A316945 -id:A316945 - cp's OEIS frontend

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)|.

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}]

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.

Original entry on oeis.org

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}]

A334246 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,k):(n,p,k) is admissible for some k}; then a(n) = |A(n)|.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 7, 13, 20, 29, 44, 66, 90, 129, 174, 232, 306, 406, 520, 675, 851, 1068, 1329, 1640, 2001, 2460, 2989, 3615, 4342, 5202, 6204, 7381, 8697, 10256, 12042, 14069, 16435, 19090, 22141, 25607, 29534

Offset: 1

Bryan Bischof, Apr 20 2020

nonn

J. H. 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.
The number n for which a(n)=1 gives the solution to Conway's puzzle.

			For n=12: {(4, 48)}.
For n=13: {(3, 36), (5, 48)}.
For n=14: {(4, 36), (3, 40), (3, 72), (5, 96), (6, 48)}.
For n=15: {(5, 36), (5, 144), (4, 72), (4, 40), (6, 96), (7, 48), (4, 96)}.

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.

If only counting unique k: A316945; if only counting unique p: A316946.

See A140437 for another sequence related to the wizards problem.

from collections import Counter
from functools import reduce
def partitions(n, I=1):
    yield (n,)
    for i in range(I, n//2 + 1):
        for p in partitions(n-i, i):
            yield (i,) + p
def p(i): #ret partitions of i, sorted by part number and product of parts
    return sorted(
        [
            (
                len(p),
                reduce(
                    (lambda x, y: x * y), p)
            )
            for p in partitions(i)
        ]
    )
def a(p_list): #returns number of pairs appearing more than once
    return len([x for x,y in Counter(p_list).most_common() if y > 1])
print(a(p(i))) # Will print the value of a(i)

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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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