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A256785
Numbers n such that digitsum(n) is a whole number when n is represented in the fractional base 1.5 = 3/2.
1
1, 5, 11, 14, 20, 21, 22, 23, 26, 29, 30, 31, 33, 34, 38, 39, 40, 41, 45, 46, 51, 52, 53, 56, 57, 58, 60, 61, 65, 69, 70, 71, 74, 78, 79, 83, 84, 85, 87, 88, 89, 90, 91, 95, 101, 105, 106, 110, 111, 112, 113, 116, 117, 118, 122, 126, 127, 132, 133, 135, 136, 140, 146, 149, 155, 159, 160, 161, 164, 165, 166, 168, 169, 173, 174, 175
OFFSET
1,2
COMMENTS
Base 1.5 requires three digits: 1, 0 and H = 0.5. For example:
1 = 1 = 1 * 1.5^0
2 = 1H = 1 * 1.5^1 + 0.5 * 1.5^0 = 1.5 + 0.5
3 = 1H0 = 1 * 1.5^2 + 0.5 * 1.5^1 = 2.25 + 0.75
4 = 1H1 = 1 * 1.5^2 + 0.5 * 1.5^1 + 1 * 1.5^0 = 2.25 + 0.75 + 1
5 = 1H0H = 1 * 1.5^3 + 0.5 * 1.5^2 + 0.5 * 1.5^0 = 3.375 + 1.125 + 0.5
6 = 1H10 = 1 * 1.5^3 + 0.5 * 1.5^2 + 1 * 1.5^1 = 3.375 + 1.125 + 1.5
7 = 1H11 = 1 * 1.5^3 + 0.5 * 1.5^2 + 1 * 1.5^1 + 1 * 1.5^0 = 3.375 + 1.125 + 1.5 + 1
The sequence above lists the n for which digsum(n,base=1.5) is a whole number.
LINKS
Matvey Borodin, Hannah Han, Kaylee Ji, Tanya Khovanova, Alexander Peng, David Sun, Isabel Tu, Jason Yang, William Yang, Kevin Zhang, Kevin Zhao, Variants of Base 3 over 2, arXiv:1901.09818 [math.NT], 2019
EXAMPLE
The sequence begins with 1, 5 and 11, because:
digsum(1,b=1.5) = 1
digsum(5,b=1.5) = 2 = digsum(1H0H) = 1 + 0.5 + 0.5
digsum(11,b=1.5) = 4 = digsum(1H11H) = 1 + 0.5 + 1 + 1 + 0.5
The digsums are all whole numbers. However, 2, 3 and 4 are excluded because:
digsum(2,b=1.5) = 1.5 = digsum(1H) = 1 + 0.5
digsum(3,b=1.5) = 1.5 = digsum(1H0) = 1 + 0.5 + 0
digsum(4,b=1.5) = 2.5 = digsum(1H1) = 1 + 0.5 + 1
The digsums are not whole numbers.
PROG
(PARI) { b=3/2; dmx=30; d=vector(dmx); nmx=1000; n=0; ni=0; while(ni<nmx, n++; nn=n; di=0; while(nn>0, di++; d[di]=nn-floor(nn/b)*b; nn\=b; ); s=0; for(i=1, di, s+=d[i]); if(floor(s)==s, ni++; write("digsum.txt", ni, " ", n)); ); }
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Anthony Sand, Apr 10 2015
STATUS
approved