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Number T(n,k) of set partitions of [n] into k blocks with equal element sum; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.
+10
17
1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 4, 0, 1, 0, 1, 7, 3, 1, 0, 1, 0, 9, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 35, 43, 0, 0, 1, 0, 1, 62, 102, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 595, 0, 68, 0, 1, 0, 1, 361, 1480, 871, 187, 17, 0, 1
EXAMPLE
T(8,1) = 1: 12345678.
T(8,2) = 7: 12348|567, 12357|468, 12456|378, 1278|3456, 1368|2457, 1458|2367, 1467|2358.
T(8,3) = 3: 1236|48|57, 138|246|57, 156|237|48.
T(8,4) = 1: 18|27|36|45.
T(9,3) = 9: 12345|69|78, 1239|456|78, 1248|357|69, 1257|348|69, 1347|258|69, 1356|249|78, 159|2346|78, 168|249|357, 159|267|348.
Triangle T(n,k) begins:
00 : 1;
01 : 0, 1;
02 : 0, 1;
03 : 0, 1, 1;
04 : 0, 1, 1;
05 : 0, 1, 0, 1;
06 : 0, 1, 0, 1;
07 : 0, 1, 4, 0, 1;
08 : 0, 1, 7, 3, 1;
09 : 0, 1, 0, 9, 0, 1;
10 : 0, 1, 0, 0, 0, 1;
11 : 0, 1, 35, 43, 0, 0, 1;
12 : 0, 1, 62, 102, 0, 0, 1;
13 : 0, 1, 0, 0, 0, 0, 0, 1;
14 : 0, 1, 0, 595, 0, 68, 0, 1;
15 : 0, 1, 361, 1480, 871, 187, 17, 0, 1;
MATHEMATICA
Needs["Combinatorica`"]; T[n_, k_] := Count[(Equal @@ (Total /@ #)&) /@ KSetPartitions[n, k], True]; Table[row = Table[T[n, k], {k, 0, Ceiling[n/2]}]; Print[n, " ", row]; row, {n, 0, 12}] // Flatten (* Jean-François Alcover, Jan 20 2017 *)
Number of ways the first n primes can be partitioned into three sets with equal sums.
+10
5
0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 423, 0, 2624, 0, 13474, 0, 0, 0, 611736, 0, 4169165, 0, 30926812, 0, 214975174, 0, 1590432628, 0, 11431365932, 0, 83946004461, 0, 0, 0, 4615654888831, 0, 35144700468737, 0, 271133285220726, 0, 2103716957561013, 0, 0, 0, 0, 0, 990170108748552983, 0, 7855344215856348141
COMMENTS
It is not true that a(2k+1) is always 0.
REFERENCES
Keith F. Lynch, Posting to Math Fun Mailing List, Sep 17 2019.
EXAMPLE
One of the three solutions for n = 10: 3 + 17 + 23 = 2 + 5 + 7 + 29 = 11 + 13 + 19.
MAPLE
s:= proc(n) option remember; `if`(n<2, 0, ithprime(n)+s(n-1)) end:
b:= proc(n, x, y) option remember; `if`(n=1, 1, (p-> (l->
add(`if`(p>l[i], 0, b(n-1, sort(subsop(i=l[i]-p, l))
[1..2][])), i=1..3))([x, y, s(n)-x-y]))(ithprime(n)))
end:
a:= n-> `if`(irem(2+s(n), 3, 'q')=0, b(n, q-2, q)/2, 0):
MATHEMATICA
s[n_] := s[n] = If[n < 2, 0, Prime[n] + s[n - 1]];
b[n_, x_, y_] := b[n, x, y] = If[n == 1, 1, Function[p, Function[l, Sum[If[ p > l[[i]], 0, b[n - 1, Sequence @@ Sort[ReplacePart[l, i -> l[[i]] - p]][[1;; 2]]]], {i, 1, 3}]][{x, y, s[n] - x - y}]][Prime[n]]];
a[n_] := If[Mod[2+s[n], 3]==0, q = Quotient[2+s[n], 3]; b[n, q-2, q]/2, 0];
PROG
(PARI)
EqSumThreeParts(v)={ my(n=#v, vs=vector(n), m=vecsum(v)/3, brk=0);
for(i=1, n-1, vs[i+1]=vs[i]+v[i]; if(vs[i]<=min(1000, m), brk=i));
my(q=Vecrev(prod(i=1, brk, 1+x^v[i]+y^v[i])));
my(recurse(k, s, p)=if(k==brk, if(s<#q, polcoef(p*q[s+1], m, y)), if(s<=vs[k], self()(k-1, s, p*(1 + y^v[k]))) + if(s>=v[k], self()(k-1, s-v[k], p)) ));
if(frac(m), 0, recurse(n-1, m, 1 + O(y*y^m))/2);
}
Number of ways the first n squares can be partitioned into three sets with equal sums.
+10
4
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 137, 211, 0, 0, 0, 3035, 0, 0, 0, 120465, 259383, 0, 0, 0, 12328889, 0, 0, 0, 673380980, 1659966694, 0, 0, 0, 69819104134, 0, 0, 0, 3761284888715, 9660240745536, 0, 0, 0, 537238185892321, 0, 0, 0, 29922345673502904
REFERENCES
Keith F. Lynch, Posting to Math Fun Mailing List, Sep 19 2019.
EXAMPLE
The unique smallest solution (for n = 13) is 1 + 9 + 25 + 36 + 81 + 121 = 16 + 49 + 64 + 144 = 4 + 100 + 169.
MAPLE
s:= proc(n) option remember; `if`(n<2, 0, n^2+s(n-1)) end:
b:= proc(n, x, y) option remember; `if`(n=1, 1, (p-> (l->
add(`if`(p>l[i], 0, b(n-1, sort(subsop(i=l[i]-p, l))
[1..2][])), i=1..3))([x, y, s(n)-x-y]))(n^2))
end:
a:= n-> `if`(irem(1+s(n), 3, 'q')=0, b(n, q-1, q)/2, 0):
MATHEMATICA
s[n_] := s[n] = If[n < 2, 0, n^2 + s[n - 1]];
b[n_, x_, y_] := b[n, x, y] = Module[{p, l}, If[n == 1, 1, p = n^2; l = {x, y, s[n] - x - y}; Sum[If[p > l[[i]], 0, b[n - 1, Sequence @@ Sort[ ReplacePart[l, i -> l[[i]] - p]][[1 ;; 2]]]], {i, 1, 3}]]];
a[n_] := Module[{q, r}, {q, r} = QuotientRemainder[1 + s[n], 3]; If[r == 0, b[n, q - 1, q]/2, 0]];
Number of ways the first n cubes can be partitioned into three sets with equal sums.
+10
3
1, 0, 0, 691, 3416, 0, 233, 1168, 0, 8857, 18157, 0, 2176512, 3628118, 0, 3204865, 8031495, 0, 79514209, 205927212, 0, 5152732369, 13493840291, 0
REFERENCES
Keith F. Lynch, Posting to Math Fun Mailing List, Sep 17 2019.
EXAMPLE
The unique smallest solution (for n = 23) is
27 + 216 + 1000 + 2197 + 5832 + 6859 + 9261 =
1 + 64 + 343 + 512 + 1728 + 4096 + 8000 + 10648 =
8 + 125 + 729 + 1331 + 2744 + 3375 + 4913 + 12167.
MAPLE
s:= proc(n) option remember; `if`(n<2, 0, n^3+s(n-1)) end:
b:= proc(n, x, y) option remember; `if`(n=1, 1, (p-> (l->
add(`if`(p>l[i], 0, b(n-1, sort(subsop(i=l[i]-p, l))
[1..2][])), i=1..3))([x, y, s(n)-x-y]))(n^3))
end:
a:= n-> `if`(irem(1+s(n), 3, 'q')=0, b(n, q-1, q)/2, 0):
MATHEMATICA
s[n_] := s[n] = If[n < 2, 0, n^3 + s[n - 1]];
b[n_, x_, y_] := b[n, x, y] = If[n == 1, 1, With[{p = n^3}, Sum[If[p > #[[i]], 0, b[n - 1, Sequence @@ Sort[ReplacePart[#, i -> #[[i]] - p]][[1 ;; 2]]]], {i, 1, 3}]]&[{x, y, s[n] - x - y}]];
a[n_] := a[n] = If[q = Quotient[1 + s[n], 3]; Mod[1 + s[n], 3] == 0, b[n, q - 1, q]/2, 0];
EXTENSIONS
a(32), a(33), a(35) recomputed and a(36)-a(38) added by Alois P. Heinz, Sep 30 2019
Number of ways the set {1,2,...,n} can be split into two subsets of which the sum of one is twice the sum of the other.
+10
1
0, 1, 1, 0, 3, 4, 0, 10, 17, 0, 46, 78, 0, 231, 401, 0, 1233, 2177, 0, 6869, 12268, 0, 39502, 71172, 0, 232686, 422076, 0, 1396669, 2547246, 0, 8512170, 15593760, 0, 52534875, 96598865, 0, 327669853, 604405633, 0, 2062171364, 3814087419, 0, 13078921499
FORMULA
a(n) is the coefficient of x^0 in Product_{k=1..n} x^(-2k)+x^k.
EXAMPLE
For n=5 we have 5/1234, 14/532 and 23/541 so a(5)=3.
MAPLE
A113035:= proc(n) local i, j, p, t; t:= NULL; for j to n do p:=1; for i to j do p:=p*(x^(-2*i)+x^(i)); od; t:=t, coeff(p, x, 0); od; t; end;
# second Maple program:
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, 3)=1, 0, b(n*(n+1)/6, n)):
MATHEMATICA
b[n_, i_] := b[n, i] = Module[{m = i(i+1)/2}, If[n > m, 0, If[n == m, 1, b[Abs[n - i], i - 1] + b[n + i, i - 1]]]];
a[n_] := If[Mod[n, 3] == 1, 0, b[n(n+1)/6, n]];
Number of ways the set {1,2,...,n} can be split into three subsets of which the sum of one is one more than the equal sums of both other subsets.
+10
1
0, 0, 0, 1, 0, 0, 5, 0, 0, 60, 0, 0, 747, 0, 0, 11076, 0, 0, 183092, 0, 0, 3238140, 0, 0, 60475317, 0, 0, 1175471401, 0, 0, 23600724220, 0, 0, 486653058995, 0, 0, 10260353188386, 0, 0, 220439819437387, 0, 0, 4813287355239594, 0, 0, 106583271423691692, 0, 0
FORMULA
a(n) is half the coefficient of xy in product(x^(-2k)+x^k(y^k+y^(-k)), k=1..n) for n>1.
EXAMPLE
For n=7 we have splittings 36/27/145, 36/127/45, 136/27/45, 135/27/46, 126/45/37 so a(7) = 5.
MAPLE
A113038:=proc(n) local i, j, p, t; t:= 0; for j from 2 to n do p:=1; for i to j do p:=p*(x^(-2*i)+x^i*(y^i+y^(-i))); od; t:=t, coeff(coeff(p, x, 1), y, 1)/2; od; t; end;
# second Maple program:
b:= proc() option remember; local i, j, t; `if`(args[1]=0, `if`(nargs=2, 1, b(args[t] $t=2..nargs)), add(`if`(args[j] -args[nargs] <0, 0, b(sort([seq(args[i] -`if`(i=j, args[nargs], 0), i=1..nargs-1)])[], args[nargs]-1)), j=1..nargs-1)) end: a:= proc(n) local m; m:= n*(n+1)/2; `if`(m>3 and irem(m, 3)=1, b(((m-1)/3)$2, (m-1)/3+1, n)/2, 0) end: seq(a(n), n=1..50); # Alois P. Heinz, Sep 03 2009
MATHEMATICA
A113038[n_] := Module[{i, j, p, t}, t = {0}; For[j = 2, j <= n, j++, p = 1; For[i = 1, i <= j, i++, p = p*(x^(-2*i) + x^i*(y^i + y^(-i))) // Expand]; t = Append[t, Coefficient[Coefficient[p, x, 1], y, 1]/2]; Print[j, " ", t[[-1]]]]; t];
Number of ways the set {1,2,...,n} can be split into three subsets of which the three sums are consecutive.
+10
1
0, 0, 1, 0, 3, 5, 0, 23, 52, 0, 254, 593, 0, 3611, 8859, 0, 55554, 142169, 0, 946871, 2466282, 0, 17095813, 45359632, 0, 323760077, 870624976, 0, 6367406592, 17307580710, 0, 129063054631, 353941332518, 0, 2682355470491, 7410591325928, 0, 56930627178287
COMMENTS
The empty subset is not allowed, otherwise we would get a(2)=1. - Alois P. Heinz, Sep 03 2009
FORMULA
a(n) is the coefficient of x^3y in product(x^(-2k)+x^k(y^k+y^(-k)), k=1..n) for n>2.
EXAMPLE
For n=5 we have splittings 4/23/15, 4/5/123, 13/5/24, so a(5)=3.
MAPLE
A113039:=proc(n) local i, j, p, t; t:= 0, 0; for j from 3 to n do p:=1; for i to j do p:=p*(x^(-2*i)+x^(i)*(y^i+y^(-i))); od; t:=t, coeff(coeff(p, x, 3), y, 1); od; t; end;
# second Maple program:
b:= proc() option remember; local i, j, t; `if` (args[1]=0, `if` (nargs=2, 1, b(args[t] $t=2..nargs)), add (`if` (args[j] -args[nargs] <0, 0, b(sort ([seq (args[i] -`if` (i=j, args[nargs], 0), i=1..nargs-1)])[], args[nargs]-1)), j=1..nargs-1)) end: a:= proc(n) local m; m:= n*(n+1)/2; `if` (n>2 and irem (m, 3)=0, b(m/3-1, m/3, m/3+1, n), 0) end: seq (a(n), n=1..42); # Alois P. Heinz, Sep 03 2009
MATHEMATICA
a[n_] := If[n <= 2, 0, Product[x^(-2k)+x^k(y^k+y^(-k)), {k, 1, n}] // SeriesCoefficient[#, {x, 0, 3}, {y, 0, 1}]&];
Number of ways the set {1,2,...,n} can be split into three subsets X, Y, Z of equal sums, where the order of X, Y, Z matters.
+10
0
0, 0, 0, 0, 6, 6, 0, 18, 54, 0, 258, 612, 0, 3570, 8880, 0, 55764, 142368, 0, 947946, 2468844, 0, 17099808, 45375498, 0, 323927184, 871038570, 0, 6369199908, 17312303760
COMMENTS
Constant term of Product_{k=1..n} (x^k+y^k+1/(x*y)^k).
EXAMPLE
For n = 1, 2, 3, 4, a(n) = 0, as n*(n+1)/2 is not divisible by 3.
For n = 5, a(5) = 6, as {1,2,3,4,5} = {1,4}U{2,3}U{5} and there are 6 permutations.
For n = 6, a(6) = 6, as {1,2,3,4,5,6} = {1,6}U{2,5}U{3,4} and there are 6 permutations.
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