A043306 Sum of all digits in all base-b representations for n, for 2 <= b <= n.
1, 3, 4, 8, 10, 16, 17, 21, 25, 35, 34, 46, 52, 60, 58, 74, 73, 91, 92, 104, 114, 136, 128, 144, 156, 168, 171, 199, 193, 223, 221, 241, 257, 281, 261, 297, 315, 339, 333, 373, 367, 409, 416, 430, 452, 498, 472, 508, 515, 547, 556, 608, 598, 638, 634, 670, 698, 756, 717, 777
Offset: 2
Examples
5 = 101_2 = 12_3 = 11_4 = 10_5. Thus a(5) = 2 + 3 + 2 + 1 = 8.
Links
- Alois P. Heinz, Table of n, a(n) for n = 2..1000
- Robin Fissum, Digit sums and the number of prime factors of the factorial n!=1.2...n, Bachelor's project in BMAT, Norwegian University of Science and Technology, 2020.
- Jan-Christoph Puchta and Jürgen Spilker, Altes und Neues zur Quersumme, Math. Semesterber., Vol. 49, No. 2 (2002), pp. 209-226; alternative link.
- Vladimir Shevelev, Compact integers and factorials, Acta Arith., Vol. 126, No. 3 (2007), pp. 195-236 (see p. 205).
Programs
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Mathematica
Table[Sum[Total[First[RealDigits[n, i]]], {i, 2, n}], {n, 2, 80}] (* Carl Najafi, Aug 16 2011 *) -
PARI
a(n) = sum(i=2, n, vecsum(digits(n, i))); \\ Michel Marcus, Jan 03 2017
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PARI
a(n) = sum(b=2, n, sumdigits(n, b)); \\ Michel Marcus, Aug 18 2017
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Python
from sympy.ntheory.digits import digits def a(n): return sum(sum(digits(n, b)[1:]) for b in range(2, n+1)) print([a(n) for n in range(2, 62)]) # Michael S. Branicky, Apr 04 2022
Formula
From Vladimir Shevelev, Jun 03 2011: (Start)
a(n) = (n-1)*n - Sum_{i=2..n} (i-1)*Sum_{r>=1} floor(n/i^r).
a(n) <= (n-1)^2*log(n+1)/log(n).
Problem: find a better upper estimate. (End)
From Amiram Eldar, Apr 16 2021: (Start)
a(n) = A014837(n) + 1.
a(n) ~ (1-Pi^2/12)*n^2 + O(n^(3/2)) (Fissum, 2020). (End)