A066578 -id:A066578 - cp's OEIS frontend

A080461 Duplicate of A066578.

1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 5, 4, 3, 2, 2, 2, 2, 2, 1, 10, 7, 5, 4, 4, 3, 3, 3, 2, 2, 10, 7, 6, 5, 4

Offset: 1

dead

A080462 Define f(k) = Floor [ k/ sum of the digits of k]. Let f(f(...(n))) = m where m is divisible by the sum of the digits of m. Then a(n)= one more than the least number of steps to obtain m.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1

Offset: 1

Amarnath Murthy, Mar 02 2003

			a(24) = 1 as 24 is divisible by 2+4. a(23) = 2, since f(23) = 4. a(14835)= 3 because f(14835) = 76, f(76) = 5, etc.

Cf. A066578.

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 06 2003

A094175 Round( n / sum of digits of n ).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 6, 4, 3, 3, 3, 2, 2, 2, 2, 10, 7, 6, 5, 4, 4, 3, 3, 3, 3, 10, 8, 6, 6, 5, 4, 4, 4, 3, 3, 10, 8, 7, 6, 6, 5, 5, 4, 4, 4, 10, 9, 7, 7, 6, 6, 5, 5, 4, 4, 10, 9, 8, 7, 6, 6, 6, 5, 5, 5, 10, 9, 8, 7, 7, 6, 6, 6, 5, 5, 10, 9, 8, 8, 7, 7, 6, 6, 6, 5, 10, 9, 8, 8, 7, 7, 6, 6, 6, 6, 100, 51, 34, 26, 21, 18, 15, 13, 12, 11, 55, 37, 28, 23, 19, 16, 15, 13, 12, 11, 40

Offset: 1

Paolo P. Lava, May 06 2004

			a(15) = round(15/(1+5)) = round(2.5) = 3.
a(17) = round(17/(1+7)) = round(2.125) = 2.
a(62) = round(62/(6+2)) = round(7.75) = 8.

Dominic McCarty, Table of n, a(n) for n = 1..10000

Cf. A066578.

Table[Floor[n/Total[IntegerDigits[n]]+1/2],{n,120}] (* Harvey P. Dale, Nov 26 2016 *)

a(n) = round(n/sumdigit(n)) where "round" is the rounded integer part of the division and "sumdigit" is the sum of the digits of the number.

Corrected and extended by Harvey P. Dale, Nov 26 2016

Offset changed to 1 by Georg Fischer, Jun 14 2024

cp's OEIS Frontend

A080461 Duplicate of A066578.

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A080462 Define f(k) = Floor [ k/ sum of the digits of k]. Let f(f(...(n))) = m where m is divisible by the sum of the digits of m. Then a(n)= one more than the least number of steps to obtain m.

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A094175 Round( n / sum of digits of n ).

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