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A327889 Triangle read by rows T(n, k) = (-1)^k * A000217(k) * F(binomial(n,k)), where F(x) = 1 if the largest decimal digit of x is 1, and 0 otherwise.

%I A327889 #13 Oct 27 2019 14:17:20
%S A327889 0,0,-1,0,0,3,0,0,0,-6,0,0,0,0,10,0,0,3,-6,0,-15,0,0,3,0,10,0,21,0,0,
%T A327889 0,0,0,0,0,-28,0,0,0,0,0,0,0,0,36,0,0,0,0,10,-15,0,0,0,-45,0,-1,0,-6,
%U A327889 0,0,0,-28,0,-45,55,0,-1,0,-6,0,0,0,0,36,0,55,-66,0,-1,0,0,0,0,0,0,0,0,0,-66,78
%N A327889 Triangle read by rows T(n, k) = (-1)^k * A000217(k) * F(binomial(n,k)), where F(x) = 1 if the largest decimal digit of x is 1, and 0 otherwise.
%C A327889 Represents alternating, normalized (linear) modification of A327853, transformed by first (decimal) digit indicator function F(x).
%C A327889 The scatterplot of the sequence represents a fractal-like structure, made out of arc-like structures. For comparison, the A327853 represents a Sierpinski's gasket (triangle), bounded by a function of A003056 (the positive inverse of triangular numbers).
%C A327889 If a number base other than decimal is used, then for a larger value of the base, the scatterplot structure will appear to be "zoomed in". The smallest base that will still represent the structure is ternary, since in binary we have F(x)=1 for all x, and the scatterplot will degrade to a simple triangle structure.
%C A327889 If we modify F(x) to look at other digits than the largest digit, then the structure appears to lose "density".
%C A327889 Why does Pascal's triangle (Sierpinski's gasket) converge to such arc-like structure when the digit indicator function F(x) is applied (in some number base)? Are there sequences other than those related to binomial coefficients, that can replicate this structure?
%H A327889 Matej Veselovac, <a href="/A327889/b327889.txt">Table of n, a(n) for n = 1..100000</a>
%H A327889 Math StackExchange, <a href="https://math.stackexchange.com/q/2585840/318073">Pattern in Pascal's triangle </a>.
%H A327889 Matej Veselovac, <a href="https://i.stack.imgur.com/BMsgg.png">Scatterplot of the sequence, for terms a(n), n=1...10^5.</a>.
%F A327889 The entries of the triangle are given by T(n, k) = (-1)^k * A000217(k) * F(binomial(n,k)), then it is read by rows, where F(x) = 1 if the largest decimal digit of x is 1, and 0 otherwise.
%e A327889 First 16 rows of the T(n, k):
%e A327889 0;
%e A327889 0, -1;
%e A327889 0,  0, 3;
%e A327889 0,  0, 0, -6;
%e A327889 0,  0, 0,  0, 10;
%e A327889 0,  0, 3, -6,  0, -15;
%e A327889 0,  0, 3,  0, 10,   0, 21;
%e A327889 0,  0, 0,  0,  0,   0,  0, -28;
%e A327889 0,  0, 0,  0,  0,   0,  0,   0, 36;
%e A327889 0,  0, 0,  0, 10, -15,  0,   0,  0,-45;
%e A327889 0, -1, 0, -6,  0,   0,  0, -28,  0,-45, 55;
%e A327889 0, -1, 0, -6,  0,   0,  0,   0, 36,  0, 55, -66;
%e A327889 0, -1, 0,  0,  0,   0,  0,   0,  0,  0,  0, -66, 78;
%e A327889 0, -1, 0,  0,  0, -15, 21, -28, 36,  0,  0,   0, 78, -91;
%e A327889 0, -1, 0,  0, 10,   0,  0,   0,  0,  0, 55,   0,  0, -91, 105;
%e A327889 0, -1, 3,  0, 10,   0,  0,   0,  0,  0,  0, -66,  0, -91, 105, -120;
%t A327889 d[n_, b_: 10] := Mod[Floor[n/b^(Floor[Log[b, n]])], b]; t[n_] := n (n + 1)/2; f[x_] := x ; r[n0_, b_: 10] := Flatten[Table[(-1)^k  Floor[f[t[k]]]*If[d[Binomial[n, k], b] == 1, 1, 0], {n, 0, n0}, {k, 0, n}]]; r[20] (* _Matej Veselovac_, Sep 29 2019 *)
%Y A327889 Cf. A001317, A007318, A003056, A000217.
%Y A327889 Cf. A327853 (original sequence, before applying the transformation).
%K A327889 sign,base,tabl,look
%O A327889 1,6
%A A327889 _Matej Veselovac_, Sep 29 2019