cp's OEIS Frontend

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.

%I A330859 #20 May 14 2020 15:58:50
%S A330859 100,101,102,103,104,105,106,107,108,109,120,121,122,123,124,125,126,
%T A330859 127,128,129,140,141,142,143,144,145,146,147,148,149,160,161,162,163,
%U A330859 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 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.
%C A330859 Due to its construction a(n) = n for n=0..109, thus the data section shows a(n) for n >= 100.
%C A330859 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 A330859 See A334387 for the difference version of the same transform.
%H A330859 Scott R. Shannon, <a href="/A330859/a330859.png">Line graph of the terms for n=0 to 1000000</a>.
%F A330859 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 A330859 Let n = 32871. Write n as a sum of its power-of-ten parts:
%e A330859 32871 = 30000+2000+800+70+1
%e A330859 Now take the sum of adjacent numbers in this sum:
%e A330859 30000+2000+800+70+1 -> (30000+2000):(2000+800):(800+70):(70+1) = 32000:2800:870:71
%e A330859 Now repeat this until a single number remains:
%e A330859 32000:2800:870:71 -> 34800:3670:941
%e A330859 34800:3670:941 -> 38470:4611
%e A330859 38470:4611 -> 43081
%e A330859 Thus a(32871) = 43081.
%e A330859 Other examples:
%e A330859 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 A330859 a(110) = 120 as 110 = 100+10+0 thus 100:10:0 -> 110:10 -> 120.
%e A330859 a(1234) = 1694 as 1234 = 1000+200+30+4 thus 1000:200:30:4 -> 1200:230:34 -> 1430:264 -> 1694.
%e A330859 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 A330859 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 A330859 Cf. A334387, A011557, A040115, A053392.
%K A330859 nonn,base
%O A330859 100,1
%A A330859 _Scott R. Shannon_, Apr 28 2020