cp's OEIS Frontend

a(A216261(n)) = n and a(m) <> n for m < A216261(n). - Reinhard Zumkeller, Mar 15 2013

If we mark with * resp. ' the graphical representations which use more resp. less segments, we have the following variants:

A063720 (6', 7', 9'), A277116 (6*, 7', 9'), A074458 (6*, 7*, 9'),

_____________ this: A006942 (6*, 7', 9*), A010371 (6*, 7*, 9*).

Sequences A234691, A234692 and variants make precise which segments are lit in each digit. These are related through the Hamming weight function A000120, e.g., A010371(n) = A000120(A234691(n)) = A000120(A234692(n)). - M. F. Hasler, Jun 17 2020

The nonnegative integers are displayed as in A006942, where a 7 is depicted by 3 segments.

Given the set S = {2, 3, 4, 5, 6, 7}, the function f defined in S as f(5) = f(6) = 3 and f(s) = 1 elsewhere, a(n) is equal to the difference between the number b(n) of S-restricted f-weighted integer compositions of n with that of n-6, i.e., b(n-6). The latter one provides the number of all those excluded cases where a nonnegative integer is displayed with leading zeros. b(n) is calculated as the sum of polynomial coefficients or extended binomial coefficients (see Equation 3 in Eger) where the index of summation is positive and it covers the numbers of possible digits that can be displayed by n segments (see first formula).

The same sequence is obtained when 7 and 9 are depicted respectively by 4 and 5 segments (A074458). - Stefano Spezia, Apr 11 2021

a(n) is the least number using n + 6 (or n + 5 for n < 5) segments on a 7-segment display, when '6' uses 6 segments. This is essentially the same as A038619 (starts with 1, 2, 6 instead of 0) and A216261 (= a(n) uses n segments: has 4 values before 0 and 22 before 20). - M. F. Hasler, Jun 17 2020

The nonnegative integers are displayed as in A063720.

Given the set S = {2, 3, 4, 5, 6, 7}, the function f defined in S as f(5) = 5 and f(s) = 1 elsewhere, a(n) is equal to the difference between the number b(n) of S-restricted f-weighted integer compositions of n with that of n-6, i.e., b(n-6). The latter one provides the number of all those excluded cases where a nonnegative integer is displayed with leading zeros. b(n) is calculated as the sum of polynomial coefficients or extended binomial coefficients (see Equation 3 in Eger) where the index of summation is positive and it covers the numbers of possible digits that can be displayed by n segments (see third formula).

The nonnegative integers are displayed as in A277116.

Given the set S = {2, 3, 4, 5, 6, 7}, the function f defined in S as f(5) = 4, f(6) = 2 and f(s) = 1 elsewhere, a(n) is equal to the difference between the number b(n) of S-restricted f-weighted integer compositions of n with that of n-6, i.e., b(n-6). The latter one provides the number of all those excluded cases where a nonnegative integer is displayed with leading zeros. b(n) is calculated as the sum of polynomial coefficients or extended binomial coefficients (see Equation 3 in Eger) where the index of summation is positive and it covers the numbers of possible digits that can be displayed by n segments (see third formula).

For n > 1, a(n) uses n + 3 segments to be displayed, when a digit '6' uses 6 segments (as in A234691, A234692 and A277116, A074458, A006942, A010371, but not in A063720). Sequence A143617 is the same but starts with 0, 8, ... and A216261 has additional terms 7 & 4 before 2 and 22 before 20. - M. F. Hasler, Jun 23 2020

The sequence begins with a(2) = 1 since at least two segments are needed to form any digit. It requires two segments to form the digit 1 and three segments to form the digit 7.

All other digits use more than 3 segments.

a(n) is undefined for n<5, as there are no numbers with 1 segment, 1 with 2 segments, 1 with 3 segments, and 2 with 4 segments. If 0 is excluded as a valid input - so the series would refer to "the n-th positive number" - then a(6) would be 111 rather than 77.

6, 7, and 9 each have two possible representations on a 7-segment display. Therefore, a number that contains at least one of each digit can be displayed in 8 different ways. We use 'variant 5' because it was the first one added to the OEIS (A006942).

This sequence is interesting because a(n) is guaranteed to exist for n>=1. For instance, printing a '6' will always result in 1 unused segment. If we want to get a number that returns m unused segments, then we can always return '66...66' with '6' appearing m times. This also provides an upper bound for the value of a(n). Despite this, the value of a(n) for a given n is nontrivial.

This sequence does not increase monotonically.

Certain digits can never appear in the sequence for n>=1:

- 8: This digit contributes 0 unused segments. If a solution contains an 8 and has more than one digit, the 8 can be removed entirely without reducing the number of unused segments, resulting in a smaller number. The only exception is when the number is a single digit, so for a(0).

- 3 and 5: Both use the same number of segments as 2. So, if a solution includes a 3 or 5, we can replace it with a 2 without changing the number of unused segments, but the overall number would be smaller.

- 9: Same as above: any 9 can be replaced by a 6. We can't extend this to 0 (same number of segments as 6 or 9), because we can't always swap a 6 or a 9 for a 0 (e.g. '600').

Among all 10 digits, 1 has the largest number of unused segments (5). As a consequence, when n = 5*p (with p>=1), a(n) = '1..1' with 1 appearing p times because if a(n) contained any other digit, it would be more than p digits long, and would thus not be optimal.

The next term a(5*p+1) necessarily has an additional digit. The smallest digit that adds only 1 unused segment is 0 but it cannot be placed in the leading position. The next best position is the second one, thus a(5*p+1) = '10...1' with p 1s and a single 0.

To construct the next term a(5*p+2), we swap the existing zero for another digit with an additional unused segment (2 being optimal as it is the smallest) and move it to the trailing position (to yield the smallest possible number). Hence a(5*p+2) = '1..12' with p 1s and a single 2. Any other approach, such as adding an extra 0 or swapping a 1 for something else, would result in a solution with more than p+1 digits and would thus not be optimal.

The next two terms can be constructed the same way, yielding a(5*p+3) = '1..14' and a(5*p+4) = '1..17' with again p+1 digits in total. We then loop back to the beginning with a(5*p+5).