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A255765 Partial sums of A255744. 5

%I #21 Nov 03 2022 08:43:09

%S 1,11,21,111,121,211,301,1111,1121,1211,1301,2111,2201,3011,3821,

%T 11111,11121,11211,11301,12111,12201,13011,13821,21111,21201,22011,

%U 22821,30111,30921,38211,45501,111111,111121,111211,111301,112111,112201,113011,113821,121111

%N Partial sums of A255744.

%C Also, this is a row of the square array A255741.

%C Is this sequence related to positive repunits? (see formula section).

%H Felix Fröhlich, <a href="/A255765/b255765.txt">Table of n, a(n) for n = 1..10000</a>

%H Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="https://arxiv.org/abs/2210.10968">Identities and periodic oscillations of divide-and-conquer recurrences splitting at half</a>, arXiv:2210.10968 [cs.DS], 2022, p. 33.

%F Question: a(2^k) = A002275(k+1), k >= 0. Is this true?

%t Accumulate@ MapAt[Floor, Array[10*9^(DigitCount[# - 1, 2, 1] - 1) &, 40], 1] (* _Michael De Vlieger_, Nov 03 2022 *)

%o (PARI) lista(nn) = {s = 1; for (n=2, nn, print1(s, ", "); s += 10*9^(hammingweight(n-1)-1););} \\ _Michel Marcus_, Mar 15 2015

%o (PARI) a(n) = sum(k=1, n, if (k==1, 1, 10*9^(hammingweight(k-1)-1))); \\ _Michel Marcus_, Mar 15 2015

%Y Cf. A002275, A005408, A151788, A147562, A151790, A151781, A151792, A151793, A255740, A255741, A255744, A255764, A255766.

%K nonn

%O 1,2

%A _Omar E. Pol_, Mar 05 2015

%E More terms from _Michel Marcus_, Mar 15 2015

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Last modified April 19 08:28 EDT 2024. Contains 371782 sequences. (Running on oeis4.)