A257042 a(n) = (3*n+7)*n^2.
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
Examples
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.
Links
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Cf. A189975.
Programs
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Magma
[(3*n+7)*n^2: n in [0..65]]; // Vincenzo Librandi, Apr 15 2015
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Maple
A257042:=n->(3*n+7)*n^2: seq(A257042(n), n=0..50); # Wesley Ivan Hurt, Apr 16 2015
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Mathematica
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 *) -
PARI
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
Formula
From Vincenzo Librandi, Apr 15 2015: (Start)
G.f.: x*(10+12*x-4*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
From Amiram Eldar, Jul 30 2024: (Start)
Sum_{n>=1} 1/a(n) = sqrt(3)*Pi/98 + Pi^2/42 + 9*log(3)/98 - 351/1372.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/84 - sqrt(3)*Pi/49 - 6*log(2)/49 + 225/1372. (End)
Extensions
More terms from Vincenzo Librandi, Apr 15 2015
Comments