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A292690
Square array whose n-th row lists all numbers having n as a substring, read by falling antidiagonals, n >= 0.
13
0, 10, 1, 20, 10, 2, 30, 11, 12, 3, 40, 12, 20, 13, 4, 50, 13, 21, 23, 14, 5, 60, 14, 22, 30, 24, 15, 6, 70, 15, 23, 31, 34, 25, 16, 7, 80, 16, 24, 32, 40, 35, 26, 17, 8, 90, 17, 25, 33, 41, 45, 36, 27, 18, 9, 100, 18, 26, 34, 42, 50, 46, 37, 28, 19, 10
OFFSET
0,2
COMMENTS
This array starts with row 0, see A293869 for the variant which starts with row 1.
LINKS
Paolo Xausa, Table of n, a(n) for n = 0..11324 (first 150 antidiagonals).
EXAMPLE
The array starts:
[ 0 10 20 30 40 50 60 70 80 90 100 101 102 ...] = A011540
[ 1 10 11 12 13 14 15 16 17 18 19 21 31 ...] = A011531
[ 2 12 20 21 22 23 24 25 26 27 28 29 32 ...] = A011532
[ 3 13 23 30 31 32 33 34 35 36 37 38 39 ...] = A011533
[ 4 14 24 34 40 41 42 43 44 45 46 47 48 ...] = A011534
[ 5 15 25 35 45 50 51 52 53 54 55 56 57 ...] = A011535
[ 6 16 26 36 46 56 60 61 62 63 64 65 66 ...] = A011536
[ 7 17 27 37 47 57 67 70 71 72 73 74 75 ...] = A011537
[ 8 18 28 38 48 58 68 78 80 81 82 83 84 ...] = A011538
[ 9 19 29 39 49 59 69 79 89 90 91 92 93 ...] = A011539
[10 100 101 102 103 104 105 106 107 108 109 110 210 ...] = A293870
[11 110 111 112 113 114 115 116 117 118 119 211 311 ...] = A293871
[12 112 120 121 122 123 124 125 126 127 128 129 212 ...] = A293872
[ ... ... ... ...]
MATHEMATICA
Block[{d = 15, q, a, s}, a = Table[q = n-1; s = IntegerString[n]; Table[While[StringFreeQ[IntegerString[++q], s]]; q, d-n], {n, 0, d-1}]; Table[a[[n+1, k-n]], {k, d}, {n, 0, k-1}]] (* Paolo Xausa, Mar 01 2024 *)
PROG
(PARI) has(n, p, m=10^#Str(p))=until(p+!p>n\=10, n%m==p&&return(1))
Mat(vectorv(12, n, a=[]; for(k=n--, oo, has(k, n)||next; a=concat(a, k); #a>12&&break); a))
for(i=1, 11, for(j=1, i, print1(%[j, i-j+1]", "))) \\ Read by antidiagonals
CROSSREFS
KEYWORD
nonn,base,tabl
AUTHOR
M. F. Hasler, Oct 18 2017
STATUS
approved