cp's OEIS Frontend

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

Since 8 <= A357970(n) <= 26 the sequence is finite and begins with offset 8.

For Fibonacci numbers containing two or more digits, just concatenate the digits.

Digit 0 ==> 6;1 ==> 2;2 ==> 5;3 ==> 5;4 ==> 4;5 ==> 5;6 ==> 6;7 ==> 3;8 ==> 7;9 ==> 6.

The number of segments in each digit 0 to 9 is [6,2,5,5,4,5,6,3,7,6].

Conjecture: Taking the least significant digit of each term is not an eventually periodic sequence.

The integers are displayed as in A006942, where a 7 is depicted by 3 segments. The negative integers are depicted by using 1 segment more for the minus sign.

Since the integer 0 is depicted by 6 segments, in order to avoid considering -0 in the case n = 7, a(7) is obtained by decreasing of a unit the result of the sum A331529(7) + A331529(6) = 12 + 7 = 19, i.e., a(7) = 19 - 1 = 18.

The same sequence is obtained when 7 and 9 are depicted respectively by 4 and 5 segments (A074458).

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).

Except for 1 and 3 every positive integer occurs; A143616 and A143617 give record values and where they occur. - Reinhard Zumkeller, Aug 27 2008

The difference between this sequence and A006942 lies in the representation chosen for the digit 7,

| | |

| (here), vs. | in A006942.

If we mark with ' the "sans serif" graphical representation which uses one segment less and with * the "heavier" version, we have the following variants:

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

___________________ A006942 (6*, 7', 9*), A010371 (6*, 7*, 9*) = this.

Sequences A234691, A234692 and variants make precise which segments are lit in each digit. They are related through the Hamming weight A000120, see formula. The sequence could be extended to negative arguments with a(-n) = a(n)+1. - M. F. Hasler, Jun 17 2020

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

A063720 (this: 6', 7', 9'), A277116 (6*, 7', 9'), A074458 (6*, 7*, 9'), _________________________ 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

Sequences A234691 and A234692 use one bit for each lit segment. See crossrefs for other variants. - M. F. Hasler, Jun 17 2020

Another version of A006942. Here the digit "6" is represented with six segments (the same as in A006942) but the digit "9" is represented with five segments instead of six segments. - Omar E. Pol, Sep 30 2016

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

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

___________________ A006942 (6*, 7', 9*), A010371 (6*, 7*, 9*).

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