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
A337875
Integers that can be written m = k*sigma(k) = q*sigma(q) where (k, q) is a primitive solution of this equation and sigma(m) is the sum of divisors of (m).
Original entry on oeis.org
336, 5952, 27776, 60480, 97536, 196560, 455168, 8062976, 15713280, 97493760, 104282640, 402604032, 1597639680, 1878818816, 2959632000, 6499584000, 15923980800, 18979440480, 33281933312, 54027792000, 102953410560, 103078428672, 103448378880
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.
10080 is not a term: 60 * sigma(60) = 70 * sigma(70) = 10080 but as 60/5 = 12 and 70/5 = 14, hence, this solution that is generated by the first example is not primitive.
For a(4): 160 * sigma(160) = 189 * sigma(189) = 60480 is the smallest example with gcd(k,q) = 1 with k = 2^5*5 = 160 and q = 3^3*7 = 189.
For a(6): 315 * sigma(315) = 351 * sigma(351) = 196560 is the smallest example with k and q both odd.
For a(18): 76230 * sigma(76230) = 80028 * sigma(80028) = 84942 * sigma(84942) = A337873(3266) = 18979440480.
-> 1) for k=76230 and q=84942; with d=11^2, k/11^2=630 and q/11^2=702.
630 * sigma(630) = 702 * sigma(702) = 1179360, hence (76230, 84942) is not a primitive solution;
-> 2) for k=80028 and q=84942; with d=13, k/13=6156 and q/13=6534.
6156 * sigma(6156) = 6534 * sigma(6534) = 104282640, hence (80028, 84942) is not a primitive solution; but
-> 3) for k=76230 and q=80028, there is no common divisor d such that k/d and q/d can satisfy (k/d)*sigma(k/d) = (q/d)*sigma(q/d), so (76239, 80028) is a primitive solution linked to m = 18979440480 that is the term a(18).
- 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)); resp = findprim(res, mx); vresp = Vec(resp); vecsort(vector(#vresp, k, vresp[k][1]*sigma(vresp[k][1])),,8);}
upto(10^12) \\ Michel Marcus, Oct 17 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
A338385
Table read by rows, in which the n-th row lists the primitive solutions (k, q), kA338384(n).
Original entry on oeis.org
18, 27, 24, 32, 56, 64, 192, 224, 400, 500, 360, 432, 540, 648, 972, 2187, 1875, 3125, 1458, 1701, 1296, 1350, 2160, 2400, 5120, 5632, 2880, 3024, 3840, 4608, 4032, 4704, 3780, 5184, 10240, 16384, 8448, 9216, 20000, 25000, 15680, 16464, 15876, 25515, 20412, 23814
Offset: 1
The table begins:
18, 27;
24, 32;
56, 64;
192, 224;
400, 500;
360, 432;
...
1st row is (18, 27) because 18 * tau(18) = 27 * tau(27) = 108 = A338384(1).
4th row is (192, 224) because 192 * tau(192) = 224 * tau(24) = 2688 = A338384(4); Note that 168 * tau(168) = 192 * tau(192) = 224 * tau(24) = 2688 = A338382(8) but (168, 192) and (168, 224) are not primitive solutions (see detailed example in A338384).
5th row is (400, 500) because 400 * tau(400) = 500 * tau(500) = 6000.
20th row is (20000, 25000) although (20000/50, 25000/50) = (400, 500) and that (400, 500) is the 5th row. Explanation: A338384(20) = 600000 = 20000*tau(20000) = 25000*tau(25000) and this pair is primitive, because for d = 50, we get 600000/(50*tau(50)) = 2000 <> (20000/50)*tau(20000/50) = (25000/50)*tau(25000/50) = 6000. To be exhaustive, the two other pairs linked with 600000: (15000, 20000) and (15000, 25000) are not primitive.
- 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.
A337876 (similar for k*sigma(k)).
A371419
Lesser member of Carmichael's variant of amicable pair: numbers k < m such that s(k) = m and s(m) = k, where s(k) = A371418(k).
Original entry on oeis.org
12, 48, 112, 160, 192, 448, 1984, 12288, 28672, 126976, 196608, 458752, 520192, 786432, 1835008, 2031616, 8126464, 8323072, 33292288, 536805376, 2147221504, 3221225472, 7516192768, 33285996544, 34359476224, 136365211648
Offset: 1
12 is a term since A371418(12) = 14 > 12, and A371418(14) = 12.
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r[n_] := n/FactorInteger[n][[1, 1]]; s[n_] := r[DivisorSigma[1, n]]; seq = {}; Do[m = s[n]; If[m > n && s[m] == n, AppendTo[seq, n]], {n, 1, 10^6}]; seq
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f(n) = {my(s = sigma(n)); if(s == 1, 1, s/factor(s)[1, 1]);}
lista(nmax) = {my(m); for(n = 1, nmax, m = f(n); if(m > n && f(m) == n, print1(n, ", ")));}
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