A337876
Table read by rows, in which the n-th row lists all the primitive solutions k, in increasing order, such that k*sigma(k) = A337875(n).
Original entry on oeis.org
12, 14, 48, 62, 112, 124, 160, 189, 192, 254, 315, 351, 448, 508, 1984, 2032, 2560, 2728, 5580, 5616, 6156, 6534, 12288, 16382, 22464, 22860, 28672, 32764, 28800, 34000, 42000, 51200, 46500, 51200, 51200, 54250, 72800, 95697, 76230, 80028, 126976, 131056, 119700, 189875
Offset: 1
The table begins:
12, 14;
48, 62;
112, 124;
160, 189;
192, 254;
315, 351;
...
1st row is (12, 14) because 12 * sigma(12) = 14 * sigma(14) = 336 = A337875(1) with p = 2 and r = 3.
2nd row is (48, 62) because 48 * sigma(48) = 62 * sigma(62) = 5952 = A337875(2) with p = 2 and r = 5.
16th row is (42000, 51200), (46500, 51200), (51200, 54250) because 42000 * sigma(42000) = 51200 * sigma(51200), 46500 * sigma(46500) = 51200 * sigma(51200) and 51200 * sigma(51200) = sigma54250 * sigma(54250) = 649999584000 = A337875(16). These 3 primitive solutions corresponding to the smallest m = 649999584000 have been found by _Michel Marcus_. The three other possible solutions (42000, 46500), (42000, 54250), (46500, 54250) are not primitive.
18th row is (76230, 80028) because 76230 * sigma(76230) = 80028 * sigma(80028) = 18979440480 = A337875(18). Note that 76230 * sigma(76230) = 80028 * sigma(80028) = 84942 * sigma(84942) = 18979440480 = A337873(3266) but (76230, 84942) and (80028, 84942) are not primitive solutions (see detailed example in A337875). These case have been found by _Jinyuan Wang_.
- Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B11, p. 101-102.
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process(x, y, resp) = {my(vresp = Vec(resp)); for (i=1, #vresp, if (x/vresp[i][1] == y/vresp[i][2], return(resp));); listput(resp, [x, y]); resp;}
findprim(res, mx) = {my(mp = Map()); my(resp = List()); for (i=1, #res, my(vx = mapget(mx, res[i])); for (j=1, #vx-1, for (k=j+1, #vx, resp = process(vx[j], vx[k], resp);););); resp;}
upto(n) = {my(m = Map(), mx = Map(), res = List(), n = sqrtint(n), resp); for(i = 1, n, my(c = i*sigma(i)); if(mapisdefined(m, c), listput(res, c); mapput(m, c, mapget(m, c) + 1); mapput(mx, c, concat(mapget(mx, c), i)), mapput(m, c, 1); mapput(mx, c, [i]);)); listsort(res, 1); res = Vec(select(x -> x <= (n+1)^2, res)); Vec(findprim(res, mx));}
upto(10^11) \\ Michel Marcus, Oct 20 2020
A337873
Numbers m such that the equation m = k*sigma(k) has more than one solution.
Original entry on oeis.org
336, 5952, 10080, 27776, 44352, 60480, 61152, 97536, 102816, 127680, 178560, 185472, 196560, 260400, 292320, 333312, 455168, 472416, 578592, 635712, 758016, 785664, 833280, 961632, 1083264, 1179360, 1189440, 1270752, 1330560, 1530816, 1717632, 1815072, 1821312, 1834560
Offset: 1
For a(1): 12 * sigma(12) = 14 * sigma(14) = 336 with p=2 and r=3.
For a(2): 48 * sigma(48) = 62 * sigma(62) = 5952 with p=2 and r=5.
For a(3): 60 * sigma(60) = 70 * sigma(70) = 10080 with 60/12 = 70/14 = 5.
a(16) = 333312 is the smallest term with 3 preimages because 336 * sigma(336) = 372 * sigma(372) = 434 * sigma(434) = 333312.
- Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B11, p. 101-102.
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m = 2*10^6; v = Table[0, {m}]; Do[i = n*DivisorSigma[1, n]; If[i <= m, v[[i]]++], {n, 1, Floor@Sqrt[m]}]; Position[v, ?(# > 1 &)] // Flatten (* _Amiram Eldar, Sep 28 2020 *)
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upto(n) = {m = Map(); res = List(); n = sqrtint(n); for(i = 1, n, c = i*sigma(i); if(mapisdefined(m, c), listput(res, c); mapput(m, c, mapget(m, c) + 1) , mapput(m, c, 1); ) ); listsort(res, 1); select(x -> x <= (n+1)^2, res) } \\ David A. Corneth, Sep 27 2020
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isok(m) = {my(nb=0); fordiv(m, d, if (d*sigma(d) == m, nb++; if (nb>1, return(1)));); return (0);} \\ Michel Marcus, Sep 29 2020
A337874
Table read by rows, in which the n-th row lists all the preimages k, in increasing order, such that k*sigma(k) = A337873(n).
Original entry on oeis.org
12, 14, 48, 62, 60, 70, 112, 124, 132, 154, 160, 189, 156, 182, 192, 254, 204, 238, 228, 266, 240, 310, 276, 322, 315, 351, 300, 350, 348, 406, 336, 372, 434, 448, 508, 444, 518, 492, 574, 516, 602, 564, 658, 528, 682, 560, 620, 636, 742
Offset: 1
The table begins:
12, 14;
48, 62;
60, 70;
112, 124;
132, 154;
160, 189;
...
1st row is (12, 14) because 12 * sigma(12) = 14 * sigma(14) = 336 = A337873(1) with p = 2 and r = 3.
2nd row is (48, 62) because 48 * sigma(48) = 62 * sigma(62) = 5952 = A337873(2) with p = 2 and r = 5.
16th row is (336, 372, 434) because 336 * sigma(336) = 372 * sigma(372) = 434 * sigma(434) = 333312 = A337873(16).
- Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B11, p. 101-102.
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m = 10^6; v = Table[{}, {m}]; Do[i = n*DivisorSigma[1, n]; If[i <= m, AppendTo[v[[i]], n]], {n, 1, Floor@Sqrt[m]}]; Select[v, Length[#] > 1 &] // Flatten (* Amiram Eldar, Oct 06 2020 *)
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upto(n) = {m = Map(); res = List(); n = sqrtint(n); w = []; for(i = 1, n, c = i*sigma(i); if(mapisdefined(m, c), listput(res, c); l = mapget(m, c); listput(l, i); mapput(m, c, l) , mapput(m, c, List(i)); ) ); listsort(res, 1); v = select(x -> x <= (n+1)^2, res); for(i = 1, #v, w = concat(w, Vec(mapget(m, v[i]))) ); w } \\ David A. Corneth, Oct 07 2020
A338384
Integers that can be written m = k*tau(k) = q*tau(q) where (k, q) is a primitive solution of this equation and tau(k) is the number of divisors of k.
Original entry on oeis.org
108, 192, 448, 2688, 6000, 8640, 12960, 17496, 18750, 20412, 32400, 86400, 112640, 120960, 138240, 169344, 181440, 245760, 304128, 600000, 658560, 714420, 857304, 979776, 1350000, 1632960, 1778112, 2073600, 2361960, 3359232, 3500000, 4561920, 7112448
Offset: 1
-> For a(1): 18 * tau(18) = 27 * tau(27) = 108.
-> For a(2): 24 * tau(24) = 32 * tau(32) = 192.
-> Why 1080 = A338382(4) is not a term? 90 * tau(90) = 135 * tau(135) = 1080 but as 90/5 = 18 and 135/5 = 27, this solution that is generated by the first example is not primitive.
-> For a(4) : 168 * tau(168) = 192 * tau(192) = 224 * tau(224) = A338382(8) = 2688.
1) for k=168 and q=192; with d=3, k/3=56 and q/3=64, with 56 * tau(56) = 64 * tau(64) = 448 = a(3), hence (168, 192) is not a primitive solution;
2) for k=168 and q=224; with d=7, k/7=24 and q/7=32, with 24 * tau(24) = 32 * tau(32) = 192 = a(2), hence (24, 32) is not a primitive solution; but
3) for k=192 and q=224, there is no common divisor d such that 192/d and 224/d can satisfy (192/d)*tau(192/d) = (224/d)*tau(224/d), so (192, 224) is a primitive solution linked to m = 2688 that is the term a(4).
- Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B12, p. 102-103.
- D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, entry 168, page 127.
Cf.
A337875 (similar for k*sigma(k))
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is(n) = {my(l, d); l = List(); d = divisors(n); for(i = 1, #d, if(d[i]*numdiv(d[i]) == n, listput(l, d[i]); ) ); forvec(x = vector(2, i, [1, #l]), if(isprimitive(l[x[1]], l[x[2]], n), return(1) ) , 2 ); 0 }
isprimitive(m, n, t) = { my(g = gcd(m, n), d = divisors(g)); for(i = 2, #d, if(m/d[i]*numdiv(m/d[i]) == t/d[i]/numdiv(d[i]) && n/d[i]*numdiv(n/d[i]) == t/d[i]/numdiv(d[i]), return(0) ) ); 1 } \\ David A. Corneth, Nov 06 2020
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