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.

655, 1064, 1258, 1461, 1642, 2361, 2464, 3382, 3442, 3835, 4738, 4925, 5275, 6208, 6550, 8291, 9274, 10640, 11197, 11548, 11593, 12508, 12580, 12915, 13706, 14610, 16420, 16625, 17184, 18232, 19641, 23610, 24640, 31714, 33820, 34420, 36226, 38350, 39826, 40722

Offset

1,1

Comments

In the first 1000 terms the primes are 8291, 11197, 11593, 72253, 315521, 1514917, 2593361, 10154231, 15878617, 17209327, 22146101, 50828863, 53107111, 67328713, 120543559, 151134019.

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.

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

Links

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Example

For 655 we have: 6 * 55 = 320, 65 * 5 = 325 and 320 + 325 = 665.
For 1064 we have: 10 * 64 = 640, 106 * 4 = 424 and 640 + 424 = 1064.
For 41464 we have: 4 * 1464 = 5856, 41 * 464 = 19024, 4146 * 4 =  16584 and 5856 + 19024 + 16584 = 41464.

Maple

with(combinat): P:=proc(q) local a, j, k, n; for n from 1 to q do a:={};
for k from 1 to ilog10(n) do a:=a union {(n mod 10^k)*trunc(n/10^k)}; od; a:=choose(a);
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;
od; end: P(10^9);

Crossrefs

Cf. A065759.

Sequence in context: A060520 A250701 A250158 * A065759 A280445 A220057

Adjacent sequences: A265734 A265735 A265736 * A265738 A265739 A265740

Keyword

nonn,base

Author

Paolo P. Lava, Dec 15 2015

Status

approved