A093847 -id:A093847 - cp's OEIS frontend

A093846 Triangle read by rows: T(n, k) = 10^(n-1) - 1 + kfloor(910^(n-1)/n), for 1 <= k <= n.

9, 54, 99, 399, 699, 999, 3249, 5499, 7749, 9999, 27999, 45999, 63999, 81999, 99999, 249999, 399999, 549999, 699999, 849999, 999999, 2285713, 3571427, 4857141, 6142855, 7428569, 8714283, 9999997, 21249999, 32499999, 43749999, 54999999, 66249999, 77499999, 88749999, 99999999

Offset: 1

Amarnath Murthy, Apr 18 2004

10^(n-1)-1 and the n-th row are n+1 numbers in arithmetic progression and the common difference is the largest such that a(n, n) has n digits. This common difference equals A061772(n).

			Triangle begins:
     9;
    54,   99;
   399,  699,  999;
  3249, 5499, 7749, 9999;
  ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A061772, A093847, A093849, A093850.

[[10^(n-1) -1 +k*Floor(9*10^(n-1)/n): k in [1..n]]: n in [1..8]]; // G. C. Greubel, Mar 22 2019

A093846 := proc(n,k) RETURN (10^(n-1)-1+k*floor(9*(10^(n-1)/n))); end; for n from 1 to 10 do for k from 1 to n do printf("%d,",A093846(n,k)); od; od; # R. J. Mathar, Jun 23 2006

Table[# -1 +k Floor[9 #/n] &[10^(n-1)], {n, 8}, {k, n}]//Flatten (* Michael De Vlieger, Jul 18 2016 *)

{T(n,k) = 10^(n-1) -1 +k*floor(9*10^(n-1)/n)}; \\ G. C. Greubel, Mar 22 2019

[[10^(n-1) -1 +k*floor(9*10^(n-1)/n) for k in (1..n)] for n in (1..8)] # G. C. Greubel, Mar 22 2019

Corrected and extended by R. J. Mathar, Jun 23 2006

Edited by David Wasserman, Mar 26 2007

A093850 Triangle T(n,k) = 10^(n-1) -1 + kfloor(910^(n-1)/(n+1)), with 1 <= r <= n, read by rows.

Original entry on oeis.org

4, 39, 69, 324, 549, 774, 2799, 4599, 6399, 8199, 24999, 39999, 54999, 69999, 84999, 228570, 357141, 485712, 614283, 742854, 871425, 2124999, 3249999, 4374999, 5499999, 6624999, 7749999, 8874999, 19999999, 29999999, 39999999, 49999999, 59999999, 69999999, 79999999, 89999999

Offset: 1

Amarnath Murthy, Apr 18 2004

The n-th row of this triangle contains n uniformly located n-digit numbers, i.e., n terms of an arithmetic progression with 10^(n-1)-1 as the term preceding the first term and (n+1)-th term is the largest possible n-digit term.
Starting with n=2, the n-th row of this triangle can be obtained by deleting the least significant digit, 9, from terms ending in 9 in the (n+1)-th row, and ignoring the main diagonal terms, of the triangle in A093846.
Floor(A093846(4,1)/10) = T(3,1) = 324, but floor(A093846(2,1)/10) = 5 and T(1,1) = 4, floor(A093846(7,1)/10) = 228571 and T(6,1) = 228570, etc. - Michael De Vlieger, Jul 18 2016

			Triangle begins with:
      4;
     39,    69;
    324,   549,   774;
   2799,  4599,  6399,  8199;
  24999, 39999, 54999, 69999, 84999;
  ....

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A093846, A093847, A061772, A093451, A093552.

Cf. A093852.

[[10^(n-1) -1 +k*Floor(9*10^(n-1)/(n+1)): k in [1..n]]: n in [1..8]]; // G. C. Greubel, Mar 21 2019

A093850 := proc(n,r)
        10^(n-1)-1+r*floor(9*10^(n-1)/(n+1)) ;
end proc:
seq(seq(A093850(n,r),r=1..n),n=1..14) ; # R. J. Mathar, Sep 28 2011

Table[# -1 +r*Floor[9*#/(n+1)] &[10^(n-1)], {n, 8}, {r, n}]//Flatten (* Michael De Vlieger, Jul 18 2016 *)

{T(n,k) = 10^(n-1) -1 +k*floor(9*10^(n-1)/(n+1))}; \\ G. C. Greubel, Mar 21 2019

[[10^(n-1) -1 +k*floor(9*10^(n-1)/(n+1)) for k in (1..n)] for n in (1..8)] # G. C. Greubel, Mar 21 2019

Second comment clarified by Michael De Vlieger, Jul 18 2016

Edited by G. C. Greubel, Mar 21 2019

A093852 a(n) = 10^(n-1) - 1 + nfloor(910^(n-1)/(n+1)).

Original entry on oeis.org

4, 69, 774, 8199, 84999, 871425, 8874999, 89999999, 909999999, 9181818179, 92499999999, 930769230759, 9357142857140, 93999999999999, 943749999999999, 9470588235294111, 94999999999999999, 952631578947368403, 9549999999999999999, 95714285714285714279

Offset: 1

Amarnath Murthy, Apr 18 2004

This sequence is the main diagonal of A093850.

			n-th row of the following triangle contains n uniformly located n-digit numbers. i.e. n terms of an arithmetic progression with 10^(n-1)-1 as the term preceding the first term and (n+1)-th term is the largest possible n-digit term.
Given the triangle defined in A093850:
...4;
..39   69;
.324  549  774;
2799 4599 6399 8199.....
then this sequence is the leading diagonal.

G. C. Greubel, Table of n, a(n) for n = 1..995

Cf. A093846, A093847, A061772, A093450, A072875.

[10^(n-1) -1 +n*Floor(9*10^(n-1)/(n+1)): n in [1..25]]; // G. C. Greubel, Mar 21 2019

A093852 := proc(n)
        r := n ;
        10^(n-1)-1+r*floor(9*10^(n-1)/(n+1)) ;
end proc:
seq(A093852(n),n=1..50) ; # R. J. Mathar, Oct 01 2011

Table[10^(n-1) -1 +n*Floor[9*10^(n-1)/(n+1)], {n,25}] (* G. C. Greubel, Mar 21 2019 *)

{a(n) = 10^(n-1) -1 +n*floor(9*10^(n-1)/(n+1))}; \\ G. C. Greubel, Mar 21 2019

[10^(n-1) -1 +n*floor(9*10^(n-1)/(n+1)) for n in (1..25)] # G. C. Greubel, Mar 21 2019

A093851 a(n) = A002283(n-1) + floor(A052268(n)/(1+n)).

Original entry on oeis.org

4, 39, 324, 2799, 24999, 228570, 2124999, 19999999, 189999999, 1818181817, 17499999999, 169230769229, 1642857142856, 15999999999999, 156249999999999, 1529411764705881, 14999999999999999, 147368421052631577, 1449999999999999999, 14285714285714285713

Offset: 1

Amarnath Murthy, Apr 18 2004

The first column r=1 of a triangle defined by T(n,r) = 10^(n-1) -1 + r*floor(9*10^(n-1)/(n+1)).

A row starts with a (virtual) 0th column of a rep-9-digit and fills the remainder with n+1 numbers in arithmetic progression with the largest step such that all numbers in the n-th row are n-digit numbers.

			The triangle starts in row n=1 as
4 9 # -1, -1+5, -1+2*5
39 69 99 # 9,9+30,9+2*30
324 549 774 999 # 99, 99+225, 99+2*225, 99+3*225
2799 4599 6399 8199 9999 # 999, 999+1800, 999+2*1800,..
...
The sequence contains the first column.

G. C. Greubel, Table of n, a(n) for n = 1..995

Cf. A061772, A093846, A093847, A093450, A093552.

[10^(n-1) -1 +Floor(9*10^(n-1)/(n+1)): n in [1..20]]; // G. C. Greubel, Apr 02 2019

A093851 := proc(n) 10^(n-1)-1+floor(9*10^(n-1)/(n+1)) ; end proc: seq(A093851(n),n=1..20) ; # R. J. Mathar, Oct 14 2010

Table[10^(n-1) -1 +Floor[9*10^(n-1)/(n+1)], {n, 1, 20}] (* G. C. Greubel, Apr 02 2019 *)

{a(n) = 10^(n-1) -1 +floor(9*10^(n-1)/(n+1))}; \\ G. C. Greubel, Apr 02 2019

[10^(n-1) -1 +floor(9*10^(n-1)/(n+1)) for n in (1..20)] # G. C. Greubel, Apr 02 2019

a(n) = 10^(n-1) -1 + floor(9*10^(n-1)/(n+1)). - G. C. Greubel, Apr 02 2019

More terms from R. J. Mathar, Oct 14 2010

cp's OEIS Frontend

A093846 Triangle read by rows: T(n, k) = 10^(n-1) - 1 + k*floor(9*10^(n-1)/n), for 1 <= k <= n.

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A093850 Triangle T(n,k) = 10^(n-1) -1 + k*floor(9*10^(n-1)/(n+1)), with 1 <= r <= n, read by rows.

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A093852 a(n) = 10^(n-1) - 1 + n*floor(9*10^(n-1)/(n+1)).

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Magma

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A093851 a(n) = A002283(n-1) + floor(A052268(n)/(1+n)).

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A093846 Triangle read by rows: T(n, k) = 10^(n-1) - 1 + kfloor(910^(n-1)/n), for 1 <= k <= n.

A093850 Triangle T(n,k) = 10^(n-1) -1 + kfloor(910^(n-1)/(n+1)), with 1 <= r <= n, read by rows.

A093852 a(n) = 10^(n-1) - 1 + nfloor(910^(n-1)/(n+1)).