A255765 Partial sums of A255744.

1, 11, 21, 111, 121, 211, 301, 1111, 1121, 1211, 1301, 2111, 2201, 3011, 3821, 11111, 11121, 11211, 11301, 12111, 12201, 13011, 13821, 21111, 21201, 22011, 22821, 30111, 30921, 38211, 45501, 111111, 111121, 111211, 111301, 112111, 112201, 113011, 113821, 121111

Offset: 1

Author

Omar E. Pol, Mar 05 2015

Keywords

• nonn

Comments

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

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

Links

• Felix Fröhlich, Table of n, a(n) for n = 1..10000
• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 33.

Crossrefs

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

Programs

Mathematica

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

PARI

lista(nn) = {s = 1; for (n=2, nn, print1(s, ", "); s += 10*9^(hammingweight(n-1)-1););} \\ Michel Marcus, Mar 15 2015

PARI

a(n) = sum(k=1, n, if (k==1, 1, 10*9^(hammingweight(k-1)-1))); \\ Michel Marcus, Mar 15 2015

Formula

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

Extensions

More terms from Michel Marcus, Mar 15 2015