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User: Thomas Quirk

Thomas Quirk has authored 2 sequences.

A204692 a(n) is the number of base-10 bouncy numbers below 10^n.

0, 0, 525, 8325, 95046, 987048, 9969183, 99932013, 999859093, 9999722967, 99999479435, 999999059545, 9999998358645, 99999997221695, 999999995423887, 9999999992645451, 99999999988439336, 999999999982190246, 9999999999973063281, 99999999999959940181, 999999999999941340896

Offset: 1

Thomas Quirk, Jan 18 2012

A bouncy number has digits that are neither monotonically increasing nor decreasing from left to right.

			a(3) = 525 because there are 525 base-10 bouncy numbers < 10^3; the smallest is A152054(1) = 101 and the largest one is A152054(525) = 989.
a(4) = 8325 because A152054(8325) = 9989 is the largest base-10 bouncy number < 10^4.

Muniru A Asiru, Table of n, a(n) for n = 1..200
Project Euler, Non-bouncy numbers Problem 113
Index entries for linear recurrences with constant coefficients, signature (21,-165,715,-1980,3762,-5082,4950,-3465,1705,-561,111,-10).

Cf. A152054.

a:=[0,0];; for n in [3..25] do a[n]:=10^n-(Binomial(n+10,10)+Binomial(n+9,9)-1-10*n); od; a; # Muniru A Asiru, Sep 28 2018

[n le 2 select 0 else (10^n - Binomial(n+10, 10) - Binomial(n+9, 9) + 10*n + 1): n in [1..25]]; // G. C. Greubel, Nov 24 2018

a:=n->`if`(n<=2,0,10^n-(binomial(n+10,10)+binomial(n+9,9)-1-10*n)); seq(a(n),n=1..25); # Muniru A Asiru, Sep 28 2018

a[n_]:= 10^n - Binomial[n+10, 10] - Binomial[n+9, 9] + 10*n + 1; Array[a, 25] (* Stefano Spezia, Sep 29 2018 *)

a(n) = if(n<3, 0, 10^n-(binomial(n+10,10)+binomial(n+9, 9)-1-10*n)); \\ Altug Alkan, Sep 27 2018

def A204692(n):
    if n <= 2:
        return 0
    else:
        return (10^n - binomial(n+10, 10) - binomial(n+9, 9) + 10*n + 1)
[A204692(n) for n in (1..25)] # G. C. Greubel, Nov 24 2018

a(1) = a(2) = 0. For n > 2, a(n) = 10^n - binomial(n+9, n) + 10*n - Sum_{i=1..n} binomial(i+9, i) = 10^n - binomial(n+10, 10) - binomial(n+9, 9) + 10*n + 1. [Corrected by Altug Alkan, Sep 28 2018]
From Chai Wah Wu, Jul 28 2023: (Start)
a(n) = 21*a(n-1) - 165*a(n-2) + 715*a(n-3) - 1980*a(n-4) + 3762*a(n-5) - 5082*a(n-6) + 4950*a(n-7) - 3465*a(n-8) + 1705*a(n-9) - 561*a(n-10) + 111*a(n-11) - 10*a(n-12) for n > 12.
G.f.: x^3*(89*x^8 - 798*x^7 + 3175*x^6 - 7350*x^5 + 10890*x^4 - 10668*x^3 + 6846*x^2 - 2700*x + 525)/((x - 1)^11*(10*x - 1)). (End)

Corrected and extended by Altug Alkan, Sep 27 2018

A176974 First exponent n to generate maximum remainder when (a + 1)^n + (a - 1)^n is divided by a^2 for variable n and a>2.

Original entry on oeis.org

1, 1, 7, 5, 3, 3, 13, 9, 5, 5, 19, 13, 7, 7, 25, 17, 9, 9, 31, 21, 11, 11, 37, 25, 13, 13, 43, 29, 15, 15, 49, 33, 17, 17, 55, 37, 19, 19, 61, 41, 21, 21, 67, 45, 23, 23, 73, 49, 25, 25, 79, 53, 27, 27, 85, 57, 29, 29, 91, 61, 31, 31, 97, 65, 33, 33, 103, 69

Offset: 3

Thomas Quirk, Apr 29 2010

Colin Barker, Table of n, a(n) for n = 3..1000
Project Euler, Problem 120: Square remainders
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A159469.

Vec(x^3*(1 + x + 7*x^2 + 5*x^3 + x^4 + x^5 - x^6 - x^7) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2) + O(x^100)) \\ Colin Barker, Oct 29 2017

From Colin Barker, Oct 29 2017: (Start)
G.f.: x^3*(1 + x + 7*x^2 + 5*x^3 + x^4 + x^5 - x^6 - x^7) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2).
a(n) = 2*a(n-4) - a(n-8) for n>10.
(End)

Corrected and extended by Ray Chandler, Oct 16 2011

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User: Thomas Quirk

A204692 a(n) is the number of base-10 bouncy numbers below 10^n.

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