A062273 a(n) is an n-digit number with digits in increasing order with 0 following 9 and this is maintained in the concatenation of any number of consecutive terms.

1, 23, 456, 7890, 12345, 678901, 2345678, 90123456, 789012345, 6789012345, 67890123456, 789012345678, 9012345678901, 23456789012345, 678901234567890, 1234567890123456, 78901234567890123, 456789012345678901, 2345678901234567890, 12345678901234567890

Offset: 1

Author

Amarnath Murthy, Jun 17 2001

Keywords

• nonn
• base
• easy

Comments

a(n) is congruent to A000217(n), mod 10; i.e., the last digit of a(n) is the same as the last digit of the n-th triangular number, base 10 (A008954). - Carl R. White, Oct 21 2009

Examples

			a(5) = 12345 as a(4) is 7890.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..999

Crossrefs

Cf. A000217, A008954.

Programs

Maple

f:= proc(n) option remember: local d,t,k;
  d:= procname(n-1) mod 10;
  t:= 0:
  for k from 1 to n do
    d:= d+1 mod 10;
    t:= t + d*10^(n-k)
  od:
t
end proc:
f(1):= 1:
map(f, [$1..30]); # Robert Israel, Apr 02 2018

Mathematica

FromDigits/@Table[Take[PadRight[{},250,Join[Range[9],{0}]],{(n(n+1))/2+ 1,((n+1)(n+2))/2}],{n,0,20}] (* Harvey P. Dale, May 15 2015 *)

PARI

a(n) = sum(i=1, n, ((n*(n-1)/2+i) % 10)*10^(n-i)); \\ Michel Marcus, May 26 2022

Python

def a(n): return sum((n*(n-1)//2+i)%10*10**(n-i) for i in range(1, n+1))
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, May 26 2022 after Michel Marcus

Formula

From Carl R. White, Oct 21 2009: (Start)

a(n) = floor( 10^(10*ceiling(n/10) + (n*(n+1)/2 mod 10)) * 1234567890/9999999999 ) mod 10^n.

The generalized form g, for any integer base b (>2), is: g(b,n) = floor( b^(b*ceiling(n/b) + (n*(n+1)/2 mod b)) * floor( b^(b+1)/(b-1)^2 - (b+1) ) / (b^b-1)) mod b^n, so here a(n) = g(10,n). (End)

a(n) = Sum_{i=1..n} ((n*(n-1)/2+i) mod 10)*10^(n-i). - Vedran Glisic, Apr 08 2011

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jun 18 2001