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A265737 Consider any concatenation of the type n = concat(a,b). Sequence lists numbers that are the sum of the products of some of such couples a and b.

%I A265737 #22 May 16 2019 08:53:15
%S A265737 655,1064,1258,1461,1642,2361,2464,3382,3442,3835,4738,4925,5275,6208,
%T A265737 6550,8291,9274,10640,11197,11548,11593,12508,12580,12915,13706,14610,
%U A265737 16420,16625,17184,18232,19641,23610,24640,31714,33820,34420,36226,38350,39826,40722
%N A265737 Consider any concatenation of the type n = concat(a,b). Sequence lists numbers that are the sum of the products of some of such couples a and b.
%C A265737 In the first 1000 terms the primes are 8291, 11197, 11593, 72253, 315521, 1514917, 2593361, 10154231, 15878617, 17209327, 22146101, 50828863, 53107111, 67328713, 120543559, 151134019.
%C A265737 Any number of the forms concat(125^z, x, 8^z, y) and concat(160, x, 625, y), where x and y are k and j zeros, with k,j>=0, and z = {1, 2, 3} is part of the sequence.
%C A265737 n is in the sequence, iff 10*n is. So the first term of sequence which is divisible by 10^n is 655*10^n. - _Altug Alkan_, Dec 17 2015
%H A265737 Paolo P. Lava, <a href="/A265737/b265737.txt">Table of n, a(n) for n = 1..1000</a>
%e A265737 For 655 we have: 6 * 55 = 320, 65 * 5 = 325 and 320 + 325 = 665.
%e A265737 For 1064 we have: 10 * 64 = 640, 106 * 4 = 424 and 640 + 424 = 1064.
%e A265737 For 41464 we have: 4 * 1464 = 5856, 41 * 464 = 19024, 4146 * 4 =  16584 and 5856 + 19024 + 16584 = 41464.
%p A265737 with(combinat): P:=proc(q) local a,j,k,n; for n from 1 to q do a:={};
%p A265737 for k from 1 to ilog10(n) do a:=a union {(n mod 10^k)*trunc(n/10^k)}; od; a:=choose(a);
%p A265737 for k from 2 to nops(a) do if n=add(a[k][j],j=1..nops(a[k])) then print(n); break; fi; od;
%p A265737 od; end: P(10^9);
%Y A265737 Cf. A065759.
%K A265737 nonn,base
%O A265737 1,1
%A A265737 _Paolo P. Lava_, Dec 15 2015