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#!/usr/bin/env python3
import time

# Timestamps whose lower digits are zero when the clock's are too HH:MM:SS "00:00:00"
# Eg., 1800000000 = 2027-01-15 00:00:00
million = 1_00_00_00
t = (int(time.time()) // million + 1) * million
while time.strftime("%H%M%S", time.localtime(t)) != '000000':
    t += million
print(f'{t} = {time.strftime("%Y-%m-%d %H:%M:%S", time.localtime(t))}')

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openpgp4fpr:1CBD499C61C7A3A7DED3935F5E21C0D40ED9EB54

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#!/usr/bin/env bash
#
# Spoiler:
# $ ./qrcodes.sh | sort | head
# 416 http://7c.dblu.me
# 416 https://j.dblu.me
# 416 http://wc.dblu.me
# 420 http://2E.dblu.me
# 420 http://73.dblu.me
#

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#!/usr/bin/env python3
import libpytunes  # https://github.com/liamks/libpytunes


if __name__ == '__main__':
    l = libpytunes.Library('iTunes Music Library.xml')
    with open('playlists.txt', 'w', encoding='utf-8') as f:
        for p in l.getPlaylistNames():
            if p not in ('Downloaded', 'Audiobooks', 'Voice Memos', 'Not One Star Rating'):
                f.write(p)

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#!/usr/bin/env python
# From https://hackernoon.com/how-to-lose-an-it-job-in-10-minutes-3d63213c8370
#
# Find cities whose names are rotated versions of other cities.
#
# Given: ['Tokyo', 'London', 'Rome', 'Donlon', 'Kyoto', 'Paris']
#
# Return:
#
#[[ 'Tokyo', 'Kyoto' ],

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#!/usr/bin/env python
# Father's Day Card Puzzle #2
# A number's persistence is the number of steps to reduce it to a single
# digit by multiplying all its digits to obain a second number, then
# multiplying all the digits of that number to obtain a third number, and so
# on until a one-digit number is obtained.
#
# For example 77 has a persistence of four because it requires four steps to
# reduce it to one digit: 77 -> 49 -> 36 -> 18 -> 8. The smallest number of
# persistence one is 10. The smallest number of persistence two is 25. The

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#!/usr/bin/env python
#
# Father's Day card puzzle #3:
# In the figure below, fill in each of the sixteen numbers from 1 to 16 such
# that the four rows and three columns add up to 29.
#
#                      (  )---(  )---(  )
#                       |             |
# (  )---(  )---(  )---(  )          (  )
#         |             |             |

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-- confluence_markup.lua
-- A Pandoc Atlassian Confluence Markup Writer
--
-- Invoke with: pandoc -to confluence_markup.lua
--
-- Based on pandoc/data/sampsdfsle.lua and zhzhxtrrk's "pandoc confluence markup writer"
-- Fixes tables and lists of lists

local image_index = 0

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def ways_to_break(amount, coins):
    """
    :param amount: the monetary value to break into some number of coins
    :param coins: a container of descending coin denominations
    :return: the number of different ways to break amount into coins
    """
    this_coin = coins[0]

    # If this is the only coin, there's one way to break it.
    if len(coins) == 1:

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Verifying I am +dblume on my passcard. https://onename.com/dblume