A301382 - OEIS

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A301382

a(1) = 1. For n > 1, a(n) is the smallest positive integer x not already in the sequence such that the product of x and its initial digit is minimal and strictly larger than the same product for any previous term.

1, 2, 3, 10, 11, 12, 13, 14, 15, 4, 17, 18, 19, 5, 6, 20, 21, 22, 23, 24, 7, 25, 26, 27, 28, 29, 8, 9, 30, 31, 32, 33, 100, 101, 34, 103, 104, 35, 106, 107, 36, 109, 110, 37, 112, 113, 38, 115, 116, 39, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

We see in the Example section that P is the smallest possible product strictly bigger than the previous one and not leading to a contradiction.

LINKS

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000

EXAMPLE

a(1) * [the first digit of a(1)] = 1 * 1 = P = 1

a(2) * [the first digit of a(2)] = 2 * 2 = P = 4

a(3) * [the first digit of a(3)] = 3 * 3 = P = 9

a(4) * [the first digit of a(4)] = 10 * 1 = P = 10

a(5) * [the first digit of a(5)] = 11 * 1 = P = 11

a(6) * [the first digit of a(6)] = 12 * 1 = P = 12

a(7) * [the first digit of a(7)] = 13 * 1 = P = 13

a(8) * [the first digit of a(8)] = 14 * 1 = P = 14

a(9) * [the first digit of a(9)] = 15 * 1 = P = 15

a(10) * [the first digit of a(10)] = 4 * 4 = P = 16

a(11) * [the first digit of a(11)] = 17 * 1 = P = 17

Etc.

PROG

(PARI) p = vector(136, k, oo); for (n=1, #p, x = n*digits(n)[1]; if (x<=#p, p[x] = min(p[x], n))); for (k=1, #p, if (p[k] != oo, print1 (p[k] ", "))) \\ Rémy Sigrist, Mar 22 2018

CROSSREFS

Sequence in context: A283436 A225558 A362089 * A288657 A055655 A371030

Adjacent sequences: A301379 A301380 A301381 * A301383 A301384 A301385

KEYWORD

nonn,base

AUTHOR

Eric Angelini and Jean-Marc Falcoz, Mar 20 2018

STATUS

approved