A248217 a(n) = 8^n - 2^n.

0, 6, 60, 504, 4080, 32736, 262080, 2097024, 16776960, 134217216, 1073740800, 8589932544, 68719472640, 549755805696, 4398046494720, 35184372056064, 281474976645120, 2251799813554176, 18014398509219840, 144115188075331584, 1152921504605798400

Offset: 0

Author

Vincenzo Librandi, Oct 04 2014

Keywords

• nonn
• easy

Comments

If 2^(n+1) is the length of the even leg of a primitive Pythagorean triangle (PPT) then it constrains the odd leg to have a length of 4^n-1 and the hypotenuse to have a length of 4^n+1. The resulting triangle has a semiperimeter of 4^n+2^n, an area of 8^n-2^n and an inradius of 2^n-1. For n > 0, a(n) is the area of such triangles. - Frank M Jackson, Sep 07 2018

Maximum anomalous cancellation multiplicity of (2n+1)-digit integers: number of (2n+1)-digit integers which can be anomalously canceled with a fixed (2n+1)-digit integer. The maximum is obtained at 88...88911...11 containing n 8's and n 1's (see Example below). Anomalous cancellation is a "canceling" of digits of a and b in the numerator and denominator of a fraction a/b which results in a fraction equal to the original, and no 0 or digits that appear different times in a and b are canceled. For example, 49/98 = 4/8, 138/184 = 3/4, 1985/5955 = 185/555, 88911/43956 = 8811/4356, but 120/340 is not because canceling the 0's is not an anomalous cancellation. - Xiaohan Zhang, Nov 21 2019

Examples

			For n=1, there are 6 numbers with 3 digits that can be anomalously canceled with 891: 297, 396, 495, 594, 693, 792. For n=2 there are 60 numbers with 88911: 12987, 13986, 14985, 15984, 16983, 17982, 21978, 22977, 23976, 24975, 25974, 26973, 27972, 28971, 31968, 32967, 33966, 34965, 35964, 36963, 37962, 38961, 41958, 42957, 43956, 44955, 45954, 46953, 47952, 48951, 51948, 52947, 53946, 54945, 55944, 56943, 57942, 58941, 61938, 62937, 63936, 64935, 65934, 66933, 67932, 68931, 71928, 72927, 73926, 74925, 75924, 76923, 77922, 78921, 82917, 83916, 84915, 85914, 86913, 87912. For n=3 504 numbers with 8889111, and no other (2n+1)-digit number has greater multiplicity. There seems to be a pattern of integer partitions in these examples, because the sum of the digits of numbers above are all multiples of 9. - _Xiaohan Zhang_, Nov 21 2019
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Eric Weisstein's World of Mathematics, Anomalous Cancellation
• Index entries for linear recurrences with constant coefficients, signature (10,-16).

Crossrefs

Cf. similar sequences listed in A248216.

Cf. A000079, A024036.

Programs

Magma

[8^n-2^n: n in [0..25]];

Mathematica

Table[8^n - 2^n, {n, 0, 25}] (* or *) CoefficientList[Series[6 x /((1 - 2 x) (1 - 8 x)), {x, 0, 30}], x]
LinearRecurrence[{10,-16},{0,6},30] (* Harvey P. Dale, Mar 29 2015 *)

PARI

a(n) = 8^n-2^n; \\ Altug Alkan, Sep 07 2018

Python

def A248217(n): return 6*binomial(pow(2,n) +1, 3)
print([A248217(n) for n in range(41)]) # G. C. Greubel, Dec 26 2024

Formula

G.f.: 6*x/((1-2*x)*(1-8*x)).

a(n) = 10*a(n-1) - 16*a(n-2).

a(n) = 2^n*(4^n-1) = A000079(n) * A024036(n) = A001018(n) - A000079(n).

E.g.f.: exp(2*x)*(-1 + exp(6*x)). - Stefano Spezia, Sep 07 2018

a(n) = 6*A016131(n-1). - R. J. Mathar, Mar 10 2022