A290025 The partial sums of 2^d(n) where d(n) is the n-th digit of the concatenated triangular numbers, and d(1)=0.
1, 3, 11, 75, 77, 78, 80, 112, 116, 118, 122, 378, 386, 450, 466, 498, 530, 562, 626, 690, 818, 1074, 1586, 1588, 1590, 1591, 1623, 1625, 1629, 1630, 1632, 1640, 1704, 1706, 1738, 1746, 1748, 1876, 1878, 1880, 2392, 2393, 2397, 2399, 2400, 2404, 2412, 2414, 2418, 2450
Offset: 1
Examples
2^d(1) + 2^d(2) + 2^d(3) = 2^0 + 2^1 + 2^3 = 11.
Programs
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Mathematica
Accumulate[2^Flatten@ Map[IntegerDigits, Array[# (# + 1)/2 &, 23, 0]]] (* Michael De Vlieger, Aug 03 2017 *)
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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 -
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[i-1] + 2^d[i]); res} \\ David A. Corneth, Aug 03 2017
Formula
a(n) = Sum_{k=1..n} 2^d(k) where d(k) = A034004(k).
Extensions
More terms from Michel Marcus, Jul 21 2017
Comments