Garrett Frandson has authored 4 sequences.
A258618
a(n) = (4*n+9)*n^2.
Original entry on oeis.org
0, 13, 68, 189, 400, 725, 1188, 1813, 2624, 3645, 4900, 6413, 8208, 10309, 12740, 15525, 18688, 22253, 26244, 30685, 35600, 41013, 46948, 53429, 60480, 68125, 76388, 85293, 94864, 105125, 116100, 127813, 140288, 153549, 167620, 182525, 198288, 214933
Offset: 0
The smallest integer that satisfies this is 240: It has 19 proper divisors (1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120) and 6 prime factors (2, 2, 2, 2, 3, 5), so d(240)=13. The square of 240, 57600, we would expect to have a difference of 68 between the number of its proper divisors and prime factors, and with respectively 80 and 12, d(57600)=68 indeed. Checking this with further integer powers of 240 will continue to generate terms in this sequence.
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[(4*n+9)*n^2: n in [0..40]]; // Vincenzo Librandi, Jun 06 2015
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Table[(4 n + 9) n^2, {n, 0, 40}] (* Vincenzo Librandi, Jun 06 2015 *)
LinearRecurrence[{4,-6,4,-1},{0,13,68,189},40] (* Harvey P. Dale, Sep 12 2020 *)
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vector(50,n,n--;(4*n+9)*n^2) \\ Derek Orr, Jun 21 2015
A258617
a(n) = (4*n+8)*n^2.
Original entry on oeis.org
0, 12, 64, 180, 384, 700, 1152, 1764, 2560, 3564, 4800, 6292, 8064, 10140, 12544, 15300, 18432, 21964, 25920, 30324, 35200, 40572, 46464, 52900, 59904, 67500, 75712, 84564, 94080, 104284, 115200, 126852, 139264, 152460, 166464, 181300, 196992, 213564
Offset: 0
The smallest integer that satisfies the (17, 5) requirement is 180: it has 17 proper divisors (1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90) and 5 prime factors (2, 2, 3, 3, 5), so d(120)=12=a(1).
The square of 180, 32400, we would expect to have a difference of 64 between the number of its proper divisors and prime factors, and with respectively 74 and 10, d(32400)=64=a(2) indeed. Checking this with further integer powers of 180 will continue to generate terms in this sequence.
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[(4*n+8)*n^2: n in [0..50]]; // Vincenzo Librandi, Jun 06 2015
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I:=[0, 12, 64, 180]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015
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Table[(4 n + 8) n^2, {n, 0, 40}] (* or *) CoefficientList[Series[4 x (3 + 4 x - x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2015 *)
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vector(50,n,n--;(4*n+8)*n^2) \\ Derek Orr, Jun 21 2015
A258402
a(n) = (n^2 + 4*n + 6) * n^2.
Original entry on oeis.org
0, 11, 72, 243, 608, 1275, 2376, 4067, 6528, 9963, 14600, 20691, 28512, 38363, 50568, 65475, 83456, 104907, 130248, 159923, 194400, 234171, 279752, 331683, 390528, 456875, 531336, 614547, 707168, 809883, 923400, 1048451, 1185792, 1336203, 1500488, 1679475
Offset: 0
The smallest integer which satisfies this is 210: It has 15 proper divisors (1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105) and 4 prime factors (2, 3, 5, 7), so d(210) = 11. The square of 210, 44100, we would expect to have a difference of 72 between the number of its proper divisors and prime factors, and with respectively 80 and 8, d(44100) = 72 indeed. Checking this with further integer powers of 210 will continue to generate terms in this sequence.
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[(n^2+4*n+6)*n^2: n in [0..40]]; // Vincenzo Librandi, Jun 06 2015
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I:=[0,11, 72,243,608]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015
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Table[(n^2 + 4n + 6) * n^2, {n, 0, 39}] (* Alonso del Arte, Jun 06 2015 *)
CoefficientList[Series[x (11 + 17 x - 7 x^2 + 3 x^3)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2015 *)
LinearRecurrence[{5,-10,10,-5,1},{0,11,72,243,608},40] (* Harvey P. Dale, May 05 2018 *)
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a(n)=(n^2+4*n+6)*n^2 \\ Charles R Greathouse IV, Jun 15 2015
A257042
a(n) = (3*n+7)*n^2.
Original entry on oeis.org
0, 10, 52, 144, 304, 550, 900, 1372, 1984, 2754, 3700, 4840, 6192, 7774, 9604, 11700, 14080, 16762, 19764, 23104, 26800, 30870, 35332, 40204, 45504, 51250, 57460, 64152, 71344, 79054, 87300, 96100, 105472, 115434, 126004, 137200, 149040, 161542, 174724
Offset: 0
The smallest integer that satisfies this is 120: it has 15 proper divisors (1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60) and 5 prime factors (2, 2, 2, 3, 5), so d(120)=10. The square of 120, 14400, we would expect to have a difference of 52 between the number of its proper divisors and prime factors, and with respectively 62 and 10, d(120)=52 indeed. Checking this with further integer powers of 120 will continue to generate terms in this sequence.
The integers which satisfy the proper-divisor-prime-factor requirement are those of A189975.
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[(3*n+7)*n^2: n in [0..65]]; // Vincenzo Librandi, Apr 15 2015
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A257042:=n->(3*n+7)*n^2: seq(A257042(n), n=0..50); # Wesley Ivan Hurt, Apr 16 2015
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Table[(3 n + 7) n^2, {n, 40}] (* or *) CoefficientList[Series[(10 + 12 x - 4 x^2) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 15 2015 *)
LinearRecurrence[{4,-6,4,-1},{0,10,52,144},40] (* Harvey P. Dale, Mar 27 2025 *)
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lista(nn) = {v = 1; while(!((numdiv(v)-1 == 15) && (bigomega(v) == 5)), v++); for (n=0, nn, vn = v^n; nb = numdiv(vn)-1-bigomega(vn); print1(nb, ", "););} \\ Michel Marcus, Apr 16 2015
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