cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A332105 a(n) is the smallest number k that can be partitioned into a set of n distinct positive even integers {e(1), e(2), ..., e(n)} where the sum of e(i)*(e(i)-1) for i = 1 to n equals k*(k-1)/2.

Original entry on oeis.org

16, 144, 80, 96, 160, 208, 256, 304, 384, 432, 544, 608, 720, 816, 832, 1040, 1168, 1264, 1360, 1568, 1664, 1808, 1984, 2080, 2256, 2480, 2704, 2800, 3104, 3248, 3520, 3744, 3968, 4112, 4464, 4688, 4880, 5200, 5472, 5744, 6016, 6336, 6608, 6800, 7248, 7568, 7888, 8080, 8528
Offset: 2

Views

Author

Dean D. Ballard, Nov 20 2020

Keywords

Comments

These numbers solve the problem of what is the required minimum even number of socks of n colors such that a random drawing of two socks has a 50% chance of matching.

Examples

			For n = 3, {6, 48, 90} is the set of even numbers with the smallest sum that has this property. With 6 socks of one color, 48 socks of another color, and 90 socks of a third color, there is exactly a 50% chance that a random draw of two socks will produce a matching pair. (6*5 + 48*47 + 90*89) = (144*143) / 2.
n = 2, sum = 16, set = {6, 10}
n = 3, sum = 144, set = {6, 48, 90}
n = 4, sum = 80, set = {2, 8, 16, 54}
n = 5, sum = 96, set = {2, 6, 8, 14, 66}
n = 6, sum = 160, set = {2, 6, 8, 10, 24, 110}
		

Crossrefs

Cf. A246750.

Programs

  • PARI
    \\ See Links in A246750 for a faster program.
    a(n)={for(k=n*(n+1)/2, oo, my(t=k*(4*k-1)); forpart(p=2*k-n*(n-1)/2, if(sum(i=1, n, (p[i]+i-1)*(2*(p[i]+i-1)-1))==t, return(4*k)), ,[n,n]))} \\ Andrew Howroyd, Nov 21 2020

Extensions

a(16)-a(50) from Andrew Howroyd, Nov 22 2020

A339271 a(n) is the smallest number k that can be partitioned into a set of n distinct positive integers {e(1), e(2), ..., e(n)} where Sum_{i=1..n} e(i)*(e(i)-1) = k*(k-1)/2.

Original entry on oeis.org

4, 13, 20, 53, 56, 92, 109, 120, 160, 200, 221, 268, 325, 389, 420, 497, 561, 616, 684, 725, 813, 901, 969, 1064, 1132, 1197, 1329, 1421, 1516, 1581, 1740, 1849, 1904, 2060, 2189, 2288, 2444, 2560, 2696, 2849, 2985, 3128, 3261, 3404, 3564, 3744, 3904, 4044, 4204, 4381, 4585, 4725
Offset: 2

Views

Author

Dean D. Ballard, Nov 29 2020

Keywords

Comments

These numbers solve the problem of what is the required minimum number of socks of n colors such that a random drawing of two socks has a 50% chance of matching. In this version the number of socks of each color is distinct, but there may be a color with only one sock.

Examples

			For n = 3, {1, 3, 9} is the set with the smallest sum that has this property. With 1 socks of one color, 3 socks of another color, and 9 socks of a third color, there is exactly a 50% chance that a random draw of two socks will produce a matching pair. (1*0 + 3*2 + 9*8) = (13*12) / 2.
n = 2, sum = 4, set = {1, 3}
n = 3, sum = 13, set = {1, 3, 9}
n = 4, sum = 20, set = {1, 2, 3, 14}
n = 5, sum = 53, set = {1, 2, 3, 11, 36}
n = 6, sum = 56, set = {1, 2, 3, 5, 6, 39}
		

Crossrefs

Cf. other variations of the problem: A246750, A332105, A339272.

Programs

  • PARI
    \\ See 'Faster PARI Program' link in A246750 for PartsByWeight.
    a(n)={local(FC=Map()); for(k=1, oo, if(PartsByWeight(n, k-n*(n-1)/2, k*(k-1)/2, (i,v)->(i+v-1)*(i+v-2)), return(k))); oo} \\ Andrew Howroyd, Nov 30 2020

Extensions

a(16)-a(24) from Michael S. Branicky, Nov 29 2020
a(25)-a(30) from Andrew Howroyd, Nov 30 2020
a(31)-a(53) from Michael S. Branicky, Dec 03 2020

A339272 a(n) is the smallest number k that can be partitioned into a set of n positive integers {e(1), e(2), ..., e(n)} where Sum_{i=1..n} e(i)*(e(i)-1) = k*(k-1)/2.

Original entry on oeis.org

4, 13, 17, 40, 24, 21, 44, 41, 41, 57, 48, 48, 68, 65, 65, 81, 85, 72, 72, 85, 89, 89, 105, 109, 109, 96, 116, 133, 113, 113, 133, 133, 140, 140, 120, 157, 153, 137, 157, 164, 164, 164, 181, 181, 181, 181, 188, 188, 201, 188, 205, 205, 225, 212, 205, 212, 212, 229, 229, 229, 229, 249, 249, 256, 236, 236, 253, 253, 253
Offset: 2

Views

Author

Dean D. Ballard, Nov 29 2020

Keywords

Comments

These numbers solve the problem of what is the required minimum number of socks of n colors such that a random drawing of two socks has a 50% chance of matching. In this version there may be a color with only one sock, and the number of socks of each color need not be distinct.
If the set solving a(n)=k has three 2's, then a(n+1)<=k, replacing them with one 3 and three 1's; this leads to many repeats in the data. Similar substitutions include {4, 5} for {1, 2, 6}, {3, 3, 4} for {1, 2, 2, 5}, and {3, 3} for {1, 1, 4}. - Dean D. Ballard and Michael S. Branicky, Nov 30 2020

Examples

			For n = 4, {1, 2, 2, 12} is the set with the smallest sum that has this property. With 1 sock of one color, 2 socks of a second color, 2 socks of a third color, and 12 socks of a fourth color, there is exactly a 50% chance that a random draw of two socks will produce a matching pair. (1*0 + 2*1 + 2*1 + 12*11) = (17*16) / 2.
n = 2, sum = 4, set = {1, 3}
n = 3, sum = 13, set = {1, 3, 9}
n = 4, sum = 17, set = {1, 2, 2, 12}
n = 5, sum = 40, set = {3, 3, 3, 3, 28}
n = 6, sum = 24, set = {1, 1, 1, 2, 2, 17}
		

Crossrefs

A246750, A332105, A339271 are other variations of the problem.

Programs

  • PARI
    \\ See 'Faster PARI Program' link in A246750 for PartsByWeight.
    a(n)={local(FC=Map()); for(k=1, oo, if(PartsByWeight(n, k, k*(k-1)/2, (i,v)->v*(v-1)), return(k))); oo} \\ Andrew Howroyd, Nov 30 2020

Extensions

a(26)-a(70) from Andrew Howroyd, Nov 30 2020
Showing 1-3 of 3 results.