A330859 The additive version of the 'Decade transform' : to obtain a(n) write n as a sum of its power-of-ten parts and then continue to calculate the sum of the adjacent parts until a single number remains.
100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 220, 221, 222
Offset: 100
Examples
Let n = 32871. Write n as a sum of its power-of-ten parts: 32871 = 30000+2000+800+70+1 Now take the sum of adjacent numbers in this sum: 30000+2000+800+70+1 -> (30000+2000):(2000+800):(800+70):(70+1) = 32000:2800:870:71 Now repeat this until a single number remains: 32000:2800:870:71 -> 34800:3670:941 34800:3670:941 -> 38470:4611 38470:4611 -> 43081 Thus a(32871) = 43081. Other examples: a(100) = 100 as 100 = 100+0+0 thus 100:0:0 -> 100:0 -> 100. The equality a(n) = n holds for n=0 to 109. a(110) = 120 as 110 = 100+10+0 thus 100:10:0 -> 110:10 -> 120. a(1234) = 1694 as 1234 = 1000+200+30+4 thus 1000:200:30:4 -> 1200:230:34 -> 1430:264 -> 1694. a(15010) = 30040 as 15010 = 10000+5000+0+10+0 thus 10000:5000:0:10:0 -> 15000:5000:10:10 -> 20000:5010:20 -> 25010:5030 -> 30040.
Links
- Scott R. Shannon, Line graph of the terms for n=0 to 1000000.
Programs
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Mathematica
a[n_] := Block[{d = IntegerDigits[n], m}, m = Length[d] - 1; Total[d Binomial[ m, Range[0, m]] 10^Range[m, 0, -1]]]; a /@ Range[100, 162] (* Giovanni Resta, May 09 2020 *)
Formula
Let d_m,d_(m-1),..,d_1,d_0 be the m decimal digits of n, then a(n) = Sum_{k=0..m} d_k*C(m,k)*10^k. - Giovanni Resta, May 09 2020
Comments