Alexander M. Miller - cp's OEIS frontend

User: Alexander M. Miller

Alexander M. Miller has authored 3 sequences.

A317254 a(n) is the smallest integer such that for all s >= a(n), there are at least n-1 different partitions of s into n parts, namely {x_{11},x_{12},...,x_{1n}}, {x_{21},x_{22},...,x_{2n}},..., and {x_{n-1,1},x_{n-1,2},...,x_{n-1,n}}, such that the products of every set are equal.

19, 23, 23, 26, 27, 29, 31, 32, 35, 36, 38, 40, 42, 44, 45, 47, 49, 50, 52, 53, 54, 55, 57, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 71, 73, 74, 75, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91, 93, 94, 95, 96, 97, 99, 100, 101

Offset: 3

Byungchul Cha, Adam Claman, Joshua Harrington, Ziyu Liu, Barbara Maldonado, Alexander M. Miller, Ann Palma, Wing Hong Tony Wong, Hongkwon V. Yi, Jul 25 2018

			a(3)=19. From s=19 onward, there are at least 2 different partitions of s into 3 parts with equal products:
s=19: {12,4,3} & {9,8,2}:
  12 + 4 + 3 = 9 + 8 + 2 =  19;
  12 * 4 * 3 = 9 * 8 * 2 = 144;
s=20: {15,3,2} & {10,9,1}:
  15 + 3 + 2 = 10 + 9 + 1 = 20;
  15 * 3 * 2 = 10 * 9 * 1 = 90;
s=21: {16,3,2} & {12,8,1}:
  16 + 3 + 2 = 12 + 8 + 1 = 21;
  16 * 3 * 2 = 12 * 8 * 1 = 96.

Byungchul Cha et al., An Investigation on Partitions with Equal Products, arXiv:1811.07451 [math.NT], 2018.
John B. Kelly, Partitions with equal products, Proc. Amer. Math. Soc. 15 (1964), 987-990.

Cf. A060277, A316945, A316946.

Do[maxsumnotwork = 0;  Do[intpart = IntegerPartitions[sum, {n}];   prod = Table[Times @@ intpart[[i]], {i, Length[intpart]}];   prodtally = Tally[prod];   repeatprod = Select[prodtally, #[[2]] >= n - 1 &];   If[repeatprod == {}, maxsumnotwork = sum], {sum, 12, 200}];  Print[n, " ", maxsumnotwork + 1], {n, 3, 60}]

A316945 A triple of positive integers (n,p,k) is admissible if there exist at least two different multisets of k positive integers, {x_1,x_2,...,x_k} and {y_1,y_2,...,y_k}, such that x_1+x_2+...+x_k=y_1+y_2+...+y_k = n and x_1x_2...x_k = y_1y_2...y_k = p. For each n, let A(n)={k:(n,p,k) is admissible for some p}, and let a(n) = |A(n)|.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 4, 4, 6, 7, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74

Offset: 1

Byungchul Cha, Adam Claman, Joshua Harrington, Ziyu Liu, Barbara Maldonado, Alexander M. Miller, Ann Palma, Wing Hong Tony Wong, Hongkwon V. Yi, Jul 20 2018

John Conway proposed an interesting math puzzle in the 1960s, which is now generally known as "Conway's wizard problem." Here is the problem.
Last night I sat behind two wizards on a bus and overheard the following:
Blue Wizard: I have a positive integer number of children, whose ages are positive integers. The sum of their ages is the number of this bus, while the product is my own age.
Red Wizard: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
Blue Wizard: No, you could not.
Red Wizard: Aha! At last, I know how old you are!
Apparently the Red Wizard had been trying to determine the Blue Wizard's age for some time. Now, what was the number of the bus?
This problem posed by Conway looks at different multisets that correspond to the same ordered triple, which motivated the study of this sequence.

			For n = 3, the only partitions of 3 are {3}, {1,2}, and {1,1,1}. Hence, there is no admissible triple (3,p,k).
For n = 4, the only partitions of 4 are {4}, {1,3}, {2,2},{1,1,2}, and {1,1,1,1}. Hence, there is no admissible triple (4,p,k).
For n = 12, the only admissible triple (12,p,k) is when p = 48 and k = 4. This is achieved by the following multisets: {1,3,4,4} and {2,2,2,6}. Thus a(12) = 1.

Jay Bennett, Riddle of the week #34: Two wizards ride a bus, Popular Mechanics. Hearst Communications, Inc., 4 Aug. 2017. 12 Jun. 2018 Accessed.
John B. Kelly, Partitions with equal products, Proc. Amer. Math. Soc. 15 (1964), 987-990.
Tanya Khovanova, Conway's Wizards, arXiv:1210.5460 [math.HO], 2012.

Cf. A060277, A316946.

Table[Count[
  Table[intpart = IntegerPartitions[sum, {n}];
   DuplicateFreeQ[
    Table[Product[intpart[[i]][[j]], {j, n}], {i,
      Length[intpart]}]], {n, sum}], False], {sum,50}]

a(n) = 0 for 1 <= n <= 11, a(12) = 1, a(13) = 2, a(14) = 4, a(15) = 4, a(16) = 6, a(17) = 7, a(18) = 7, and a(n) = n - 10 for n >= 19.

A316946 A triple of positive integers (n,p,k) is admissible if there exist at least two different multisets of k positive integers, {x_1,x_2,...,x_k} and {y_1,y_2,...,y_k}, such that x_1+x_2+...+x_k = y_1+y_2+...+y_k = n and x_1x_2...x_k = y_1y_2...y_k = p. For each n, let A(n) = {p:(n,p,k) is admissible for some k}, and let a(n) = |A(n)|.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 6, 10, 14, 19, 26, 33, 43, 54, 68, 87, 106, 129, 157, 187, 226, 269, 319, 378, 445, 521, 610, 712, 825, 952, 1099, 1261, 1443, 1655, 1889, 2148, 2440, 2769, 3135, 3542, 4000, 4494, 5049, 5661, 6346, 7099, 7938, 8857, 9862, 10972, 12190, 13532, 15000, 16611, 18366

Offset: 1

Byungchul Cha, Adam Claman, Joshua Harrington, Ziyu Liu, Barbara Maldonado, Alexander M. Miller, Ann Palma, Wing Hong Tony Wong, Hongkwon V. Yi, Jul 20 2018

nonn

John Conway proposed an interesting math puzzle in the 1960s, which is now generally known as the "Conway's wizard problem." Here is the problem.
Last night I sat behind two wizards on a bus and overheard the following:
Blue Wizard: I have a positive integer number of children, whose ages are positive integers. The sum of their ages is the number of this bus, while the product is my own age.
Red Wizard: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
Blue Wizard: No, you could not.
Red Wizard: Aha! At last, I know how old you are!
Apparently the Red Wizard had been trying to determine the Blue Wizard's age for some time. Now, what was the number of the bus?
This problem posed by Conway looks at different multisets that correspond to the same ordered triple, which motivated the study of this sequence.

			a(15) = 6 since A(15) = {36,40,48,72,96,144}:
p = 36 [9, 2, 2, 1, 1], [6, 6, 1, 1, 1]
p = 40 [10, 2, 2, 1], [8, 5, 1, 1]
p = 48 [6, 2, 2, 2, 1, 1, 1], [4, 4, 3, 1, 1, 1, 1]
p = 72 [9, 2, 2, 2], [8, 3, 3, 1], [6, 6, 2, 1],
p = 96 [8, 3, 2, 2], [6, 4, 4, 1], [6, 2, 2, 2, 2, 1], [4, 4, 3, 2, 1, 1]
p = 144 [6, 3, 2, 2, 2], [4, 4, 3, 3, 1].

Jay Bennett, Riddle of the week #34: Two wizards ride a bus, Popular Mechanics. Hearst Communications, Inc., 4 Aug. 2017. 12 Jun. 2018 Accessed.
John B. Kelly, Partitions with equal products, Proc. Amer. Math. Soc. 15 (1964), 987-990.

Cf. A060277, A316945.

Do[repeats = {};  Do[intpart = IntegerPartitions[sum, {n}];   prod = Tally[Table[Times @@ intpart[[i]], {i, Length[intpart]}]];   repeatprod = Select[prod, #[[2]] > 1 &];   If[repeatprod != {},    repeats = Join[repeats, Transpose[repeatprod][[1]]]], {n, 3,    sum - 8}]; output = DeleteDuplicates[repeats];  Print[sum, " ", Length[output]], {sum, 12, 100}]

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User: Alexander M. Miller

A317254 a(n) is the smallest integer such that for all s >= a(n), there are at least n-1 different partitions of s into n parts, namely {x_{11},x_{12},...,x_{1n}}, {x_{21},x_{22},...,x_{2n}},..., and {x_{n-1,1},x_{n-1,2},...,x_{n-1,n}}, such that the products of every set are equal.

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