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 1-5 of 5 results.

A103277 Smallest i such that there exists j such that i = x + y + z, j = x*y*z has exactly n solutions in positive integers x <= y <= z.

Original entry on oeis.org

3, 13, 39, 118, 185, 400, 511, 1022, 1287, 2574, 4279, 8558, 11777, 24377, 23554, 46111, 99085, 165490
Offset: 1

Views

Author

David W. Wilson, Jan 27 2005

Keywords

Comments

Least number k such that there exists n partitions of k into 3 parts each having the same product.
The greatest number k such that there exists n partitions of k into 3 parts each having the same product: 18, 102, 492, 1752, ...
The number of members in each "class" of the set having n partitions into 3 parts each having the same product: 12, 54, 147, 397, ....

Examples

			3 = 1+1+1 & 1*1*1 = 1.
13 = 6+6+1 = 9+2+2 & 6*6*1 = 9*2*2 = 36.
39 = 20+15+4 = 24+10+5 = 25+8+6 & 20*15*4 = 24*10*5 = 25*8*6 = 1200.
118 = 54+50+14 = 63+40+15 = 70+30+18 = 72+25+21 & 54*50*14 = 63*40*15 = 70*30*18 = 72*25*21 = 37800.
185 = 90+84+11 = 110+63+12 = 126+44+15 = 132+35+18 = 135+28+22 & 90*84*11 = 110*63*12 = 126*44*15 = 132*35*18 = 135*28*22 = 83160.
400 = 196+180+24 = 245+128+27 = 252+120+28 = 270+98+32 = 280+84+36 = 288+70+42 & 196*180*24 = 245*128*27 = 252*120*28 = 270*98*32 = 280*84*36 = 288*70*42 = 846720.
511 = 260+216+35 = 280+195+36 = 315+156+40 = 325+144+42 = 336+130+45 = 360+91+60 = 364+75+72 & 260*216*35 = 280*195*36 = 315*156*40 = 325*144*42 = 336*130*45 = 360*91*60 = 364*75*72 = 1965600.
1022 = 520+432+70 = 560+390+72 = 630+312+80 = 650+288+84 = 672+260+90 = 675+256+91 = 720+182+120 = 728+150+144 & 520*432*70 = 560*390*72 = 630*312*80 = 650*288*84 = 672*260*90 = 675*256*91 = 720*182*120 = 728*150*144 = 15724800.
1287 = 600+588+99 = 648+539+100 = 720+462+105 = 770+405+112 = 825+336+126 = 840+315+132 = 880+245+162 = 882+240+165 = 891+200+196 & 600*588*99 = 648*539*100 = 720*462*105 = 770*405*112 = 825*336*126 = 840*315*132 = 880*245*162 = 882*240*165 = 891*200*196 = 34927200.
From _Donovan Johnson_, Mar 29 2010: (Start)
2574 = 198+1176+1200 = 200+1078+1296 = 210+924+1440 = 224+810+1540 = 231+768+1575 = 252+672+1650 = 264+630+1680 = 324+490+1760 = 330+480+1764 = 392+400+1782 & 198*1176*1200 = 200*1078*1296 = 210*924*1440 = 224*810*1540 = 231*768*1575 = 252*672*1650 = 264*630*1680 = 324*490*1760 = 330*480*1764 = 392*400*1782 = 279417600.
4279 = 378+1925+1976 = 380+1820+2079 = 385+1710+2184 = 399+1540+2340 = 429+1330+2520 = 440+1274+2565 = 504+1045+2730 = 532+975+2772 = 550+936+2793 = 637+792+2850 = 684+735+2860 & 378*1925*1976 = 380*1820*2079 = 385*1710*2184 = 399*1540*2340 = 429*1330*2520 = 440*1274*2565 = 504*1045*2730 = 532*975*2772 = 550*936*2793 = 637*792*2850 = 684*735*2860 = 1437836400.
8558 = 756+3850+3952 = 760+3640+4158 = 770+3420+4368 = 798+3080+4680 = 858+2660+5040 = 880+2548+5130 = 896+2475+5187 = 1008+2090+5460 = 1064+1950+5544 = 1100+1872+5586 = 1274+1584+5700 = 1368+1470+5720 & 756*3850*3952 = 760*3640*4158 = 770*3420*4368 = 798*3080*4680 = 858*2660*5040 = 880*2548*5130 = 896*2475*5187 = 1008*2090*5460 = 1064*1950*5544 = 1100*1872*5586 = 1274*1584*5700 = 1368*1470*5720 = 11502691200.
11777 = 171+5600+6006 = 175+4914+6688 = 198+3675+7904 = 224+3003+8550 = 228+2925+8624 = 240+2717+8820 = 245+2640+8892 = 385+1512+9880 = 416+1386+9975 = 462+1235+10080 = 540+1045+10192 = 600+936+10241 = 637+880+10260 & 171*5600*6006 = 175*4914*6688 = 198*3675*7904 = 224*3003*8550 = 228*2925*8624 = 240*2717*8820 = 245*2640*8892 = 385*1512*9880 = 416*1386*9975 = 462*1235*10080 = 540*1045*10192 = 600*936*10241 = 637*880*10260 = 5751345600.
24377 = 1196+11400+11781 = 1197+11220+11960 = 1232+9690+13455 = 1254+9200+13923 = 1360+7722+15295 = 1520+6435+16422 = 1547+6270+16560 = 1748+5304+17325 = 1890+4807+17680 = 1932+4680+17765 = 2244+3933+18200 = 2261+3900+18216 = 2448+3575+18354 = 2907+2990+18480 & 1196*11400*11781 = 1197*11220*11960 = 1232*9690*13455 = 1254*9200*13923 = 1360*7722*15295 = 1520*6435*16422 = 1547*6270*16560 = 1748*5304*17325 = 1890*4807*17680 = 1932*4680*17765 = 2244*3933*18200 = 2261*3900*18216 = 2448*3575*18354 = 2907*2990*18480 = 160626866400.
23554 = 342+11200+12012 = 350+9828+13376 = 351+9728+13475 = 396+7350+15808 = 448+6006+17100 = 456+5850+17248 = 480+5434+17640 = 490+5280+17784 = 665+3584+19305 = 770+3024+19760 = 832+2772+19950 = 924+2470+20160 = 1080+2090+20384 = 1200+1872+20482 = 1274+1760+20520 & 342*11200*12012 = 350*9828*13376 = 351*9728*13475 = 396*7350*15808 = 448*6006*17100 = 456*5850*17248 = 480*5434*17640 = 490*5280*17784 = 665*3584*19305 = 770*3024*19760 = 832*2772*19950 = 924*2470*20160 = 1080*2090*20384 = 1200*1872*20482 = 1274*1760*20520 = 46010764800.
(End)
From _Duncan Moore_, Sep 02 2017: (Start)
46111 = 4446+20160+21505 = 4455+19760+21896 = 4576+17595+23940 = 4680+16560+24871 = 4725+16192+25194 = 4807+15600+25704 = 4928+14858+26325 = 5100+13984+27027 = 5187+13600+27324 = 5520+12376+28215 = 5610+12096+28405 = 5712+11799+28600 = 6270+10465+29376 = 7360+8721+30030 = 7735+8280+30096 = 7904+8100+30107 & 4446*20160*21505 = 4455*19760*21896 = 4576*17595*23940 = 4680*16560*24871 = 4725*16192*25194 = 4807*15600*25704 = 4928*14858*26325 = 5100*13984*27027 = 5187*13600*27324 = 5520*12376*28215 = 5610*12096*28405 = 5712*11799*28600 = 6270*10465*29376 = 7360*8721*30030 = 7735*8280*30096 = 7904*8100*30107 = 1927522396800.
99085 = 3770+47120+48195 = 3780+45240+50065 = 3952+37758+57375 = 3978+37107+58000 = 4176+33250+61659 = 4199+32886+62000 = 4216+32625+62244 = 4495+29070+65520 = 4500+29016+65569 = 4914+25296+68875 = 5320+22620+71145 = 7280+15390+76415 = 7395+15120+76570 = 7905+14040+77140 = 8370+13195+77520 = 9367+11718+78000 = 9945+11020+78120 & 3770*47120*48195 = 3780*45240*50065 = 3952*37758*57375 = 3978*37107*58000 = 4176*33250*61659 = 4199*32886*62000 = 4216*32625*62244 = 4495*29070*65520 = 4500*29016*65569 = 4914*25296*68875 = 5320*22620*71145 = 7280*15390*76415 = 7395*15120*76570 = 7905*14040*77140 = 8370*13195*77520 = 9367*11718*78000 = 9945*11020*78120 = 8561475468000.
165490 = 14000+72488+79002 = 14022+71500+79968 = 14080+69615+81795 = 14280+65520+85690 = 14432+63308+87750 = 14820+59040+91630 = 14896+58344+92250 = 16236+49504+99750 = 16380+48790+100320 = 16830+46740+101920 = 17290+44880+103320 = 17589+43776+104125 = 18720+40180+106590 = 19152+39000+107338 = 20090+36720+108680 = 21648+33592+110250 = 23940+30030+111520 = 25840+27720+111930 & 14000*72488*79002 = 14022*71500*79968 = 14080*69615*81795 = 14280*65520*85690 = 14432*63308*87750 = 14820*59040*91630 = 14896*58344*92250 = 16236*49504*99750 = 16380*48790*100320 = 16830*46740*101920 = 17290*44880*103320 = 17589*43776*104125 = 18720*40180*106590 = 19152*39000*107338 = 20090*36720*108680 = 21648*33592*110250 = 23940*30030*111520 = 25840*27720*111930 = 80173757664000
(End)

Crossrefs

Cf. A060277, A119028, A119646, A317578.
See A103278 for least j associated with i = A103277(n).

Programs

  • Mathematica
    tanya[n_] := tanya[n] = Max[Length /@ Split[ Sort[Times @@@ Partition[Last /@ Flatten[ FindInstance[a + b + c == n && a >= b >= c > 0, {a, b, c}, Integers, Round[n^2/12]]], 3]]]];

Extensions

Additional comments and examples from Joseph Biberstine (jrbibers(AT)indiana.edu) and Robert G. Wilson v, Jul 27 2006
Edited by N. J. A. Sloane, Apr 29 2007
a(10)-a(15) from Donovan Johnson, Mar 29 2010
a(16)-a(18) from Duncan Moore, Sep 02 2017

A317578 Number T(n,k) of distinct integers that are product of the parts of exactly k partitions of n into 3 positive parts; triangle T(n,k), n>=3, k>=1, read by rows.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 12, 1, 12, 2, 19, 19, 1, 22, 1, 27, 28, 1, 31, 1, 31, 3, 38, 1, 42, 1, 46, 1, 50, 1, 50, 3, 57, 2, 51, 7, 64, 3, 71, 2, 70, 5, 77, 4, 85, 3, 86, 5, 84, 9, 104, 2, 104, 5, 108, 6, 108, 8, 1, 123, 5, 122, 9, 119, 14, 136, 9, 147, 7
Offset: 3

Views

Author

Alois P. Heinz, Jul 31 2018

Keywords

Examples

			T(13,2) = 1: only 36 is product of the parts of exactly 2 partitions of 13 into 3 positive parts: [6,6,1], [9,2,2].
T(14,2) = 2: 40 ([8,5,1], [10,2,2]) and 72 ([6,6,2], [8,3,3]).
T(39,3) = 1: 1200 ([20,15,4], [24,10,5], [25,8,6]).
T(49,3) = 2: 3024 ([24,18,7], [27,14,8], [28,12,9]) and 3600 ([20,20,9], [24,15,10], [25,12,12]).
Triangle T(n,k) begins:
   1;
   1;
   2;
   3;
   4;
   5;
   7;
   8;
  10;
  12;
  12, 1;
  12, 2;
  19;
  19, 1;
  22, 1;

Links

Crossrefs

Cf. A001399, A060277, A069905, A103277, A211540.
Row sums give A306403.
Column k=1 gives A306435.

Programs

  • Maple
    b:= proc(n) option remember; local m, c, i, j, h, w;
      m, c:= proc() 0 end, 0; forget(m);
      for i to iquo(n, 3) do for j from i to iquo(n-i, 2) do
        h:= i*j*(n-j-i);
        w:= m(h); w:= w+1; m(h):= w;
        c:= c+x^w-x^(w-1)
      od od; c
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
seq(T(n), n=3..100);
  • Mathematica
    b[n_] := b[n] = Module[{m, c, i, j, h, w} , m[_] = 0; c = 0; For[i = 1, i <= Quotient[n, 3], i++, For[j = i, j <= Quotient[n - i, 2], j++, h = i*j*(n-j-i); w = m[h]; w++; m[h] = w; c = c + x^w - x^(w-1)]]; c];
T[n_] := CoefficientList[b[n], x] // Rest;
T /@ Range[3, 100] // Flatten (* Jean-François Alcover, Jun 13 2021, after Alois P. Heinz *)

Formula

Sum_{k>=1} k * T(n,k) = A001399(n-3) = A069905(n) = A211540(n+2).
Sum_{k>=2} T(n,k) = A060277(n).
min { n >= 0 : T(n,k) > 0 } = A103277(k).

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

Views

Author

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

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf. A060277, A316946.

Programs

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

Formula

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

Views

Author

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

Keywords

Comments

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.

Examples

			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].

Links

Crossrefs

Cf. A060277, A316945.

Programs

  • Mathematica
    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

Views

Author

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

Keywords

Examples

			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.

Links

Crossrefs

Cf. A060277, A316945, A316946.

Programs

  • Mathematica
    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}]
Showing 1-5 of 5 results.

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