%I
%S 1,3,11,75,77,78,80,112,116,118,122,378,386,450,466,498,530,562,626,
%T 690,818,1074,1586,1588,1590,1591,1623,1625,1629,1630,1632,1640,1704,
%U 1706,1738,1746,1748,1876,1878,1880,2392,2393,2397,2399,2400,2404,2412,2414,2418,2450
%N The partial sums of 2^d(n) where d(n) is the nth digit of the concatenated triangular numbers, and d(1)=0.
%C The differences between consecutive terms are <= 2^9. So the sequence contains arbitrarily long arithmetic progressions. The sequence of powers of 2 does not contain progressions, however. This is a result of the fact that 2^n satisfies the recurrence relation a(n+1)=2a(n).
%F a(n) = Sum_{k=1..n} 2^d(k) where d(k) = A034004(k).
%e 2^d(1) + 2^d(2) + 2^d(3) = 2^0 + 2^1 + 2^3 = 11.
%t Accumulate[2^Flatten@ Map[IntegerDigits, Array[# (# + 1)/2 &, 23, 0]]] (* _Michael De Vlieger_, Aug 03 2017 *)
%o (PARI) lista(nn) = {print1(cur=1, ", "); for(n=1, nn, d = digits(n*(n+1)/2); for(i=1, #d, cur += 2^d[i]; print1(cur, ", ");););} \\ _Michel Marcus_, Jul 21 2017
%o (PARI) first(n) = {my(d = [0], i = 1, t = 2, res = vector(n)); res[1] = 1; while(#d < n, d = concat(d, digits(i)); i+=t; t++); for(i=2, n, res[i] = res[i1] + 2^d[i]); res} \\ _David A. Corneth_, Aug 03 2017
%Y Cf. A000217, A034004.
%K nonn,base,easy
%O 1,2
%A _Joseph Wheat_, Jul 17 2017
%E More terms from _Michel Marcus_, Jul 21 2017
