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a(n) is obtained by first replacing with zeros all other digits except the leftmost (the most significant) in the greedy A001563-base representation of n (A276326), then converting back to decimal. Used to compute A276335.

"Carry-free sum" in this context means that when the digits of summands (written in factorial base, see A007623) are lined up (right-justified), then summing up of each column will not result in carries to any columns left of that column, that is, the sum of digits of the k-th column from the right (with the rightmost as column 1) over all the summands is the same as the k-th digit of n, thus at most k. Among other things, this implies that in any solution, at most one of the summands may be odd. Moreover, such an odd summand is present if and only if n is odd.

a(n) is the number of set partitions of the multiset that contains d copies of each number k, collected over all k in which digit-positions (the rightmost being k=1) there is a nonzero digit d in true factorial base representation of n, where also digits > 9 are allowed.

Distinct terms are the distinct terms in A050322, that is, A045782. - David A. Corneth & Antti Karttunen, Aug 10 2018

From Mats Granvik, Jun 14 2013: (Start)

The logarithmic integral li(x) = exponential integral Ei(log(x)).

The generating function for tau A000005, the number of divisors of n is: Sum_{n >= 1} a(n) x^n = Sum_{k > 0} x^k/(1 - x^k). Another way to write the generating function for tau A000005 is Sum_{n>=1} A000005(n) x^n = Sum_{a=1..Infinity} Sum_{b>=1} x^(a*b).

If we instead think of the integral with the same form, evaluate at x = exp(1) = 2.7182818284... = A001113 and set the integration limits to zero and sqrt(log(n)), we get for n >= 0:

Logarithmic integral li(n) = Integral_{a = 0..sqrt(log(n))} Integral_{b=0..sqrt(log(n))} exp(1)^(a*b) + EulerGamma + log(log(n)). (End)

li(2)-1 is the minimum [known to date, for n>1] of |li(n) - PrimePi(n)|. - Jean-François Alcover, Jul 10 2013

The modern logarithmic integral function li(x) = Integral_{t=0..x} (1/log(t)) replaced the Li(x) = Integral_{t=2..x} (1/log(t)) which was sometimes used because it avoids the singularity at x=1. This constant is the offset between the two functions: li(2) = li(x) - Li(x) = Integral_{t=0..2} (1/log(t)). - Stanislav Sykora, May 09 2015

Denominators in the power series expansion of the higher order exponential integral E(x,4,1) - ((gamma^4/24+Pi^2*gamma^2/24+zeta(3)*gamma/3+Pi^4/160) + (gamma^3/6+ Pi^2*gamma/12+ zeta(3)/3)*log(x) + (gamma^2/4+ Pi^2/24)*log(x)^2 + (gamma/6)*log(x)^3 + log(x)^4/24), n>0. See A163931 for information on the E(x,m,n). - Johannes W. Meijer, Oct 16 2009

This is a subsequence of Euler's difference table A068106 and of A047920 (in a different ordering), since 0! = 1 was left out here. - Georg Fischer, Mar 23 2019

Row sums = A001563: (1, 4, 18, 96, 600, 4320, ...). A130477(n,k) * A130478(n,k) = A130493(n,k). Example: take dot products of rows with equal numbers of terms in A130477 and A130478, (1, 3, 8, 12) dot (24, 8, 3, 2) = (24, 24, 24, 24).

A graph with n edges is graceful if its vertices can be labeled with distinct integers in the range 0,1,...,n in such a way that when the edges are labeled with the absolute differences between the labels of their end-vertices, the n edges have the distinct labels 1,2,...,n.

Decree that (row 1) = (1), (row 2) = (2, 4), (row 3) = (3,5,8,16), (row 4) = (6,9,12,17,20,32,64). Let r(n) = A001563(n+3), so that r(r) = r(n-1) + r(n-2) + r(n-3) + r(n-4) with r(1) =1, r(2) = 2, r(3) = 4, r(4) = 7. Row n of the array, for n >= 5, consists of the numbers, in increasing order, defined as follows: all 4*x from x in row n-1, together with all 1 + 4*x from x in row n-2, together with all 2 + 4*x from x in row n-3, together with all 3 + 4*x for x in row n-4. Thus, the number of numbers in row n is r(n), a tetranacci number. Every positive integer occurs exactly once in the array, so that the resulting sequence is a permutation of the positive integers.