[seqfan] Re: A sum visible in the 1st integer

[Jack Brennen]

I get a sequence (not hand-computed):

T = 10,100,101,90,91,80,81,70,71,60,61,50,51,40,41,30,31,20,21,102,
82,62,42,200,201,103,72,52,32,104,63,300,301,210,105,53,211,106,43,
107,302,108,220,221,109,110,112,92,73,400,401,310,113,83,54,114,74,
311,230,231,115,64,232,116,500,501,402,240,241,117,403,118,312,119,
242,243,120,121,123,93,65,124,84,410,125,75,250,251,126,600,601,502,
320,252,321,127,503,253,254,128,411,322,129,330,331,260,261,130,131,
262,412,263,332,132,134,94,504,135,85,340,341,264,265,136,76,137,
602,413,138,510,139,420,270,271,140,141,342,142,272,511,421,143,145,
95,422,273,423,146,86,274,343,147,700,701,603,350,351,275,276,148,
604,280,87,149,512,352,150,151,430,353,281,152,354,153,282,605,154,
156,96,360,361,283,513,284,431,285,362,157,800,801,702,514,158,703,
432,159,610,160,161,520,286,287,162,433,163,363,364,290,97,291,164,
292,704,365,165,167,98,168,802,611,521,169,705,293,612,440,441,370,
371,294,522,372,170,171,613,373,442,295,443,172,523,296,374,375,297,
298,173,450,451,376,174,380,381,706,175,2000,2001,176,178,900,901,
803,524,179,804,452,382,180,181,710,182,614,2010,183,530,383,531,
453,184,454,185,384,460,461,385,386,2011,186,2020,2021,187,189,902,
711,615,190,191,805,387,192,712,532,193,620,621,194,533,2030,390,
391,806,2031,195,462,463,196,392,197,2032,198,1000,1001,903,622,464,
2040,465,1002,807,1003,713,470,471,393,623,394,534,1004,624,2041,
395,472,540,473,474,396,397,2042,2043,1005,541,398,1006,475,2050,
542,2051,476,1007,3000,3001,2052,3002,1008,2053,2054,1009,1010,1011,
904,543,1012,810,811,714,3010,1013,715,2060,625,1014,630,631,544,
1015,550,551,480,481,716,1016,482,632,483,552,3011,2061,553,2062,
484,485,3012,1017,3020,2063,3021,1018,2064,2065,1019,1020,1021,905,
486,2070,720,721,633,3022,1022,812,634,2071,635,1023,722,554,1024,
640,487,1025,560,561,490,491,813,562,492,723,493,641,563,2072,564,
1026,494,565,1027,3030,3031,2073,495,496,3032,1028,2074,3040,497,
2075,2076,1029,1030,1031,906,3041,2080,814,498,1032,815,3042,1033,
724,3043,1034,642,2081,725,2082,643,1035,570,571,644,2083,572,573,
2084,4000,4001,3050,574,1036,4002,2085,3051,2086,2087,1037,3052,...

It matches yours up to the point where you re-used 43.

Seems fairly chaotic.

- Jack


On 2/4/2016 8:29 AM, Eric Angelini wrote:
> Hello SqFans,
> Same idea as my prior post (that got not a single reply -- sigh!)
>
> T = 10,100,101,90,91,80,81,70,71,60,61,50,51,40,41,30,31,20,21,102,
> 82,62,42,200,201,103,72,52,32,104,63,300,301,210,105,53,211,106,43,
> 107,302,108,220,221,109,110,112,92,73,43,113,83,54,114,74,310,115,
> 64,230,231,116,500,501,400,401,302,117,402,232,...
>
> The rules for expanding T are:
> # Say "a" and "b" are two consecutive terms of S;
> # Add the last digit of "a" to the first digit of "b"
> # The result must be a substring of "a"
> # "b" cannot be a term already in S.
>
> We see here that
> 0+1 = 1 which is visible in 10
> 0+1 = 1 which is visible in 100
> 1+9 = 10 which is visible in 101
> 9+0 = 9 which is visible in 90
> etc.
> The term after 21 (end of first row) cannot be 10 (though 1+1 = 2
> which is visible in 21) because 10 was already used before. So 102,
> being the next available integer, has been selected.
>
> T is NOT a permutation of the integers > 0, of course.
>
> This is tricky to compute by hand. Needing a lot of backtracking.
> Hope no typos were left.
>
> Best,
> É.
>
>
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