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 A353183 Number of numbers < 10^n in which more than half of the digits are the same. 2
 9, 18, 270, 603, 8307, 19737, 265257, 653742, 8672022, 21893256, 288028728, 739651770, 9675345546, 25164110070, 327788101782, 860977172187, 11178969569667, 29595164756157, 383284574197677, 1021259144052675, 13198843891723059, 35357274978994503, 456176418630573735, 1227566989710948393 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS Zhining Yang, Table of n, a(n) for n = 1..300 Project Euler, Problem 788. Dominating Numbers FORMULA 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). EXAMPLE 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}. MATHEMATICA 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 *) PROG (Python) 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)]) (Python) 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)]) CROSSREFS Cf. A353181, A353182 (first differences). Sequence in context: A050685 A278588 A133361 * A005400 A134115 A071587 Adjacent sequences: A353180 A353181 A353182 * A353184 A353185 A353186 KEYWORD nonn,base,easy AUTHOR Zhining Yang, Apr 29 2022 STATUS approved

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Last modified December 4 16:44 EST 2023. Contains 367563 sequences. (Running on oeis4.)