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A290278 a(n) = Sum_{k=0..n} (A007953(5*k) - A007953(k)). 1

%I #17 Aug 07 2017 03:32:17

%S 0,4,3,6,4,6,3,4,0,0,4,12,15,22,24,30,31,36,36,40,39,42,40,42,39,40,

%T 36,36,31,30,33,40,42,48,49,54,54,58,57,60,58,60,57,58,54,54,49,48,42,

%U 40,42,48,49,54,54,58,57,60,58,60,57,58,54,54,49,48,42,40

%N a(n) = Sum_{k=0..n} (A007953(5*k) - A007953(k)).

%C This sequence is closely related to A289411.

%C The scatterplots of this sequence and of A289411 have similarites, including the same type of symmetry on the first 10^k terms for k > 0.

%C For b > 1, let d_b be the digital sum in base b:

%C - we have for example d_10 = A007953,

%C - also, d_b(b*n) = d_b(n) for any n >= 0,

%C - and d_b(n + m) = d_b(n) + d_b(m) iff n and m can be added without carry in base b,

%C - hence if i divides b and k > 0 and 0 <= n <= b^k-1, then d_b(i * n) + d_b(i * (b^k-1 - n)) = k*(b-1) (as i * n and i * (b^k-1 - n) can be added without carry in base b),

%C - if i and j divides b and k > 0 and 0 <= n <= b^k-1, then d_b(i * n) - d_b(j * n) = d_b(j * (b^k-1 - n)) - d_b(i * (b^k-1 - n)) (this implies the conjecture about the symmetry of A289411),

%C - also, if i and j divides b and k > 0, Sum_{m=0..b^k-1} (d_b(i * m) - d_b(j * m)) = 0.

%C For b > 1, i > 0 and j > 0 such that neither i nor j are divisible by b, let G(b,i,j) be the function defined by n -> Sum_{k=0..n} (d_b(i*k) - d_b(j*k)):

%C - we have for example G(10,5,1) = a (this sequence),

%C - G(b,i,i) = 0, G(b,i,j) = -G(b,j,i), G(b,i,j) + G(b,j,k) = G(b,i,k),

%C - if i and j divide b and k > 0 and 0 <= n <= b^k-2, then G(b,i,j)(n) = G(b,i,j)(b^k-2 - n) (in other words, the sequence G(b,i,j) restricted to the first b^k-1 terms is symmetrical), and G(b,i,j)(b^k-2) = 0 (in other words, G(b,i,j) has infinitely many zeros).

%H Rémy Sigrist, <a href="/A290278/b290278.txt">Table of n, a(n) for n = 0..10000</a>

%t Block[{nn = 68, k = 5, s}, s = Table[Total@ IntegerDigits[k n] - Total@ IntegerDigits@ n, {n, 0, nn}]; Table[Total@ Take[s, n], {n, nn}]] (* _Michael De Vlieger_, Jul 31 2017 *)

%o (PARI) s = 0; for (n=0, 67, s += sum digits(5*n) - sum digits(n); print1 (s ", "))

%Y Cf. A289411.

%K nonn,base,look

%O 0,2

%A _Rémy Sigrist_, Jul 25 2017

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