A381187 -id:A381187 - cp's OEIS frontend

A380887 a(n) is the smallest positive integer s that can be partitioned into n positive integers whose product is s * 100^(n-1).

1, 400, 525, 644, 759, 864, 972, 1089, 1188, 1296, 1403, 1508, 1612, 1722, 1827, 1932, 2040, 2145, 2250, 2354, 2457, 2565, 2668, 2772, 2880, 2988, 3087, 3192, 3294, 3399, 3498, 3604, 3705, 3810, 3915, 4018, 4116, 4221, 4323, 4425, 4536, 4635, 4732, 4836, 4940

Offset: 1

Markus Sigg, Feb 07 2025

nonn

The AM-GM inequality shows a(n) >= 100 * n^(n/(n-1)). This bound gives a(2) >= 400, a(3) >= 520, a(4) >= 635, a(5) >= 748, a(6) >= 859.
a(n) is 100 times the smallest price r such that n prices exist whose sum and product both are equal to r. For example (see Guardian article link) 7.11 = 1.20 + 1.25 + 1.50 + 3.16 = 1.20 * 1.25 * 1.50 * 3.16.
Upper bounds for the next terms a(31)-a(40) are 3498, 3604, 3705, 3810, 3915, 4018, 4116, 4221, 4323, 4425. - Karl-Heinz Hofmann, Mar 26 2025
a(n) is well-defined: For n > 1, the sum of the numbers 1, ..., 1, k+1, k*(k+n-1), where the first n-2 numbers are 1 and k = 100^(n-1), is an example (possibly the largest one) of a positive integer s with the required properties. - Markus Sigg, Mar 30 2025
Better upper bounds can be given for specific values of n: Let s > 1 be an integer number and n = 2^s-s. Then the n-tuple of n-s times the number 100 and s times the number 200 has the required properties, hence a(2^s-s) <= 100*(n-s) + 200*s = 100 * 2^s. For s = 6, together with the lower bound from above, this gives 6229 <= a(58) <= 6400. - Markus Sigg, Apr 22 2025
For every n, the n-tuple (100, ..., 100, 10100, n + 99) has the required properties, hence a(n) <= 101*n + 9999. - Markus Sigg, May 30 2025

			a(2) = 400 because 200 + 200 = 400 and 200 * 200 = 400 * 100^1 and no positive integer smaller than 400 exists with the requested properties.
For a(3) the sum is 525 = 150 + 175 + 200.
For a(4) it is 644 = 125 + 160 + 175 + 184.
For a(5) it is 759 = 125 + 125 + 160 + 165 + 184.

David A. Corneth, Table of n, a(n) for n = 1..300
Alex Bellos, Can you solve it? Sexy maths, 2. The 7-Eleven, The Guardian, 3 Feb 2025.
David A. Corneth, PARI program
Markus Sigg, PARI program suitable for calculating a(n) for up to about n = 30.

Cf. A381187, A381619, A381621, A382547.

PARI
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\\ See Sigg link
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PARI
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\\ See Corneth link
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a(8)-a(9) from Hugo Pfoertner, Feb 13 2025
a(10)-a(12) from Hugo Pfoertner, Feb 16 2025
a(13) from Karl-Heinz Hofmann, Mar 02 2025
a(14)-a(30) from Markus Sigg, Mar 27 2025
a(31)-a(32) from Markus Sigg, Apr 23 2025
a(33)-a(45) from Jinyuan Wang, May 01 2025

A381619 Sorted list of sums of 3 prices in minor currency units for a currency that has a 2-decimal minor unit, such that the riddle "sum of prices equals product of prices" has a solution, with prices expressed as floating point numbers with 2 decimals.

Original entry on oeis.org

525, 540, 546, 549, 555, 561, 567, 570, 585, 588, 600, 612, 630, 642, 660, 660, 663, 675, 726, 735, 744, 750, 759, 765, 783, 792, 798, 810, 819, 825, 840, 840, 891, 897, 900, 930, 945, 957, 966, 966, 975, 981, 996, 1050, 1050, 1071, 1080, 1092, 1125, 1134, 1155, 1155, 1170

Offset: 1

Hugo Pfoertner and Markus Sigg, Mar 02 2025

The sequence has 622 terms. See linked files for all solutions.

A natural number s occurs k times in the list if there exist k multisets {x,y,z} of natural numbers with s = x + y + z and 10000*s = x*y*z.

			a(1) = 525 because 1.50 + 1.75 + 2.00 = 1.50*1.75*2.00 = 5.25 is the solution with minimum sum;
a(15) = a(16) = 660 because there are 2 solutions:
  0.80 + 2.50 + 3.30 = 0.80*2.50*3.30 = 6.60 and
  1.10 + 1.50 + 4.00 = 1.10*1.50*4.00 = 6.60;
a(31) = a(32) = 840:
  0.60 + 2.80 + 5.00 = 0.60*2.80*5.00 = 8.40 and
  1.00 + 1.40 + 6.00 = 1.00*1.40*6.00 = 8.40;
a(622) = 100030002 is the largest term:
  0.01 + 100.01 + 1000200.00 = 0.01*100.01*1000200.00 = 1000300.02.

Hugo Pfoertner, Table of n, a(n) for n = 1..622
Hugo Pfoertner, Solution triples sorted by sum. (2025)
Hugo Pfoertner, Solution triples sorted by smallest price. (2025)
Eric Snyder and others, Finding solutions of sum a_i = product a_i = n, where the a_i are "price rationals", question in Mathematics StackExchange, Jun 29, 2022.

Cf. A380887, A381187, A381620, A381621, A382510.

A381621 Sorted list of sums of 4 prices in minor currency units for a currency that has a 2-decimal minor unit, such that the riddle "sum of prices equals product of prices" has a solution, with prices expressed as floating point numbers with 2 decimals.

Original entry on oeis.org

644, 651, 660, 663, 665, 672, 675, 675, 678, 680, 684, 684, 686, 689, 693, 693, 702, 705, 707, 707, 708, 711, 713, 714, 714, 720, 720, 720, 725, 726, 728, 728, 729, 735, 735, 735, 737, 747, 750, 752, 756, 756, 756, 756, 762, 765, 765, 765, 765, 767, 770, 770, 774, 774, 774, 777

Offset: 1

Hugo Pfoertner, Mar 04 2025

The sequence is finite with 22640 terms.

A natural number s occurs k times in the list if there exist k multisets {a,b,c,d} of natural numbers with s = a + b + c + d and 100^3*s = a*b*c*d.

			a(1) = 644 = A380887(4) because 1.25 + 1.60 + 1.75 + 1.84 = 1.25*1.60*1.75*1.84 = 6.44 is the solution with minimum sum;
a(7) = a(8) = 675 because there are 2 solutions:
  1.00 + 1.50 + 2.00 + 2.25 = 1.00*1.50*2.00*2.25 = 6.75 and
  1.20 + 1.25 + 1.80 + 2.50 = 1.20*1.25*1.80*2.50 = 6.75;
a(22) = 711 corresponds to the solution of the puzzle "The 7-Eleven" quoted from "The Guardian" in A380887.
a(22640) = 1000004000003 is the largest term, corresponding to the quadruple of prices [0.01, 0.01, 10000.01, 10000030000.00].

Hugo Pfoertner, Table of n, a(n) for n = 1..10000
Hugo Pfoertner, Solution quadruples sorted by sum, up to sum 10^5. (2025)
Hugo Pfoertner, Complete list of 22640 solution quadruples. (2025)

Cf. A380887, A381187, A381619, A382508, A382510.

A381620 a(n) is the number of solutions to the problem described in A381619 with smallest price equal to n.

Original entry on oeis.org

50, 50, 12, 30, 25, 26, 8, 30, 9, 25, 5, 15, 2, 4, 13, 5, 2, 8, 3, 30, 3, 3, 1, 8, 25, 4, 1, 4, 2, 12, 0, 10, 2, 1, 5, 5, 0, 1, 0, 15, 2, 4, 1, 3, 8, 2, 1, 2, 0, 15, 1, 2, 0, 1, 2, 2, 0, 1, 1, 15, 1, 0, 2, 0, 3, 3, 1, 2, 2, 5, 1, 6, 1, 2, 9, 3, 1, 0, 0, 5, 1, 4

Offset: 1

Hugo Pfoertner, Mar 12 2025

			a(13) = 2 because there are 2 triples {x,y,z} satisfying 100^2*(x+y+z)=x*y*z with x=13:
  {13, 770, 783000} and {13, 800, 20325};
a(23) = 1: {23, 435, 416000} is the only triple with smallest term 23: 10000*(23+435+916000) = 23*435*916000 = 9164580000 = 10000*A381619(578).

Hugo Pfoertner, Table of n, a(n) for n = 1..150
Hugo Pfoertner, Solution triples sorted by smallest price. (2025)

Cf. A380887, A381187, A381619.

A382547 a(n) is the smallest positive integer s that can be partitioned into n distinct positive integers whose product is s * 100^(n-1), or 0 if no such s exists.

Original entry on oeis.org

1, 405, 525, 644, 762, 882, 1038, 1155, 1302, 1428, 1638, 1863, 2079, 2187, 2457, 2673, 3078, 3213, 3402, 3861, 4374, 5103, 5103, 6174

Offset: 1

Markus Sigg, Mar 31 2025

a(n) >= A380887(n) in case of a(n) > 0.
There are only finitely many positive a(n): If x_1 < ... < x_n are positive integers with the required properties, then x_k >= k, and (n-1)! * x_n <= x_1 * ... * x_n = 100^(n-1) * (x_1 + ... + x_n) <= 100^(n-1) * n * x_n gives (n-1)! <= 100^(n-1) * n, hence n <= 274. In fact, n <= 273 must hold, see Mathematics StackExchange link. A more elaborate argumentation in the same discussions shows n <= 269.
By restricting the search space, solution tuples have been found for 25 <= n <= 42. These tuples are not guaranteed to have the smallest possible sum and thus only give upper bounds for a(n). For example, the tuple (1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 20, 25, 30, 32, 40, 50, 60, 64, 75, 80, 100, 120, 125, 128, 150, 160, 200, 225, 250, 400, 625, 800, 1000, 1250, 2500, 3125, 5000, 6250, 12500, 78125, 1953125) shows a(42) <= 2066715.
Programs used for A380887 can be adapted for this sequence.

			a(2) = 405 because 180 + 225 = 405 and 180 * 225 = 405 * 100^1 and no positive integer smaller than 405 exists with the requested properties.

Mathematics StackExchange user Servaes, What is the maximum number of distinct prices whose product equals their sum?, answer dated May 29 2025.

Cf. A380887, A381187, A383026.

dfs(rs, rp, i, r, tp) = if(r==1, return(rs==rp&&setsearch(d, rs)>i)); if((rs/r)^r<=rp, return(0)); for(j=i+1, oo, if(tp>rp, return(0)); if(rp%d[j]==0, if(dfs(rs-d[j], rp/d[j], j, r-1, tp/d[j]), return(1))); tp*=d[j+r]/d[j]);
a(n) = if(n>1, my(p); for(s=100*n, oo, d=divisors(p=s*100^(n-1)); if(dfs(s, p, 0, n, prod(i=1, n, d[i])), return(s))), 1); \\ Jinyuan Wang, May 14 2025

a(17)-a(21) from Markus Sigg, Apr 21 2025

a(22)-a(24) from Jinyuan Wang, May 14 2025

A383026 Triangle T(n,k) read by rows whose n-th row is the lexicographically first n-tuple of ordered distinct positive integers with sum A382547(n) and product A382547(n) * 100^(n-1), or an n-tuple of zeros when A382547(n) = 0.

Original entry on oeis.org

1, 180, 225, 150, 175, 200, 125, 160, 175, 184, 125, 127, 150, 160, 200, 100, 125, 140, 150, 175, 192, 80, 100, 125, 150, 160, 173, 250, 80, 100, 110, 125, 140, 150, 200, 250, 50, 100, 112, 125, 150, 155, 160, 200, 250, 50, 80, 100, 125, 128, 150, 170, 175, 200, 250

Offset: 1

Markus Sigg, Apr 13 2025

Because A382547(n) > 0 for only finitely many n, the triangle has only finitely many nonzero rows.

			Triangle begins:
    1,
  180, 225,
  150, 175, 200,
  125, 160, 175, 184,
  125, 127, 150, 160, 200,
  100, 125, 140, 150, 175, 192,
   80, 100, 125, 150, 160, 173, 250,
   80, 100, 110, 125, 140, 150, 200, 250,
   50, 100, 112, 125, 150, 155, 160, 200, 250,
   50,  80, 100, 125, 128, 150, 170, 175, 200, 250,
   50,  65,  75, 100, 125, 128, 150, 175, 200, 250, 320,
   25,  50,  80, 100, 125, 128, 150, 200, 225, 230, 250, 300,
  ...
For n = 6 there are three 6-tuples with sum A382547(6) = 882 and product 100^5 * 882, namely (100, 125, 140, 150, 175, 192), (100, 125, 147, 150, 160, 200), (112, 120, 125, 150, 175, 200). The first of these is the lexicographically smallest and thus is row 6 of the triangle.

Markus Sigg, Table of n, a(n) for n = 1..231, rows 1..21, flattened.

Cf. A382547, A380887, A381187.

cp's OEIS Frontend

A380887 a(n) is the smallest positive integer s that can be partitioned into n positive integers whose product is s * 100^(n-1).

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A381619 Sorted list of sums of 3 prices in minor currency units for a currency that has a 2-decimal minor unit, such that the riddle "sum of prices equals product of prices" has a solution, with prices expressed as floating point numbers with 2 decimals.

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A381621 Sorted list of sums of 4 prices in minor currency units for a currency that has a 2-decimal minor unit, such that the riddle "sum of prices equals product of prices" has a solution, with prices expressed as floating point numbers with 2 decimals.

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A381620 a(n) is the number of solutions to the problem described in A381619 with smallest price equal to n.

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A382547 a(n) is the smallest positive integer s that can be partitioned into n distinct positive integers whose product is s * 100^(n-1), or 0 if no such s exists.

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Examples

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Programs

PARI

Extensions

A383026 Triangle T(n,k) read by rows whose n-th row is the lexicographically first n-tuple of ordered distinct positive integers with sum A382547(n) and product A382547(n) * 100^(n-1), or an n-tuple of zeros when A382547(n) = 0.

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