) \\ Charles R Greathouse IV, Mar 09 2014
\\ See Corneth link
from sympy import divisors
from sympy.combinatorics.subsets import Subset
for n in range(1,10**3):
d = divisors(n)
s = sum(d)
if not s % 2 and max(d) <= s/2:
for x in range(1,2**len(d)):
if sum(Subset.unrank_binary(x,d).subset) == s/2:
print(n,end=', ')
break
# Chai Wah Wu, Aug 13 2014
from sympy import divisors import numpy as np A083207 = [] for n in range(2,10**3): d = divisors(n) s = sum(d) if not s % 2 and 2*n <= s: d.remove(n) s2, ld = int(s/2-n), len(d) z = np.zeros((ld+1,s2+1),dtype=int) for i in range(1,ld+1): y = min(d[i-1],s2+1) z[i,range(y)] = z[i-1,range(y)] z[i,range(y,s2+1)] = np.maximum(z[i-1,range(y,s2+1)],z[i-1,range(0,s2+1-y)]+y) if z[i,s2] == s2: A083207.append(n) break # Chai Wah Wu, Aug 19 2014
def is_Zumkeller(n):
s = sigma(n)
if not (2.divides(s) and n*2 <= s): return False
S = s // 2 - n
R = (m for m in divisors(n) if m <= S)
return any(sum(c) == S for c in Combinations(R))
A083207_list = lambda lim: [n for n in (1..lim) if is_Zumkeller(n)]
print(A083207_list(272)) # Peter Luschny, Sep 03 2018
sigma(49) = sigma(8) + sigma(41) that is 57 = 15 + 42. sigma(93) = sigma(31) + sigma(62) that is 128 = 32 + 96. In more than one way: sigma(117) = sigma(41) + sigma(76) = sigma(52) + sigma(65) = sigma(56) + sigma(61) that is 182 = 42 + 140 = 98 + 84 = 120 + 62.
a211223 n = a211223_list !! (n-1) a211223_list = map (+ 1) $ findIndices (> 0) a211225_list -- Reinhard Zumkeller, Jan 06 2013
with(numtheory); A211223:=proc(q) local i,n; for n from 1 to q do for i from 1 to trunc(n/2) do if sigma(i)+sigma(n-i)=sigma(n) then print(n); break; fi; od; od; end: A211223(10000);
sigmaPartitionQ[n_] := With[{s = DivisorSigma[1, n], ip = IntegerPartitions[ n, {2}]}, MemberQ[ip, {x_, y_} /; s == DivisorSigma[ 1, x] + DivisorSigma[ 1, y]]]; Select[Range[160], sigmaPartitionQ] (* Jean-François Alcover, Aug 19 2013 *)
is(n)=my(t=sigma(n));for(i=1,n\2,if(sigma(i)+sigma(n-i)==t, return(1))) \\ Charles R Greathouse IV, May 04 2012
a(3)=1 because sigma(3)=sigma(1)+sigma(2)=4; a(32)=2 because sigma(32)=sigma(4)+sigma(28)=sigma(14)+sigma(18)=63; a(117)=3 because sigma(117)=sigma(41)+sigma(76)=sigma(52)+sigma(65)=sigma(56)+sigma(61)=182; etc.
a211225 n = length $ filter (== a000203 n) $ zipWith (+) us' vs where (us,vs@(v:_)) = splitAt (fromInteger $ (n - 1) `div` 2) a000203_list us' = if even n then v : reverse us else reverse us -- Reinhard Zumkeller, Jan 06 2013
with(numtheory); A211225:=proc(q) local b,i,n; for n from 1 to q do b:=0; for i from 1 to trunc(n/2) do if sigma(i)+sigma(n-i)=sigma(n) then b:=b+1; fi; od; print(b) od; end: A211225(1000);
a[n_] := With[{s = DivisorSigma[1, n]}, Sum[Boole[s == DivisorSigma[1, x] + DivisorSigma[1, n-x]], {x, 1, Quotient[n, 2]}]];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, May 04 2023 *)
a(n)=my(t=sigma(n)); sum(i=1, n\2, sigma(i)+sigma(n-i)==t) \\ Charles R Greathouse IV, May 07 2012
Number 27720 is in the sequence because sigma(27720)/4 = 28080 = 360 + 27720 = 20 + 60 + 280 + 2310 + 4620 + 6930 + 13860 = 9 + 30 + 420 + 1540 + 1980 + 2772 + 3080 + 3465 + 5544 + 9240 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 10 + 11 + 12 + 14 + 15 + 18 + 21 + 22 + 24 + 28 + 33 + 35 + 36 + 40 + 42 + 44 + 45 + 55 + 56 + 63 + 66 + 70+ 72 + 77 + 84 + 88 + 90 + 99 + 105 + 110 + 120 + 126 + 132 + 140 + 154 + 165 + 168 + 180 + 198 + 210 + 220 + 231 + 252 + 264+ 308 + 315 + 330 + 385 + 396 + 440 + 462 + 495 + 504 + 616 + 630 + 660 + 693 + 770 + 792 + 840 + 924 + 990 + 1155 + 1260 + 1320 + 1386 + 1848 + 2520 + 3960 (summands are all divisors of 27720).
with(numtheory);with(combstruct); A204831:=proc(i) local S,R,Stop,Comb,c,d,k,m,n,s; for n from 1 to i do s:=sigma(n); c:=op(divisors(n)); if (modp(s,4)=0 and 4*n<=s) then S:=1/4*s-n; R:=select(m->m<=S,[c]); Stop:=false; Comb:=iterstructs(Combination(R)); while not (finished(Comb) or Stop) do Stop:=add(d,d=nextstruct(Comb))=S; od; if Stop then print(n); fi; fi; od; end: A204831(100000); # Paolo P. Lava, Jan 24 2012
a(7)=933; 933 has 7 partitions of two numbers, x and y, for which sigma(933) = sigma(x) + sigma(y) = 1248. In fact: sigma(311) + sigma(622) = 312 + 936 = 1248; sigma(322) + sigma(611) = 576 + 672 = 1248; sigma(370) + sigma(563) = 684 + 564 = 1248; sigma(391) + sigma(542) = 432 + 816 = 1248; sigma(398) + sigma(535) = 600 + 648 = 1248; sigma(407) + sigma(526) = 456 + 792 = 1248; sigma(442) + sigma(491) = 756 + 492 = 1248; Furthermore 933 is the minimum number to have this property.
with(numtheory); A211224:=proc(q) local a,b,i,j,n,v; v:=array(1..10000); for n from 1 to 10000 do v[n]:=0; od; a:=0; for n from 1 to q do b:=0; for i from 1 to trunc(n/2) do if sigma(i)+sigma(n-i)=sigma(n) then b:=b+1; fi; od; if b=a+1 then a:=b; print(n); j:=1; while v[b+j]>0 do a:=b+j; print(v[b+j]); j:=j+1; od; else if b>a+1 then if v[b]=0 then v[b]:=n; fi; fi; fi; od; end: A211224(1000);
ct(n)=my(t=sigma(n));sum(i=1,n\2,sigma(i)+sigma(n-i)==t) v=vector(100);for(n=1,1e5,t=ct(n);if(t&&t<=#v&&!v[t],v[t]=n));v \\ Charles R Greathouse IV, May 04 2012
sigma*(127)=sigma*(45)+sigma*(82) that is 212=86+126. In more than one way: sigma*(133)=sigma*(50)+sigma*(83)=sigma*(52)+sigma*(81) that is 204=80+124=94+110.
with(numtheory); A210732:=proc(q) local a,b,c,i,j,k,n; for n from 3 to q do a:=0; for k from 2 to n-1 do if abs((n mod k)-k/2)<1 then a:=a+k; fi; od; for i from 1 to trunc(n/2) do b:=0; c:=0; for k from 2 to i-1 do if abs((i mod k)-k/2)<1 then b:=b+k; fi; od; for k from 2 to n-i-1 do if abs(((n-i) mod k)-k/2)<1 then c:=c+k; fi; od; if a=b+c then print(n); break; fi; od; od; end: A210732(10000);
One of the proper divisors of 1080 is 120 and sigma(1080) - 3*120 = 3600 - 360 = 3240 = 3*1080. One of the proper divisors of 17850 is 6 and sigma(17850) - 3*6 = 53568 - 18 = 53550 = 3*17850.
with(numtheory): P:=proc(q,h) local a,b,c,k; c:=0; a:=sort([op(divisors(q))]); for k from 1 to nops(a)-1 do if sigma(q)-h*a[k]=h*q then c:=1; break; fi; od; if c=1 then q; fi; end: seq(P(i,3),i=1..7200);
k=3; Select[Range[7128], (t = DivisorSigma[1, #]/k - #; # > t > 0 && IntegerQ[t] && Mod[#, t] == 0) &] (* Giovanni Resta, Aug 25 2017 *)
Triangular number 780 is in sequence because sigma(780)/3 = 784 = 4+780 = 2+5+6+10+12+13+15+20+26+30+39+52+60+65+78+156+195 = 1+3+130+260+390 (summands are all divisors of 780).
Comments