A353181 -id:A353181 - cp's OEIS frontend

A353182 Number of n-digit numbers in which more than half of the digits are the same.

9, 9, 252, 333, 7704, 11430, 245520, 388485, 8018280, 13221234, 266135472, 451623042, 8935693776, 15488764524, 302623991712, 533189070405, 10317992397480, 18416195186490, 353689409441520, 637974569854998, 12177584747670384, 22158431087271444, 420819143651579232, 771390571080374658

Offset: 1

Zhining Yang, Apr 29 2022

a(n) is the number of terms between 10^(n-1) and 10^n in A353181.

			a(2)=9 because there are 9 numbers whose digits are the same, more than half of the length 2.

Zhining Yang, Table of n, a(n) for n = 1..300
Project Euler, Problem 788, Dominating Numbers

Cf. A353181, A353183 (partial sums).

a[n_]:=Sum[Binomial[n,k]9^(n-k+1),{k,Floor[n/2]+1,n}]; Array[a,24] (* Stefano Spezia, Apr 29 2022 *)

import math
def a(n):
  return (sum(math.comb(n,i)*9**(n-i+1) for i in range(n//2+1,n+1)))
print([a(n) for n in range(1, 31)])

def a(n):
    r=[0, 9, 9]
    for i in range(n):
        r.append(-((5400+8280*i+2880*i*i)*r[i]+(-360-828*i-288*i*i)*r[i+1]+(-228-310*i-80*i*i)*r[i+2])//(21+31*i+8*i*i))
    return r[n]
print([a(i) for i in range(1, 21)]) # Xianwen Wang, May 02 2022

a(n) = Sum_{k=floor(n/2)+1..n} C(n,k)*9^(n-k+1) (thanks to Zhao Hui Du for help in the derivation of this formula).

a(n+3) = (-(5400+8280*n+2880*n^2)*a(n)+(360+828*n+288*n^2)*a(n+1)+(228+310*n+80*n^2)*a(n+2))/(21+31*n+8*n^2) (thanks to Xianwen Wang for help in the derivation of this formula).

A353183 Number of numbers < 10^n in which more than half of the digits are the same.

Original entry on oeis.org

9, 18, 270, 603, 8307, 19737, 265257, 653742, 8672022, 21893256, 288028728, 739651770, 9675345546, 25164110070, 327788101782, 860977172187, 11178969569667, 29595164756157, 383284574197677, 1021259144052675, 13198843891723059, 35357274978994503, 456176418630573735, 1227566989710948393

Offset: 1

Zhining Yang, Apr 29 2022

			a(2) = 18 because there are 18 numbers less than 10^2 in which more than half of the digits are the same: {1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99}.

Zhining Yang, Table of n, a(n) for n = 1..300
Project Euler, Problem 788. Dominating Numbers

Cf. A353181, A353182 (first differences).

a[n_]:=Sum[Sum[Binomial[m,k]9^(m-k+1),{k,Floor[m/2]+1,m}],{m,1,n}]; Array[a,24] (* Stefano Spezia, Apr 29 2022 *)

import math
def a(n):
    return(sum(sum(math.comb(m,i)*9**(m-i+1) for i in range(m//2+1, m+1)) for m in range(1, n+1)))
print([a(i) for i in range(1, 21)])

def a(n):
    r=[0, 9, 18, 270, 603]
    for i in range(n):
        r.append(-((1440+720*i)*r[i]+(-3024-1152*i)*r[1+i]+(1668+448*i)*r[2+i]+(-28-4*i)*r[3+i]+(-61-13*i)*r[4+i])//(5+i))
    return r[n]
print([a(i) for i in range(1, 21)])

a(n) = Sum_{m=1..n} Sum_{k=floor(m/2)+1..m} C(m,k)*9^(m-k+1).
a(n+4) = ((16560 + 14040*n + 2880*n^2)*a(n) - (18036 + 15444*n + 3168*n^2)*a(n+1) + (858 + 934*n + 208*n^2)*a(n+2) + (678 + 517*n + 88*n^2)*a(n+3))/(60 + 47*n + 8*n^2).
a(n+5) = -((1440 + 720*n)*a(n) + (-3024 - 1152*n)*a(n+1) + (1668 + 448*n)*a(n+2) + (-28 - 4*n)*a(n+3) + (-61 - 13*n)*a(n+4))/(5+n).

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A353182 Number of n-digit numbers in which more than half of the digits are the same.

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A353183 Number of numbers < 10^n in which more than half of the digits are the same.

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Author

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Programs

Mathematica

Python

Python

Formula