A010719 -id:A010719 - cp's OEIS frontend

A159469 Maximum remainder when (k + 1)^n + (k - 1)^n is divided by k^2 for variable n and k > 2.

6, 8, 20, 24, 42, 48, 72, 80, 110, 120, 156, 168, 210, 224, 272, 288, 342, 360, 420, 440, 506, 528, 600, 624, 702, 728, 812, 840, 930, 960, 1056, 1088, 1190, 1224, 1332, 1368, 1482, 1520, 1640, 1680, 1806, 1848, 1980, 2024, 2162, 2208, 2352, 2400, 2550, 2600

Offset: 3

Gaurav Kumar, Apr 13 2009

			For n = 3, maxr => 3*3 - 3 = 6 since 3 is odd.
For n = 4, maxr => 4*4 - 2*4 = 8 since 4 is even.

Iain Fox, Table of n, a(n) for n = 3..10000
Project Euler, Problem 120: Square remainders
Iain Fox, Proof for recursive definition and generating function
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A050187.

LinearRecurrence[{1,2,-2,-1,1},{6,8,20,24,42},50] (* Harvey P. Dale, Apr 18 2018 *)

a(n) = if (n % 2, n^2 - n, n^2 - 2*n); \\ Michel Marcus, Aug 26 2013

first(n) = Vec(x^3*(-6-2*x)/((x+1)^2*(x-1)^3) + O(x^(n+3))) \\ Iain Fox, Nov 26 2017

maxr(n) = n*n - 2*n if n is even, and n*n - n if n is odd.
G.f.: x^3*(-6-2*x)/((x+1)^2*(x-1)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009 (proved by Iain Fox, Nov 26 2017)
a(n) = 2*A050187(n). - R. J. Mathar, Aug 08 2009 (proved by Iain Fox, Nov 27 2017)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 7. - Colin Barker, Oct 29 2017 (proved by Iain Fox, Nov 26 2017)
a(n) = n^2 - n*(3 + (-1)^n)/2. - Iain Fox, Nov 26 2017
From Iain Fox, Nov 27 2017: (Start)
a(n) = A000290(n) - A022998(n).
a(n) = 2*A093005(n-2) + A168273(n-1).
a(n) = (4*(A152749(n-2)) + A091574(n-1) - A010719(n-1))/3.
E.g.f.: x*(exp(x)*x - sinh(x)).
(End)

A176323 Decimal expansion of (10+sqrt(110))/4.

Original entry on oeis.org

5, 1, 2, 2, 0, 2, 2, 1, 2, 0, 4, 2, 5, 3, 7, 8, 8, 6, 7, 4, 7, 8, 6, 3, 3, 7, 8, 4, 1, 9, 9, 8, 4, 3, 9, 9, 6, 1, 8, 8, 1, 7, 9, 6, 4, 4, 2, 0, 3, 7, 5, 9, 9, 6, 2, 1, 8, 9, 3, 8, 9, 4, 0, 8, 9, 5, 0, 0, 1, 4, 8, 1, 3, 7, 5, 2, 7, 5, 1, 7, 2, 8, 5, 4, 8, 4, 6, 3, 2, 2, 2, 3, 3, 2, 9, 8, 6, 0, 4, 4, 5, 0, 9, 9, 1

Offset: 1

Klaus Brockhaus, Apr 15 2010

Continued fraction expansion of (10+sqrt(110))/4 is A010719.

			5.12202212042537886747...

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A176221 (decimal expansion of sqrt(110)), A010719 (repeat 5, 8).

SetDefaultRealField(RealField(100)); (10+Sqrt(110))/4; // G. C. Greubel, Dec 05 2019

evalf( (10+sqrt(110))/4, 100); # G. C. Greubel, Dec 05 2019

RealDigits[(10+Sqrt[110])/4,10,120][[1]] (* Harvey P. Dale, Jun 09 2016 *)

default(realprecision, 100); (10+sqrt(110))/4 \\ G. C. Greubel, Dec 05 2019

numerical_approx((10+sqrt(110))/4, digits=100) # G. C. Greubel, Dec 05 2019

A136188 Digital roots of the Fermat numbers in A000215(n).

Original entry on oeis.org

3, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8

Offset: 0

Ant King, Dec 24 2007

As 2^(2^n)+1=5 (mod 9) for odd values of n and 2^(2^n)+1=8 (mod 9) for even values of n>0, it follows that the digital roots of the Fermat numbers form a cyclic sequence, with the 5's corresponding to odd values of n and the 8's to even values of n.

Decimal expansion of 71/198. - Enrique Pérez Herrero, Nov 13 2021

			2^(2^3) + 1 = 257. This has digital root 5 and hence a(3) = 5.

I. Izmirli, On Some Properties of Digital Roots, Advances in Pure Mathematics, Vol. 4 No. 6 (2014), Article ID:47285.
Eric Weisstein's World of Mathematics, Digital Root.
Eric Weisstein's World of Mathematics, Fermat Number.
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A000215, A010888, A135928.

Essentially the same as A010719.

FermatNumber[n_]:=2^(2^n)+1;DigitalRoot[n_]:=FixedPoint[Plus@@IntegerDigits[ # ]&,n];DigitalRoot/@(FermatNumber[ # ] &/@Range[0,25])

a(n)=if(n,if(n%2,5,8),3) \\ Charles R Greathouse IV, May 01 2016

a(n) = A010888(A000215(n)).

A176455 Decimal expansion of (20+2*sqrt(110))/5.

Original entry on oeis.org

8, 1, 9, 5, 2, 3, 5, 3, 9, 2, 6, 8, 0, 6, 0, 6, 1, 8, 7, 9, 6, 5, 8, 1, 4, 0, 5, 4, 7, 1, 9, 7, 5, 0, 3, 9, 3, 9, 0, 1, 0, 8, 7, 4, 3, 0, 7, 2, 6, 0, 1, 5, 9, 3, 9, 5, 0, 3, 0, 2, 3, 0, 5, 4, 3, 2, 0, 0, 2, 3, 7, 0, 2, 0, 0, 4, 4, 0, 2, 7, 6, 5, 6, 7, 7, 5, 4, 1, 1, 5, 5, 7, 3, 2, 7, 7, 7, 6, 7, 1, 2, 1, 5, 8, 5

Offset: 1

Klaus Brockhaus, Apr 20 2010

Continued fraction expansion of (20+2*sqrt(110))/5 is A010719 preceded by 8.

			(20+2*sqrt(110))/5 = 8.19523539268060618796...

Cf. A176221 (decimal expansion of sqrt(110)), A010719 (repeat 5, 8).

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A159469 Maximum remainder when (k + 1)^n + (k - 1)^n is divided by k^2 for variable n and k > 2.

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A176323 Decimal expansion of (10+sqrt(110))/4.

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A136188 Digital roots of the Fermat numbers in A000215(n).

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A176455 Decimal expansion of (20+2*sqrt(110))/5.

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