A306443 -id:A306443 - cp's OEIS frontend

A022894 Number of solutions to c(1)prime(1) +...+ c(2n+1)prime(2n+1) = 0, where c(i) = +-1 for i > 1, c(1) = 1.

0, 1, 1, 2, 5, 13, 39, 122, 392, 1286, 4341, 14860, 51085, 178402, 634511, 2260918, 8067237, 29031202, 105250449, 383579285, 1404666447, 5171065198, 19141008044, 71124987313, 263548339462, 983424096451, 3684422350470, 13818161525284, 51938115653565

Offset: 0

Clark Kimberling

c(1)*prime(1) + ... + c(2n)*prime(2n) = 0 has no solution, because the l.h.s. has an odd number of odd terms and the r.h.s. is even.

			a(1) = 1 because 2 + 3 - 5 = 0,
a(2) = 1 because 2 - 3 + 5 + 7 - 11 = 0,
a(3) = 2 because
  2 + 3 - 5 - 7 + 11 + 13 - 17 =
  2 + 3 - 5 + 7 - 11 - 13 + 17 = 0.
a(4) = 5 because
  2 - 3 - 5 + 7 + 11 + 13 + 17 - 19 - 23 =
  2 - 3 + 5 - 7 + 11 + 13 - 17 + 19 - 23 =
  2 - 3 + 5 + 7 - 11 - 13 + 17 + 19 - 23 =
  2 - 3 + 5 + 7 - 11 + 13 - 17 - 19 + 23 =
  2 + 3 + 5 - 7 - 11 - 13 + 17 - 19 + 23 = 0
and there are no others up through the ninth prime.

Ray Chandler, Table of n, a(n) for n = 0..1000 (first 101 terms from T. D. Noe)

Cf. A113040, A215036, A083309 (sums of odd primes).
Cf. A022895, A022896 (r.h.s. = 1 & 2, using all primes), A083309 and A022897 - A022899 (using primes >= 3), A022900 - A022902 (using primes >=5), A022903, A022904, A022920 (using primes >= 7); A261061 - A261063 & A261045 (r.h.s. = -1); A261057, A261059, A261060 & A261044 (r.h.s. = -2).
Bisection (odd part) of A306443.

sp:= proc(n) sp(n):= `if`(n=1, 0, ithprime(n)+sp(n-1)) end:
b := proc(n,i) option remember; `if`(n>sp(i), 0, `if`(i=1, 1,
        b(n+ithprime(i), i-1)+ b(abs(n-ithprime(i)), i-1)))
     end:
a:= n-> b(2, 2*n+1):
seq(a(n), n=0..40);  # Alois P. Heinz, Aug 05 2012

Do[a = Table[ Prime[i], {i, 1, n} ]; c = 0; k = 2^(n - 1); While[k < 2^n, If[ Apply[ Plus, a*(-1)^(IntegerDigits[k, 2] + 1)] == 0, c++ ]; k++ ]; Print[c], {n, 1, 32, 2} ]

A022894={a(n, s=0-prime(1), p=1)=if(n<=s, if(s==p, n==s, a(abs(n-p), s-p, precprime(p-1))+a(n+p, s-p, precprime(p-1))), if(s<=0, a(abs(s), max(sum(i=p+1, p+(p>1)+2*n, prime(i)),1), prime(p+(p>1)+2*n))))} \\ M. F. Hasler, Aug 09 2015

Conjecture: limit_{n->oo} a(n)^(1/n) = 4. - Vaclav Kotesovec, Jun 05 2019

a(n) is the constant term in expansion of (1/2) * Product_{k=1..2*n+1} (x^prime(k) + 1/x^prime(k)). - Ilya Gutkovskiy, Jan 25 2024

Edited by Robert G. Wilson v, Jan 29 2002
More terms from T. D. Noe, Jan 16 2007
Edited by M. F. Hasler, Aug 09 2015

A113040 Number of solutions to +-p(1)+-p(2)+-...+-p(2n)=1 where p(i) is the i-th prime.

Original entry on oeis.org

1, 1, 3, 6, 16, 45, 138, 439, 1417, 4698, 16021, 55146, 190274, 671224, 2404289, 8535117, 30635869, 110496946, 401422210, 1467402238, 5393176633, 19883249002, 73856531314, 273602448261, 1017563027699, 3803902663467, 14266523388813, 53564969402478

Offset: 1

Floor van Lamoen, Oct 12 2005

nonn

+-p(1)+-p(2)+-...+-p(2n+1)=1 has no solutions because the l.h.s. is even.

			2 + 3 + 5 - 7 + 11 - 13 = - 2 + 3 + 5 - 7 - 11 + 13 = - 2 + 3 - 5 + 7 + 11 - 13 = 1 so a(3) = 3.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..1000 (first 130 terms from Alois P. Heinz)

Cf. A083309, A022894 - A022904, A022920, A261057 - A261063 and A261044 - A261045; A215036.

Bisection (even part) of A306443.

A113040:=proc(n) local i,j,p,t; t:= NULL; for j from 2 to 2*n by 2 do p:=1; for i to j do p:=p*(x^(-ithprime(i))+x^(ithprime(i))); od; t:=t,coeff(p,x,1); od; t; end;
# second Maple program:
sp:= proc(n) sp(n):= `if`(n=0, 0, ithprime(n)+sp(n-1)) end:
b := proc(n,i) option remember; `if`(n>sp(i), 0, `if`(i=0, 1,
        b(n+ithprime(i), i-1)+ b(abs(n-ithprime(i)), i-1)))
     end:
a:= n-> b(1, 2*n):
seq(a(n), n=1..40);  # Alois P. Heinz, Aug 05 2012

sp[n_] := If[n == 0, 0, Prime[n]+sp[n-1]]; b[n_, i_] := b[n, i] =If[n > sp[i], 0, If[i == 0, 1, b[n+Prime[i], i-1] + b[Abs[n-Prime[i]], i-1]]]; a[n_] := b[1, 2*n]; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Nov 11 2015, after Alois P. Heinz *)

a(n) = A022895(2n) + A261061(n). - M. F. Hasler, Aug 09 2015
Conjecture: limit_{n->infinity} a(n)^(1/n) = 4. - Vaclav Kotesovec, Jun 05 2019
a(n) = [x^1] Product_{k=1..2*n} (x^prime(k) + 1/x^prime(k)). - Ilya Gutkovskiy, Jan 25 2024

A069918 Number of ways of partitioning the set {1...n} into two subsets whose sums are as nearly equal as possible.

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 4, 7, 23, 40, 35, 62, 221, 397, 361, 657, 2410, 4441, 4110, 7636, 28460, 53222, 49910, 93846, 353743, 668273, 632602, 1199892, 4559828, 8679280, 8273610, 15796439, 60400688, 115633260, 110826888, 212681976, 817175698, 1571588177, 1512776590

Offset: 1

Robert G. Wilson v, Apr 24 2002

nonn

If n mod 4 = 0 or 3, a(n) is the number of solutions to +- 1 +- 2 +- 3 +- ... +- n = 0 or 1; if n mod 4 = 1 or 2, a(n) is half this number.

			If the triangular number T_n (see A000217) is even then the two totals must be equal, otherwise the two totals differ by one.
a(6) = 5: T6 = 21 and is odd. There are five sets such that the sum of one side is equal to the other side +/- 1. They are 5+6 = 1+2+3+4, 4+6 = 1+2+3+5, 1+4+6 = 2+3+5, 1+3+6 = 2+4+5 and 2+3+6 = 1+4+5.

Alois P. Heinz, Table of n, a(n) for n = 1..1000
S. R. Finch, Signum equations and extremal coefficients [broken link]
Steven R. Finch, Signum equations and extremal coefficients, February 7, 2009. [Cached copy, with permission of the author]
Brian Hayes, The Easiest Hard Problem, American Scientist, Vol. 90, No. 2, March-April 2002, pp. 113-117.
Recreational Mathematics Newsletter, Kolstad, PROBLEM 4: Subset Sums [broken link]
Dave Rusin, More Number Theory [broken link]

Cf. A060005, A058377, A025591, A063865, A063866, A063867, A306443.

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, b(n-1, n-1) +b(n+1, n-1), b(n, n-1)):
seq(a(n), n=1..60);  # Alois P. Heinz, Nov 02 2011

Needs["DiscreteMath`Combinatorica`"]; f[n_] := f[n] = Block[{s = Sort[Plus @@@ Subsets[n]], k = n(n + 1)/2}, If[ EvenQ[k], Count[s, k/2]/2, (Count[s, Floor[k/2]] + Count[s, Ceiling[k/2]]) /2]]; Table[ f[n], {n, 1, 22}]
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[ Which[ Mod[n, 4] == 0 || Mod[n, 4] == 3, f[n, 0]/2, Mod[n, 4] == 1 || Mod[n, 4] == 2, f[n, 1]], {n, 1, 40}]

If n mod 4 = 0 or 3 then the two subsets have the same sum and a(n) = A025591(n); if n mod 4 = 1 or 2 then the two subsets have sums which differ by 1 and a(n) = A025591(n)/2. - Henry Bottomley, May 08 2002

More terms from Henry Bottomley, May 08 2002

Comment corrected by Steven Finch, Feb 01 2009

A307877 Number of ways of partitioning the set of the first n positive squares into two subsets whose sums differ at most by 1.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 2, 1, 1, 5, 2, 1, 5, 13, 43, 43, 57, 193, 274, 239, 430, 1552, 3245, 2904, 5419, 18628, 31048, 27813, 50213, 188536, 372710, 348082, 649300, 2376996, 4197425, 3913496, 7287183, 27465147, 53072709, 50030553, 93696497, 351329160, 650125358

Offset: 0

Alois P. Heinz, Jun 04 2019

nonn

			a(6) = 2: 1,9,36/4,16,25; 1,4,16,25/9,36.
a(7) = 1: 1,4,16,49/9,25,36.

Alois P. Heinz, Table of n, a(n) for n = 0..200
Wikipedia, Partition problem

Cf. A000290, A069918, A083527, A158092, A306443, A308682.

s:= proc(n) s(n):= `if`(n=0, 1, n^2+s(n-1)) end:
b:= proc(n, i) option remember; `if`(i=0, `if`(n<=1, 1, 0),
      `if`(n>s(i), 0, (p-> b(n+p, i-1)+b(abs(n-p), i-1))(i^2)))
    end:
a:= n-> ceil(b(0, n)/2):
seq(a(n), n=0..45);

s[n_] := s[n] = If[n == 0, 1, n^2 + s[n - 1]];
b[n_, i_] := b[n, i] = If[i == 0, If[n <= 1, 1, 0], If[n > s[i], 0, Function[p, b[n + p, i - 1] + b[Abs[n - p], i - 1]][i^2]]];
a[n_] := Ceiling[b[0, n]/2];
a /@ Range[0, 45] (* Jean-François Alcover, Dec 07 2020, after Alois P. Heinz *)

a(n) = A083527(n) if n == 0 or 3 (mod 4).

A331479 Table read by rows: row n lists the numbers m such that the first n primes can be partitioned into m subsets all of which have the same sum.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 4, 1, 1, 2, 1, 1, 2, 4, 1, 3, 1, 2, 4, 5, 1, 3, 1, 2, 4, 1, 3, 1, 2, 4, 1, 7, 1, 2, 1, 3, 1, 2, 4, 5, 1, 3, 1, 2, 4, 1, 3, 1, 2, 4, 5, 8, 1, 3, 9, 1, 2, 4, 5, 8, 1, 3, 1, 2, 4, 7, 1, 3, 1, 2, 4, 1, 3, 1, 2, 4, 8

Offset: 1

Jon E. Schoenfield, Jan 17 2020

Consider the following one-dimensional bin packing problem: given n items whose sizes are the first n primes, list the numbers m such that all the items can be packed into m bins of identical capacity, with each bin packed completely full. The resulting list is row n.

If a row contains a number m, it necessarily contains all divisors of m.

			In bin-packing terms, for n=19, the sum of the 19 item sizes, i.e., the sum of the first n primes, is 2 + 3 + ... + 67 = 568, whose divisors begin 1, 2, 4, 8, ...; the bin capacity must be at least 67 (the size of the largest item), and 568/67 < 9, so the number of bins m cannot exceed 8. However, the 19 items cannot be packed into 8 bins: the bin capacity would be 568/8 = 71 (which, as an odd sum, would require that each bin containing only odd-sized items -- i.e., every bin other than the one containing the item of size 2 -- contain an odd number of items, hence at least 3 items, but there are only 19 items in total). So the remaining values of m are 1 (i.e., packing all 19 items in a single bin), 2 (e.g., 568/2 = 284 = 67 + 61 + 59 + 53 + 41 + 3 = 47 + 43 + 37 + 31 + 29 + 23 + 19 + 17 + 13 + 11 + 7 + 5 + 2), and 4 (e.g., 568/4 = 142 = 67 + 61 + 11 + 3 = 59 + 53 + 23 + 7 = 47 + 43 + 37 + 13 + 2 = 41 + 31 + 29 + 19 + 17 + 5), so row 19 consists of the numbers 1, 2, and 4.
.                                       Numbers m such that
             Sum of  Divisors m of sum  1st n primes can be
      n-th   1st n       such that      partitioned into m
   n  prime  primes  m <= sum/prime(n)  subsets w/same sum
  --  -----  ------  -----------------  -------------------
   1     2       2   1                  1;
   2     3       5   1                  1;
   3     5      10   1, 2               1, 2;
   4     7      17   1                  1;
   5    11      28   1, 2               1, 2;
   6    13      41   1                  1;
   7    17      58   1, 2               1, 2;
   8    19      77   1                  1;
   9    23     100   1, 2, 4            1, 2;
  10    29     129   1, 3               1, 3;
  11    31     160   1, 2, 4, 5         1, 2, 4;
  12    37     197   1                  1;
  13    41     238   1, 2               1, 2;
  14    43     281   1                  1;
  15    47     328   1, 2, 4            1, 2, 4;
  16    53     381   1, 3               1, 3;
  17    59     440   1, 2, 4, 5         1, 2, 4, 5;
  18    61     501   1, 3               1, 3;
  19    67     568   1, 2, 4, 8         1, 2, 4;
  20    71     639   1, 3, 9            1, 3;
  21    73     712   1, 2, 4, 8         1, 2, 4;
  22    79     791   1, 7               1, 7;
  23    83     874   1, 2               1, 2;
  24    89     963   1, 3, 9            1, 3;
  25    97    1060   1, 2, 4, 5, 10     1, 2, 4, 5;
  26   101    1161   1, 3, 9            1, 3;
  27   103    1264   1, 2, 4, 8         1, 2, 4;
  28   107    1371   1, 3               1, 3;
  29   109    1480   1, 2, 4, 5, 8, 10  1, 2, 4, 5, 8;
  30   113    1593   1, 3, 9            1, 3, 9;
  31   127    1720   1, 2, 4, 5, 8, 10  1, 2, 4, 5, 8;
  32   131    1851   1, 3               1, 3;
  33   137    1988   1, 2, 4, 7, 14     1, 2, 4, 7;
  34   139    2127   1, 3               1, 3;
  35   149    2276   1, 2, 4            1, 2, 4;
  36   151    2427   1, 3               1, 3;
  37   157    2584   1, 2, 4, 8         1, 2, 4, 8;

Cf. A000040, A007504, A053050, A306443, A327449.

cp's OEIS Frontend

A022894 Number of solutions to c(1)*prime(1) +...+ c(2n+1)*prime(2n+1) = 0, where c(i) = +-1 for i > 1, c(1) = 1.

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A113040 Number of solutions to +-p(1)+-p(2)+-...+-p(2n)=1 where p(i) is the i-th prime.

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A069918 Number of ways of partitioning the set {1...n} into two subsets whose sums are as nearly equal as possible.

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A307877 Number of ways of partitioning the set of the first n positive squares into two subsets whose sums differ at most by 1.

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Maple

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A331479 Table read by rows: row n lists the numbers m such that the first n primes can be partitioned into m subsets all of which have the same sum.

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A022894 Number of solutions to c(1)prime(1) +...+ c(2n+1)prime(2n+1) = 0, where c(i) = +-1 for i > 1, c(1) = 1.