A330859 - OEIS

%I #20 May 14 2020 15:58:50

%S 100,101,102,103,104,105,106,107,108,109,120,121,122,123,124,125,126,

%T 127,128,129,140,141,142,143,144,145,146,147,148,149,160,161,162,163,

%U 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

%N 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.

%C Due to its construction a(n) = n for n=0..109, thus the data section shows a(n) for n >= 100.

%C To obtain the additive version of the 'Decade transform' of n first 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. See the Examples for details.

%C See A334387 for the difference version of the same transform.

%H Scott R. Shannon, <a href="/A330859/a330859.png">Line graph of the terms for n=0 to 1000000</a>.

%F 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

%e Let n = 32871. Write n as a sum of its power-of-ten parts:

%e 32871 = 30000+2000+800+70+1

%e Now take the sum of adjacent numbers in this sum:

%e 30000+2000+800+70+1 -> (30000+2000):(2000+800):(800+70):(70+1) = 32000:2800:870:71

%e Now repeat this until a single number remains:

%e 32000:2800:870:71 -> 34800:3670:941

%e 34800:3670:941 -> 38470:4611

%e 38470:4611 -> 43081

%e Thus a(32871) = 43081.

%e Other examples:

%e 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.

%e a(110) = 120 as 110 = 100+10+0 thus 100:10:0 -> 110:10 -> 120.

%e a(1234) = 1694 as 1234 = 1000+200+30+4 thus 1000:200:30:4 -> 1200:230:34 -> 1430:264 -> 1694.

%e 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.

%t 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 *)

%Y Cf. A334387, A011557, A040115, A053392.

%K nonn,base

%O 100,1

%A _Scott R. Shannon_, Apr 28 2020